W12-W13. Пределы и непрерывность

1. Краткое содержание

В этой главе понятие limit (предела) переносится с последовательностей на функции и даётся строгое определение continuous (непрерывных) функций. Рассматриваются разные типы пределов, теоремы о пределах и приложения к поведению функций.

1.1 Предел функции

Изучая \(y = f(x)\), часто интересуются поведением около точки \(c\), даже если \(c\) не входит в область определения. Типичный случай — подстановка в \(c\) даёт деление на нуль.

Пример: пусть \(f(x) = \frac{x^2 - 1}{x - 1}\). При \(x = 1\) знаменатель обнуляется, значение не определено; при \(x \neq 1\) упрощают:

\[f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} = x + 1, \quad \forall x \neq 1.\]

График — прямая \(y = x + 1\) без точки \((1, 2)\). Хотя \(f(1)\) не задано, значения \(f(x)\) можно сделать сколь угодно близкими к \(2\), выбирая \(x\) достаточно близкими к \(1\).

1.1.1 Определение Коши: \(\varepsilon\)–\(\delta\) для предела

Epsilon-delta definition (определение предела по Коши в терминах \(\varepsilon\) и \(\delta\)) формализует идею предела функции.

Пусть \(f(x)\) определена на некотором открытом интервале вокруг \(c\), за исключением, возможно, самой точки \(c\). Говорят, что предел \(f(x)\) при \(x \to c\) равен \(L \in \mathbb{R}\), и пишут

\[\lim_{x \to c} f(x) = L,\]

если для любого \(\varepsilon > 0\) существует \(\delta = \delta(\varepsilon) > 0\) такое, что

\[|f(x) - L| < \varepsilon \quad \text{при} \quad 0 < |x - c| < \delta.\]

Иными словами, как только \(x\) попадает в достаточно малую проколотую окрестность \(c\), значения \(f(x)\) оказываются в \(\varepsilon\)-окрестности \(L\).

Важно: точка \(c\) должна быть предельной (limit point) для области определения \(f\). Точка \(c\) — предельная для множества \(A\), если в любой открытой окрестности \(c\) есть точка из \(A\), отличная от \(c\).

Пример: для \(A = (1, 2) \cup \{3\}\) множество предельных точек — отрезок \([1, 2]\). Число \(3\) предельной точкой не является: есть окрестность \(3\), не содержащая других точек \(A\).

1.1.2 How to Find Delta for a Given Epsilon

Finding the appropriate \(\delta > 0\) for a given \(\varepsilon > 0\) can be accomplished in two steps:

1. Solve the inequality: Solve \(|f(x) - L| < \varepsilon\) to find an open interval \((a, b)\) containing \(c\) on which the inequality holds for all \(x \neq c\).
2. Find delta: Find a value of \(\delta > 0\) that places the open \(\delta\)-interval \((c - \delta, c + \delta)\) centered at \(c\) inside the interval \((a, b)\).

Example 1: Show that \(\lim_{x \to 1} (5x - 3) = 2\).

Set \(f(x) = 5x - 3\), \(c = 1\), and \(L = 2\). For any given \(\varepsilon > 0\), we need to find \(\delta > 0\) such that if \(x \neq 1\) and \(0 < |x - 1| < \delta\), then \(|f(x) - 2| < \varepsilon\).

We have: \[|f(x) - 2| < \varepsilon \implies |(5x - 3) - 2| = |5x - 5| < \varepsilon \implies 5|x - 1| < \varepsilon \implies |x - 1| < \frac{\varepsilon}{5}\]

Thus, we can take \(\delta = \frac{\varepsilon}{5}\). Then if \(0 < |x - 1| < \delta = \frac{\varepsilon}{5}\): \[|(5x - 3) - 2| = |5x - 5| = 5|x - 1| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon\]

This proves that \(\lim_{x \to 1} (5x - 3) = 2\). Note that \(\delta = \frac{\varepsilon}{5}\) is not the only value that works—any smaller positive \(\delta\) will also work.

Example 2: Prove basic limits:

1. \(\lim_{x \to c} x = c\): Let \(\varepsilon > 0\) be given. We need \(|x - c| < \varepsilon\) whenever \(0 < |x - c| < \delta\). If we take \(\delta = \varepsilon\), then \(|x - c| < \varepsilon\) immediately holds.
2. \(\lim_{x \to c} k = k\) (where \(k\) is a constant): Let \(\varepsilon > 0\) be given. We need \(|k - k| < \varepsilon\) whenever \(0 < |x - c| < \delta\). Since \(k - k = 0\), we always have \(|k - k| < \varepsilon\), so any positive \(\delta\) works.

Example 3: Prove that \(\lim_{x \to 5} \sqrt{x - 1} = 2\).

For any given \(\varepsilon > 0\), we need to find \(\delta > 0\) such that: \[|\sqrt{x - 1} - 2| < \varepsilon \quad \text{whenever} \quad 0 < |x - 5| < \delta\]

For all \(x \neq 5\): \[|\sqrt{x - 1} - 2| < \varepsilon \implies -\varepsilon < \sqrt{x - 1} - 2 < \varepsilon \implies 2 - \varepsilon < \sqrt{x - 1} < 2 + \varepsilon\] \[\implies (2 - \varepsilon)^2 < x - 1 < (2 + \varepsilon)^2 \implies 1 + (2 - \varepsilon)^2 < x < 1 + (2 + \varepsilon)^2\]

We want to place the interval \((5 - \delta, 5 + \delta)\) inside \((1 + (2 - \varepsilon)^2, 1 + (2 + \varepsilon)^2)\).

For \(\varepsilon < 4\), take \(\delta = \min\{4 - (2 - \varepsilon)^2, -4 + (2 + \varepsilon)^2\} = \min\{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\}\). Then \(0 < |x - 5| < \delta\) will place \(x\) in the required interval.

For \(\varepsilon \geq 4\), take \(\delta = \min\{5, -4 + (2 + \varepsilon)^2\}\).

Example 4: Let \(f(x) = \begin{cases} x^2, & x \neq 2 \\ 1, & x = 2 \end{cases}\). Prove that \(\lim_{x \to 2} f(x) = 4\).

Note that even though \(f(2) = 1 \neq 4\), the limit can still be 4 because the limit only cares about values near \(c = 2\), not at \(c = 2\) itself.

For \(x \neq 2\), we need to solve \(|x^2 - 4| < \varepsilon\): \[|x^2 - 4| < \varepsilon \implies 4 - \varepsilon < x^2 < 4 + \varepsilon\]

For \(\varepsilon < 4\): \(\sqrt{4 - \varepsilon} < x < \sqrt{4 + \varepsilon}\).

Take \(\delta = \min\{2 - \sqrt{4 - \varepsilon}, \sqrt{4 + \varepsilon} - 2\}\). Then \(0 < |x - 2| < \delta\) will place \(x\) in the required interval.

For \(\varepsilon \geq 4\): Take \(\delta = \min\{2, \sqrt{4 + \varepsilon} - 2\}\).

1.1.3 Heine’s Definition of Limit

An alternative definition of limit uses sequences.

Heine’s definition: Let \(f: A \to \mathbb{R}\), and let \(c\) be a limit point of \(A\). We say that \(f\) has a limit \(L\) at \(c\) if, for any sequence \(\{x_n\}\) with \(x_n \to c\) as \(n \to \infty\) and \(x_n \neq c\) for all \(n \in \mathbb{N}\), we have \(f(x_n) \to L\) as \(n \to \infty\).

Theorem (Equivalence of Definitions): The Cauchy definition and the Heine definition are equivalent. That is: \[\lim_{x \to c} f(x) = L \iff \lim_{n \to \infty} f(x_n) = L\] for every sequence \(\{x_n\} \subset A\) such that \(x_n \neq c\) for all \(n \in \mathbb{N}\) and \(x_n \to c\) as \(n \to \infty\).

Theorem (Uniqueness of the Limit): If \(f\) has a limit at \(c\), then this limit is unique.

Proof: Suppose \(f\) has two different limits \(L_1 \neq L_2\). Let \(\{x_n\} \subset A \setminus \{c\}\) be a sequence that converges to \(c\). Then the sequence \(\{f(x_n)\}\) converges to two different limits \(L_1\) and \(L_2\). This contradicts the uniqueness of the limit of a sequence. Therefore, the limit must be unique.

1.2 Limit Theorems

1.2.1 Limit Laws

If \(L\), \(M\), \(c\), and \(k\) are real numbers with \(\lim_{x \to c} f(x) = L\) and \(\lim_{x \to c} g(x) = M\), then:

1. Sum Rule: \(\lim_{x \to c} (f(x) + g(x)) = L + M\)
2. Difference Rule: \(\lim_{x \to c} (f(x) - g(x)) = L - M\)
3. Constant Multiple Rule: \(\lim_{x \to c} (kf(x)) = kL\)
4. Product Rule: \(\lim_{x \to c} (f(x) \cdot g(x)) = L \cdot M\)
5. Quotient Rule: \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}\), provided \(M \neq 0\)
6. Power Rule: \(\lim_{x \to c} [f(x)]^n = L^n\), \(n\) a positive integer
7. Root Rule: \(\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}\), \(n\) a positive integer (if \(n\) is even, assume \(f(x) \geq 0\))

Proof of Sum Rule (using Heine’s definition): Let \(\{x_n\}\) be a sequence in \(A\) that converges to \(c\) with \(x_n \neq c\) for every \(n\). Then \(\lim_{n \to \infty} f(x_n) = L\) and \(\lim_{n \to \infty} g(x_n) = M\). By the limit laws for sequences: \[\lim_{n \to \infty} (f(x_n) + g(x_n)) = \lim_{n \to \infty} (f + g)(x_n) = L + M\] This means \(\lim_{x \to c} (f(x) + g(x)) = L + M\).

Proof of Sum Rule (using epsilon-delta definition): Let \(\varepsilon > 0\) be given. Since \(\lim_{x \to c} f(x) = L\), there exists \(\delta_1 > 0\) such that: \[|f(x) - L| < \frac{\varepsilon}{2} \quad \text{whenever} \quad 0 < |x - c| < \delta_1\]

Similarly, since \(\lim_{x \to c} g(x) = M\), there exists \(\delta_2 > 0\) such that: \[|g(x) - M| < \frac{\varepsilon}{2} \quad \text{whenever} \quad 0 < |x - c| < \delta_2\]

Let \(\delta = \min\{\delta_1, \delta_2\}\). If \(0 < |x - c| < \delta\), then both inequalities hold, and by the triangle inequality: \[|f(x) + g(x) - (L + M)| \leq |f(x) - L| + |g(x) - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon\]

This shows that \(\lim_{x \to c} (f(x) + g(x)) = L + M\). The other limit laws can be proved similarly.

1.2.2 Limits of Polynomials and Rational Functions

Theorem:

1. Limits of Polynomials: If \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\), then: \[\lim_{x \to c} P(x) = P(c) = a_n c^n + a_{n-1} c^{n-1} + \cdots + a_1 c + a_0\]
2. Limits of Rational Functions: If \(P(x)\) and \(Q(x)\) are polynomials and \(Q(c) \neq 0\), then: \[\lim_{x \to c} \frac{P(x)}{Q(x)} = \frac{P(c)}{Q(c)}\]

1.2.3 Indeterminate Forms

Indeterminate forms are expressions that appear when taking limits where direct substitution results in an ambiguous value, such as: \[\frac{0}{0}, \quad \frac{\infty}{\infty}, \quad 0 \cdot \infty, \quad \infty - \infty, \quad 1^{\infty}, \quad 0^0, \quad \infty^0\]

Example 1 (Eliminating Common Factors - \(\frac{0}{0}\) form): \[\lim_{x \to 1} \frac{x^2 + x - 2}{x^2 - x} = \lim_{x \to 1} \frac{(x-1)(x+2)}{x(x-1)} = \lim_{x \to 1} \frac{x+2}{x} = \frac{3}{1} = 3\]

Example 2 (Rationalization - \(\frac{0}{0}\) form): \[\lim_{x \to 0} \frac{\sqrt{x^2 + 100} - 10}{x} = \lim_{x \to 0} \frac{\sqrt{x^2 + 100} - 10}{x} \cdot \frac{\sqrt{x^2 + 100} + 10}{\sqrt{x^2 + 100} + 10}\] \[= \lim_{x \to 0} \frac{x^2 + 100 - 100}{x(\sqrt{x^2 + 100} + 10)} = \lim_{x \to 0} \frac{x^2}{x(\sqrt{x^2 + 100} + 10)} = \lim_{x \to 0} \frac{x}{\sqrt{x^2 + 100} + 10} = 0\]

1.2.4 The Sandwich Theorem

Theorem (Sandwich Theorem): Suppose that \(g(x) \leq f(x) \leq h(x)\) for all \(x\) in some open interval containing \(c\), except possibly at \(x = c\) itself. If \(\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} f(x) = L\).

The function \(f\) is “sandwiched” between \(g\) and \(h\), and since both \(g\) and \(h\) approach the same limit \(L\), the function \(f\) must also approach \(L\).

Example 1: How does \(f(x) = x^2 \sin(1/x^2)\) behave near \(x = 0\)?

For any \(x \neq 0\): \[-1 \leq \sin(1/x^2) \leq 1 \implies -x^2 \leq x^2 \sin(1/x^2) \leq x^2\]

Since \(\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0\), by the Sandwich Theorem: \[\lim_{x \to 0} x^2 \sin(1/x^2) = 0\]

Example 2: We can prove using the Sandwich Theorem that: \[\lim_{\theta \to 0} \sin \theta = 0, \quad \lim_{\theta \to 0} \cos \theta = 1\]

For all \(\theta \in \mathbb{R}\): \[-|\theta| \leq \sin \theta \leq |\theta|, \quad \text{and} \quad 0 \leq 1 - \cos \theta \leq |\theta|\]

Since \(\lim_{\theta \to 0} -|\theta| = \lim_{\theta \to 0} |\theta| = 0\), we get the desired limits by the Sandwich Theorem.

1.3 One-Sided Limits

One-sided limits extend the limit concept to situations where \(x\) approaches \(c\) from only one direction: from the left (\(x < c\)) or from the right (\(x > c\)).

Definition (One-Sided Limits): Let \(f: A \to \mathbb{R}\), and let \(c\) be a limit point of \(A\).

1. Right-hand limit: Assume \(A\) contains an interval \((c, d)\) to the right of \(c\). We say \(f(x)\) has right-hand limit \(L\) at \(c\), written: \[\lim_{x \to c^+} f(x) = L\] if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad c < x < c + \delta\]
2. Left-hand limit: Assume \(A\) contains an interval \((b, c)\) to the left of \(c\). We say \(f\) has left-hand limit \(L\) at \(c\), written: \[\lim_{x \to c^-} f(x) = L\] if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad c - \delta < x < c\]

Example: Show that \(\lim_{x \to 0^+} \sqrt{x} = 0\).

Let \(\varepsilon > 0\) be given. We want to find \(\delta > 0\) such that: \[|\sqrt{x} - 0| < \varepsilon \quad \text{whenever} \quad 0 < x < \delta\]

Squaring both sides: \(x < \varepsilon^2\) whenever \(0 < x < \delta\).

If we choose \(\delta = \varepsilon^2\), then: \[\sqrt{x} < \varepsilon \quad \text{whenever} \quad 0 < x < \delta = \varepsilon^2\]

This proves \(\lim_{x \to 0^+} \sqrt{x} = 0\).

Theorem (Relationship Between Two-Sided and One-Sided Limits): Suppose \(f\) is defined on an open interval containing \(c\), except perhaps at \(c\) itself. Then \(f(x)\) has a limit as \(x \to c\) if and only if it has both limits from the left and right at \(c\), and these limits are equal: \[\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = L \quad \text{and} \quad \lim_{x \to c^+} f(x) = L\]

Example: Consider \(f(x) = \frac{x - c}{|x - c|}\), defined on \(\mathbb{R} \setminus \{c\}\).

• For \(x > c\): \(f(x) = \frac{x-c}{x-c} = 1\), so \(\lim_{x \to c^+} f(x) = 1\)
• For \(x < c\): \(f(x) = \frac{x-c}{-(x-c)} = -1\), so \(\lim_{x \to c^-} f(x) = -1\)

Since the one-sided limits are different, \(\lim_{x \to c} f(x)\) does not exist.

Example (finding parameter for limit existence): Consider: \[f(x) = \begin{cases} ax^2 + 1, & \text{if } x \leq 1 \\ -5x^3 + 3a, & \text{if } x > 1 \end{cases}\]

Find \(a \in \mathbb{R}\) such that \(\lim_{x \to 1} f(x)\) exists.

We have: \[\lim_{x \to 1^-} f(x) = a + 1, \quad \lim_{x \to 1^+} f(x) = 3a - 5\]

The limit exists if \(a + 1 = 3a - 5\), which gives \(2a = 6\), so \(a = 3\).

1.3.1 Important Limit: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\)

Theorem: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) (where \(\theta\) is in radians).

Proof: Consider a unit circle with angle \(\theta\) where \(0 < \theta < \frac{\pi}{2}\). From the geometry: \[\text{area}(\triangle OAP) < \text{area(sector } OAP) < \text{area}(\triangle OAT)\]

We can express these areas as: \[\text{area}(\triangle OAP) = \frac{1}{2}(1)(\sin \theta) = \frac{1}{2} \sin \theta\] \[\text{area(sector } OAP) = \frac{1}{2}r^2 \theta = \frac{1}{2}\theta \quad \text{(only for } \theta \text{ in radians)}\] \[\text{area}(\triangle OAT) = \frac{1}{2}(1)(\tan \theta) = \frac{1}{2} \tan \theta\]

Thus: \[\frac{1}{2} \sin \theta < \frac{1}{2} \theta < \frac{1}{2} \tan \theta\]

Since \(0 < \theta < \frac{\pi}{2}\), we have \(\sin \theta > 0\), so dividing by \(\frac{1}{2} \sin \theta\): \[1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta} \implies \cos \theta < \frac{\sin \theta}{\theta} < 1\]

Since \(\lim_{\theta \to 0^+} \cos \theta = 1\), by the Sandwich Theorem: \[\lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1\]

For the left-hand limit, note that \(\sin \theta\) and \(\theta\) are both odd functions, so \(f(\theta) = \frac{\sin \theta}{\theta}\) is an even function. By symmetry: \[\lim_{\theta \to 0^-} \frac{\sin \theta}{\theta} = \lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1\]

Therefore, \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\).

1.4 Limits at Infinity, Infinite Limits, and Asymptotes

1.4.1 Limits at Infinity

When studying function behavior as the magnitude of \(x\) becomes increasingly large, we use limits at infinity.

Definition:

1. We say \(f(x)\) has limit \(L\) as \(x\) approaches \(\infty\), written \(\lim_{x \to \infty} f(x) = L\), if for every \(\varepsilon > 0\), there exists \(M\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad x > M\]
2. We say \(f(x)\) has limit \(L\) as \(x\) approaches \(-\infty\), written \(\lim_{x \to -\infty} f(x) = L\), if for every \(\varepsilon > 0\), there exists \(N\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad x < N\]

Example 1: Let \(k \in \mathbb{R}\) be a constant. Then \(\lim_{x \to \pm\infty} k = k\).

For \(x \to \infty\): Let \(\varepsilon > 0\) be given. We need \(|k - k| < \varepsilon\) for all \(x > M\). Since \(|k - k| = 0 < \varepsilon\) always, any \(M\) works. Similarly for \(x \to -\infty\).

Example 2: Show that \(\lim_{x \to \pm\infty} \frac{1}{x} = 0\).

For \(x \to \infty\): Let \(\varepsilon > 0\) be given. We need: \[\left|\frac{1}{x} - 0\right| = \left|\frac{1}{x}\right| < \varepsilon \quad \text{whenever} \quad x > M\]

This holds if \(M = \frac{1}{\varepsilon}\) (or any larger number). Similarly, for \(x \to -\infty\), take \(N = -\frac{1}{\varepsilon}\).

Example 3: Show that \(\lim_{x \to -\infty} e^x = 0\).

Let \(\varepsilon > 0\) be given. We need \(|e^x - 0| < \varepsilon\) whenever \(x < N\). \[|e^x - 0| < \varepsilon \implies e^x < \varepsilon \implies x < \ln \varepsilon\]

Taking \(N = \ln \varepsilon\), we have \(|e^x - 0| < \varepsilon\) whenever \(x < \ln \varepsilon\).

Remark: All limit laws (sum, difference, product, quotient, etc.) apply when we replace \(\lim_{x \to c}\) by \(\lim_{x \to \infty}\) or \(\lim_{x \to -\infty}\). The Sandwich Theorem also holds for limits at infinity.

Examples:

1. \(\displaystyle \lim_{x \to \infty} \left(-1 + \frac{2}{x}\right) = -1 + 2 \cdot 0 = -1\)
2. \(\displaystyle \lim_{x \to \infty} \frac{3x^2 + 2x - 5}{-2x^2 + x - 1} = \lim_{x \to \infty} \frac{3 + \frac{2}{x} - \frac{5}{x^2}}{-2 + \frac{1}{x} - \frac{1}{x^2}} = \frac{3 + 0 - 0}{-2 + 0 - 0} = -\frac{3}{2}\)
3. \(\displaystyle \lim_{x \to -\infty} \frac{3x - 5}{2x^2 + x - 1} = \lim_{x \to -\infty} \frac{\frac{3}{x} - \frac{5}{x^2}}{2 + \frac{1}{x} - \frac{1}{x^2}} = \frac{0 - 0}{2 + 0 - 0} = 0\)
4. \(\displaystyle \lim_{x \to \infty} x \sin(1/x)\): Let \(t = 1/x\). As \(x \to \infty\), \(t \to 0^+\). Thus: \[\lim_{x \to \infty} x \sin(1/x) = \lim_{t \to 0^+} \frac{\sin t}{t} = 1\]
5. For \(L = \lim_{x \to \infty} (\sqrt{9x^2 - 13} - 3x)\) (indeterminate form \(\infty - \infty\)): \[L = \lim_{x \to \infty} \frac{(\sqrt{9x^2 - 13} - 3x)(\sqrt{9x^2 - 13} + 3x)}{\sqrt{9x^2 - 13} + 3x} = \lim_{x \to \infty} \frac{9x^2 - 13 - 9x^2}{\sqrt{9x^2 - 13} + 3x}\] \[= \lim_{x \to \infty} \frac{-13}{\sqrt{9x^2 - 13} + 3x} = \lim_{x \to \infty} \frac{-\frac{13}{x}}{\sqrt{9 - \frac{13}{x^2}} + 3} = \frac{0}{3 + 3} = 0\]

1.4.2 Infinite Limits

Definition:

1. We say \(f(x)\) approaches \(\infty\) as \(x\) approaches \(c\), written \(\lim_{x \to c} f(x) = \infty\), if for every positive real number \(B\) there exists \(\delta > 0\) such that: \[f(x) > B \quad \text{whenever} \quad 0 < |x - c| < \delta\]
2. We say \(f(x)\) approaches \(-\infty\) as \(x\) approaches \(c\), written \(\lim_{x \to c} f(x) = -\infty\), if for every negative real number \(-B\) (\(B > 0\)) there exists \(\delta > 0\) such that: \[f(x) < -B \quad \text{whenever} \quad 0 < |x - c| < \delta\]

Example 1: Prove that \(\lim_{x \to 0} \frac{1}{x^2} = \infty\).

Given \(B > 0\), we want \(\delta > 0\) such that: \[\frac{1}{x^2} > B \quad \text{whenever} \quad 0 < |x| < \delta\]

We have: \[\frac{1}{x^2} > B \iff x^2 < \frac{1}{B} \iff |x| < \frac{1}{\sqrt{B}}\]

Choosing \(\delta = \frac{1}{\sqrt{B}}\): if \(0 < |x| < \delta\), then \(\frac{1}{x^2} > \frac{1}{\delta^2} = B\).

More Examples:

• \(\displaystyle \lim_{x \to 2} \frac{(x-2)^2}{x^2-4} = \lim_{x \to 2} \frac{(x-2)^2}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x-2}{x+2} = 0\)
• \(\displaystyle \lim_{x \to 2^+} \frac{x-3}{x^2-4} = \lim_{x \to 2^+} \frac{x-3}{(x-2)(x+2)} = -\infty\) (numerator approaches \(-1\), denominator approaches \(0^+\))
• \(\displaystyle \lim_{x \to 2^-} \frac{x-3}{x^2-4} = \lim_{x \to 2^-} \frac{x-3}{(x-2)(x+2)} = \infty\) (numerator approaches \(-1\), denominator approaches \(0^-\))
• \(\displaystyle \lim_{x \to 2} \frac{x-3}{x^2-4}\) does not exist (one-sided limits differ)
• \(\displaystyle \lim_{x \to \pm\infty} \frac{2x^5 - 6x^4 + 1}{3x^2 + x - 7} = \lim_{x \to \pm\infty} x^3 \frac{2 - \frac{6}{x} + \frac{1}{x^5}}{3 + \frac{1}{x} - \frac{7}{x^2}} = \pm\infty\)

1.4.3 Horizontal Asymptotes

If the distance between the graph of a function and a fixed line approaches zero as a point on the graph moves increasingly far from the origin, we say the line is an asymptote of the graph.

Definition (Horizontal Asymptote): A line \(y = b\) is a horizontal asymptote of the graph of \(y = f(x)\) if: \[\text{either } \lim_{x \to \infty} f(x) = b \quad \text{or } \lim_{x \to -\infty} f(x) = b\]

Example: Find the horizontal asymptotes of \(f(x) = \frac{x^3 - 2}{|x|^3 + 1}\).

For \(x > 0\): \[\lim_{x \to \infty} \frac{x^3 - 2}{|x|^3 + 1} = \lim_{x \to \infty} \frac{x^3 - 2}{x^3 + 1} = \lim_{x \to \infty} \frac{1 - \frac{2}{x^3}}{1 + \frac{1}{x^3}} = 1\]

For \(x < 0\): \[\lim_{x \to -\infty} \frac{x^3 - 2}{|x|^3 + 1} = \lim_{x \to -\infty} \frac{x^3 - 2}{(-x)^3 + 1} = \lim_{x \to -\infty} \frac{x^3 - 2}{-x^3 + 1} = \lim_{x \to -\infty} \frac{1 - \frac{2}{x^3}}{-1 + \frac{1}{x^3}} = -1\]

The horizontal asymptotes are \(y = -1\) and \(y = 1\).

1.4.4 Vertical Asymptotes

Definition (Vertical Asymptote): A line \(x = a\) is a vertical asymptote of the graph of \(y = f(x)\) if: \[\text{either } \lim_{x \to a^+} f(x) = \pm\infty \quad \text{or } \lim_{x \to a^-} f(x) = \pm\infty\]

Example 1: Find asymptotes of \(f(x) = \frac{x+3}{x+2}\).

Domain: \(D_f = \mathbb{R} \setminus \{-2\}\). We can write \(f(x) = 1 + \frac{1}{x+2}\).

• Horizontal: \(\lim_{x \to \pm\infty} f(x) = 1\), so \(y = 1\) is a horizontal asymptote
• Vertical: \(\lim_{x \to -2^+} f(x) = \infty\) and \(\lim_{x \to -2^-} f(x) = -\infty\), so \(x = -2\) is a vertical asymptote

Example 2: Find asymptotes of \(f(x) = -\frac{8}{x^2 - 4}\).

Domain: \(D_f = \mathbb{R} \setminus \{-2, 2\}\).

• Horizontal: \(\lim_{x \to \pm\infty} f(x) = 0\), so \(y = 0\) is a horizontal asymptote
• Vertical:
  • \(\lim_{x \to 2^+} f(x) = -\infty\), \(\lim_{x \to 2^-} f(x) = \infty\), so \(x = 2\) is a vertical asymptote
  • \(\lim_{x \to -2^+} f(x) = \infty\), \(\lim_{x \to -2^-} f(x) = -\infty\), so \(x = -2\) is a vertical asymptote

1.4.5 Oblique (Slant) Asymptotes

Definition (Oblique Asymptote): A function \(f(x)\) is asymptotic to the straight line \(y = \alpha x + \beta\) (\(\alpha \neq 0\)) if: \[\text{either } \lim_{x \to \infty} (f(x) - (\alpha x + \beta)) = 0 \quad \text{or } \lim_{x \to -\infty} (f(x) - (\alpha x + \beta)) = 0\]

Example: Find the oblique asymptote of \(f(x) = \frac{x^2 - 3}{2x - 4}\).

Using polynomial division: \[f(x) = \frac{x^2 - 3}{2x - 4} = \frac{1}{2}x + 1 + \frac{1}{2x - 4}\]

Therefore: \[\lim_{x \to \pm\infty} \left(f(x) - \left(\frac{1}{2}x + 1\right)\right) = \lim_{x \to \pm\infty} \frac{1}{2x - 4} = 0\]

The line \(y = \frac{1}{2}x + 1\) is an oblique asymptote.

1.5 Continuity

1.5.1 Continuity at a Point

The continuity is one of the most important properties of a function. Intuitively, a function is continuous if we can draw its graph without lifting the pen from paper.

Definition: Let \(c\) be either an interior point or an endpoint of an interval in the domain of \(f\).

1. The function \(f\) is continuous at \(c\) if: \[\lim_{x \to c} f(x) = f(c)\]
2. The function \(f\) is right-continuous at \(c\) if: \[\lim_{x \to c^+} f(x) = f(c)\]
3. The function \(f\) is left-continuous at \(c\) if: \[\lim_{x \to c^-} f(x) = f(c)\]

Note that for \(f\) to be continuous at \(c\), three conditions must hold:

• \(f(c)\) is defined
• \(\lim_{x \to c} f(x)\) exists
• \(\lim_{x \to c} f(x) = f(c)\)

A function is continuous over a closed interval \([a, b]\) if it is right-continuous at \(a\), left-continuous at \(b\), and continuous at all interior points.

If a function is not continuous at point \(c\) in its domain, we say \(f\) is discontinuous at \(c\).

1.5.2 Continuous Functions

A continuous function is one that is continuous at every point in its domain. If a function is discontinuous at one or more points, it is a discontinuous function.

Theorem (Properties of Continuous Functions): If functions \(f\) and \(g\) are continuous at \(x = c\), then the following are also continuous at \(x = c\):

1. Sums: \(f + g\)
2. Differences: \(f - g\)
3. Constant multiples: \(\alpha f\) for any \(\alpha \in \mathbb{R}\)
4. Products: \(fg\)
5. Quotients: \(f/g\), provided \(g(c) \neq 0\)
6. Powers: \(f^n\), \(n\) a positive integer
7. Roots: \(\sqrt[n]{f}\), provided it is defined on an interval containing \(c\)

Examples of continuous functions:

1. Identity and constant functions: \(f(x) = x\) and \(f(x) = k\) are continuous everywhere.
2. Reciprocal function: \(f(x) = \frac{1}{x}\) is continuous on its domain \(\mathbb{R} \setminus \{0\}\).
3. Polynomials: Every polynomial \(P(x) = a_n x^n + \cdots + a_1 x + a_0\) is continuous everywhere because \(\lim_{x \to c} P(x) = P(c)\).
4. Rational functions: If \(P(x)\) and \(Q(x)\) are polynomials, then \(\frac{P(x)}{Q(x)}\) is continuous wherever \(Q(x) \neq 0\).
5. Absolute value: \(f(x) = |x|\) is continuous everywhere. For \(x > 0\), \(f(x) = x\) (polynomial). For \(x < 0\), \(f(x) = -x\) (polynomial). At \(x = 0\), \(\lim_{x \to 0} |x| = 0 = |0|\).
6. Trigonometric functions: \(\sin x\) and \(\cos x\) are continuous everywhere. All six trigonometric functions are continuous wherever they are defined. For example, \(\tan x\) is continuous on \(\cdots \cup (-\frac{\pi}{2}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup \cdots\).

1.6 Continuity of Compositions of Functions

Remark (Inverse Functions): When a continuous function defined on an interval has an inverse, the inverse function is itself continuous over its domain. This is because the graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y = x\). As an example, all inverse trigonometric functions are continuous over their domains.

Theorem (Compositions of Continuous Functions): If \(f\) is continuous at \(c\) and \(g\) is continuous at \(f(c)\), then the composition \(g \circ f\) is continuous at \(c\), and: \[\lim_{x \to c} (g \circ f)(x) = g(f(c))\]

The continuity of compositions holds for any finite number of functions, provided each is continuous where it is applied.

Examples:

1. \(y = \sqrt{P_n(x)}\) where \(P_n(x)\) is a polynomial. The square root function is continuous on \([0, \infty)\). The given function is the composition of the polynomial with the square root function, so it’s continuous on its natural domain (where \(P_n(x) \geq 0\)).
2. \(y = \frac{x^{2/3}}{1 + x^4}\). The numerator is the cube root of the identity function squared; the denominator is an everywhere-positive polynomial. The quotient is continuous everywhere.
3. \(y = \left|\frac{x-2}{x^2-16}\right|\). The quotient \(\frac{x-2}{x^2-16}\) is continuous for \(x \in \mathbb{R} \setminus \{-4, 4\}\), and the function is the composition with the continuous absolute value function, so it’s continuous on \(\mathbb{R} \setminus \{-4, 4\}\).

Theorem (Limits of Continuous Functions): If \(\lim_{x \to c} f(x) = b\) and \(g\) is continuous at \(b\), then: \[\lim_{x \to c} g(f(x)) = g(b) = g\left(\lim_{x \to c} f(x)\right)\]

Proof: Let \(\varepsilon > 0\) be given. Since \(g\) is continuous at \(b\), \(\lim_{y \to b} g(y) = g(b)\), so there exists \(\delta_1 > 0\) such that: \[|g(y) - g(b)| < \varepsilon \quad \text{whenever} \quad |y - b| < \delta_1\]

Note this inequality also holds when \(y = b\).

Since \(\lim_{x \to c} f(x) = b\), setting \(y = f(x)\), there exists \(\delta > 0\) such that: \[|f(x) - b| < \delta_1 \quad \text{whenever} \quad 0 < |x - c| < \delta\]

This implies \(|g(f(x)) - g(b)| < \varepsilon\) whenever \(0 < |x - c| < \delta\).

By the definition of limit, \(\lim_{x \to c} g(f(x)) = g(b)\).

Examples:

1. \(\displaystyle \lim_{x \to 0} \sin\left(\frac{\pi}{2} \cos(\tan x)\right) = \sin\left(\frac{\pi}{2} \cos(0)\right) = \sin\left(\frac{\pi}{2}\right) = 1\)
2. \(\displaystyle \lim_{x \to 1} \cos^{-1}(\ln(\sqrt{x})) = \cos^{-1}(\ln(1)) = \cos^{-1}(0) = \frac{\pi}{2}\)
3. \(\displaystyle \lim_{x \to 0} \sec\left(e^x + \pi \tan\left(\frac{\pi}{4\sec x}\right) - 1\right) = \sec\left(1 + \pi \tan\left(\frac{\pi}{4}\right) - 1\right) = \sec(\pi) = -1\)

1.7 Intermediate Value Theorem

A function has the Intermediate Value Property if whenever it takes on two values, it also takes on all values in between.

Theorem (Intermediate Value Theorem): If \(f\) is a continuous function on a closed interval \([a, b]\), and if \(y_0\) is any value between \(f(a)\) and \(f(b)\), then \(y_0 = f(c)\) for some \(c\) in \([a, b]\).

Geometrically, any horizontal line \(y = y_0\) between \(f(a)\) and \(f(b)\) will cross the curve \(y = f(x)\) at least once over \([a, b]\).

The proof depends on the completeness property of real numbers. Continuity on the interval is essential—if \(f\) is discontinuous at even one point, the theorem may fail.

Consequence for Graphing (Connectedness): The graph of a function continuous on an interval cannot have any breaks—it will be a single, unbroken curve.

Consequence for Root Finding: A solution of \(f(x) = 0\) is called a root or zero of \(f\). The Intermediate Value Theorem tells us that if \(f\) is continuous and changes sign on an interval, then the interval contains a zero of the function.

Example: Show that there is a root of \(\sqrt{2x + 5} = 4 - x^2\) between 0 and 2.

The equation is equivalent to \(f(x) = 0\) where: \[f(x) = \sqrt{2x + 5} - 4 + x^2\]

The function \(f\) is continuous on \([0, 2]\). We have: \[f(0) = \sqrt{5} - 4 \approx 2.236 - 4 = -1.764 < 0\] \[f(2) = \sqrt{9} - 4 + 4 = 3 > 0\]

Since \(f(0) < 0 < f(2)\), by the Intermediate Value Theorem, there exists \(c \in [0, 2]\) such that \(f(c) = 0\), meaning there is a root in \([0, 2]\).

1.8 Continuous Extension to a Point

If \(f(c)\) is not defined but \(\lim_{x \to c} f(x) = L\) exists, we can define a new function: \[F(x) = \begin{cases} f(x), & \text{if } x \text{ is in the domain of } f \\ L, & \text{if } x = c \end{cases}\]

The function \(F\) is continuous at \(x = c\) and is called the continuous extension of \(f\) to \(x = c\).

Example: The function \(f(x) = \frac{x^2 + x - 6}{x^2 - 4}\), \(x \neq 2\), has a continuous extension to \(x = 2\).

We have: \[f(x) = \frac{x^2 + x - 6}{x^2 - 4} = \frac{(x-2)(x+3)}{(x-2)(x+2)} = \frac{x+3}{x+2} \quad \text{for } x \neq 2\]

Thus \(\lim_{x \to 2} f(x) = \frac{5}{4}\), and the continuous extension is: \[F(x) = \frac{x+3}{x+2}\]

1.9 The Number \(e\) and Related Limits

We have seen from sequences that: \[\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e = 2.71828\ldots\]

Theorem: For real \(x\): \[\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e\]

Corollary 1: This is equivalent to: \[\lim_{x \to 0} (1 + x)^{1/x} = e\]

Corollary 2: For any \(a \in \mathbb{R}\): \[\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a\]

Proof: Let \(t = a/x\). When \(x \to \infty\), \(t \to 0\). Then: \[\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = \lim_{t \to 0} (1 + t)^{a/t} = \left(\lim_{t \to 0} (1 + t)^{1/t}\right)^a = e^a\]

Lemma: If \(\lim_{x \to c} g(x) = a\) and \(\lim_{x \to c} h(x) = b\), then: \[\lim_{x \to c} (g(x))^{h(x)} = a^b\]

Corollary (for indeterminate form \(1^{\infty}\)): Suppose \(\lim_{x \to c} g(x) = 1\) and \(\lim_{x \to c} h(x) = \infty\). Then: \[\lim_{x \to c} (g(x))^{h(x)} = \lim_{x \to c} \left([1 + (g(x) - 1)]^{\frac{1}{g(x)-1}}\right)^{(g(x)-1)h(x)} = e^{\lim_{x \to c} [h(x)(g(x)-1)]}\]

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2. Определения

• Limit of a function (\(\varepsilon\)-\(\delta\) definition): \(\lim_{x \to c} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon\) whenever \(0 < |x - c| < \delta\).
• Limit point: A point \(c\) is a limit point of a set \(A\) if each neighborhood of \(c\) contains at least one point of \(A\) different from \(c\).
• Heine’s definition of limit: \(f\) has limit \(L\) at \(c\) if for any sequence \(x_n \to c\) with \(x_n \neq c\), we have \(f(x_n) \to L\).
• Right-hand limit: \(\lim_{x \to c^+} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon\) whenever \(c < x < c + \delta\).
• Left-hand limit: \(\lim_{x \to c^-} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon\) whenever \(c - \delta < x < c\).
• Limit at infinity: \(\lim_{x \to \infty} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(M\) such that \(|f(x) - L| < \varepsilon\) whenever \(x > M\).
• Infinite limit: \(\lim_{x \to c} f(x) = \infty\) means for every \(B > 0\), there exists \(\delta > 0\) such that \(f(x) > B\) whenever \(0 < |x - c| < \delta\).
• Horizontal asymptote: A line \(y = b\) is a horizontal asymptote of \(y = f(x)\) if \(\lim_{x \to \infty} f(x) = b\) or \(\lim_{x \to -\infty} f(x) = b\).
• Vertical asymptote: A line \(x = a\) is a vertical asymptote of \(y = f(x)\) if \(\lim_{x \to a^+} f(x) = \pm\infty\) or \(\lim_{x \to a^-} f(x) = \pm\infty\).
• Oblique (slant) asymptote: The line \(y = \alpha x + \beta\) (\(\alpha \neq 0\)) is an oblique asymptote if \(\lim_{x \to \pm\infty} (f(x) - (\alpha x + \beta)) = 0\).
• Continuous at a point: A function \(f\) is continuous at \(c\) if \(\lim_{x \to c} f(x) = f(c)\).
• Right-continuous: \(f\) is right-continuous at \(c\) if \(\lim_{x \to c^+} f(x) = f(c)\).
• Left-continuous: \(f\) is left-continuous at \(c\) if \(\lim_{x \to c^-} f(x) = f(c)\).
• Continuous function: A function continuous at every point in its domain.
• Continuous extension: If \(\lim_{x \to c} f(x) = L\) but \(f(c)\) is undefined, defining \(F(c) = L\) creates a continuous extension of \(f\) at \(c\).
• Intermediate Value Property: If \(f\) takes values \(f(a)\) and \(f(b)\), it takes all values between them.
• Indeterminate form: Expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(1^{\infty}\), \(0^0\), \(\infty^0\) that arise in limit evaluation where direct substitution is ambiguous.

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3. Формулы

• Basic limits: \(\lim_{x \to c} x = c\), \(\lim_{x \to c} k = k\) (constant)
• Sum rule: \(\lim_{x \to c} (f(x) + g(x)) = \lim_{x \to c} f(x) + \lim_{x \to c} g(x)\)
• Difference rule: \(\lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} f(x) - \lim_{x \to c} g(x)\)
• Product rule: \(\lim_{x \to c} (f(x) \cdot g(x)) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)\)
• Quotient rule: \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}\), provided denominator \(\neq 0\)
• Power rule: \(\lim_{x \to c} [f(x)]^n = [\lim_{x \to c} f(x)]^n\)
• Root rule: \(\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to c} f(x)}\) (with appropriate domain restrictions)
• Polynomial limit: If \(P(x)\) is a polynomial, then \(\lim_{x \to c} P(x) = P(c)\)
• Rational function limit: \(\lim_{x \to c} \frac{P(x)}{Q(x)} = \frac{P(c)}{Q(c)}\) if \(Q(c) \neq 0\)
• Important trigonometric limit: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) (\(\theta\) in radians)
• Related trigonometric limits: \(\lim_{\theta \to 0} \sin \theta = 0\), \(\lim_{\theta \to 0} \cos \theta = 1\)
• Limit with exponential: \(\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e\)
• Equivalent form: \(\lim_{x \to 0} (1 + x)^{1/x} = e\)
• General exponential limit: \(\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a\)
• Exponential-logarithmic limit: \(\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a\) (for \(a > 0\))
• Indeterminate form \(1^{\infty}\): If \(\lim_{x \to c} g(x) = 1\) and \(\lim_{x \to c} h(x) = \infty\), then \(\lim_{x \to c} (g(x))^{h(x)} = e^{\lim_{x \to c} [h(x)(g(x)-1)]}\)
• Limits at infinity: \(\lim_{x \to \pm\infty} \frac{1}{x} = 0\), \(\lim_{x \to -\infty} e^x = 0\)
• Composition of continuous functions: If \(f\) is continuous at \(c\) and \(g\) is continuous at \(f(c)\), then \(\lim_{x \to c} g(f(x)) = g(f(c))\)

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4. Примеры

4.1. Limit with Cube Root (Лаба 12, Задание 1a)

Evaluate \(\displaystyle \lim_{x\to0} \frac{\sqrt[3]{x+1} - 1}{x}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Use the algebraic identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) by multiplying by the conjugate expression for cube roots.

1. Identify the conjugate factor: For \(\sqrt[3]{x+1} - 1\), we multiply by \(\frac{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\)
2. Apply the identity: Using \(a - b = \frac{a^3 - b^3}{a^2 + ab + b^2}\) with \(a = \sqrt[3]{x+1}\) and \(b = 1\): \[\sqrt[3]{x+1} - 1 = \frac{(x+1) - 1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1} = \frac{x}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\]
3. Substitute into the limit: \[\lim_{x\to0} \frac{\sqrt[3]{x+1} - 1}{x} = \lim_{x\to0} \frac{x}{x \cdot \left[(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1\right]}\] \[= \lim_{x\to0} \frac{1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\]
4. Evaluate by direct substitution: \[= \frac{1}{(\sqrt[3]{1})^2 + \sqrt[3]{1} + 1} = \frac{1}{1 + 1 + 1} = \frac{1}{3}\]

Answer: \(\dfrac{1}{3}\)

4.2. Limit of Polynomial Rational Function (Лаба 12, Задание 1b)

Evaluate \(\displaystyle \lim_{x\to-1} \frac{x^3 - x^2 - 5x - 3}{(x+1)^2}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Since substituting \(x = -1\) gives \(\frac{0}{0}\), factor the numerator to cancel common factors.

1. Verify the indeterminate form: Numerator at \(x = -1\): \((-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0\) Denominator at \(x = -1\): \((0)^2 = 0\) This is \(\frac{0}{0}\), so we need to factor.

2. Factor the numerator: Since \(x = -1\) is a root, \((x+1)\) is a factor. Perform polynomial division: \[x^3 - x^2 - 5x - 3 = (x+1)(x^2 - 2x - 3)\]

   Factor further: \(x^2 - 2x - 3 = (x-3)(x+1)\)

   So: \(x^3 - x^2 - 5x - 3 = (x+1)^2(x-3)\)

3. Simplify the expression: \[\frac{x^3 - x^2 - 5x - 3}{(x+1)^2} = \frac{(x+1)^2(x-3)}{(x+1)^2} = x - 3 \quad (x \neq -1)\]

4. Evaluate the limit: \[\lim_{x\to-1} (x - 3) = -1 - 3 = -4\]

Answer: \(-4\)

4.3. Trigonometric Limit with Sine and Cosine (Лаба 12, Задание 1c)

Evaluate \(\displaystyle \lim_{x\to0} \frac{x \sin x}{2 - 2 \cos x}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Use the identity \(1 - \cos x = 2\sin^2(x/2)\) and known limits.

1. Factor the denominator: \[\lim_{x\to0} \frac{x \sin x}{2(1 - \cos x)}\]
2. Use the identity \(1 - \cos x = 2\sin^2(x/2)\): \[= \lim_{x\to0} \frac{x \sin x}{2 \cdot 2\sin^2(x/2)} = \lim_{x\to0} \frac{x \sin x}{4\sin^2(x/2)}\]
3. Rewrite using standard limit forms: \[= \lim_{x\to0} \frac{x}{\sin(x/2)} \cdot \frac{\sin x}{4\sin(x/2)}\] \[= \lim_{x\to0} \frac{x}{x/2} \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{\sin x}{x} \cdot \frac{x}{4\sin(x/2)}\] \[= \lim_{x\to0} 2 \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{\sin x}{x} \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{1}{2}\]
4. Alternatively, using L’Hôpital’s rule or direct substitution: \[= \lim_{x\to0} \frac{x \sin x}{4\sin^2(x/2)} = \lim_{x\to0} \frac{x \cdot 2\sin(x/2)\cos(x/2)}{4\sin^2(x/2)}\] \[= \lim_{x\to0} \frac{x \cos(x/2)}{2\sin(x/2)} = \lim_{x\to0} \frac{x/2}{\sin(x/2)} \cdot \cos(x/2) = 1 \cdot 1 = 1\]

Answer: \(1\)

4.4. One-Sided Limits of a Rational Function (Лаба 12, Задание 2)

Given \(f(x) = \frac{x^2 - 3x + 2}{x^3 - 4x}\), find the following limits:

1. \(\displaystyle \lim_{x\to-2^+} f(x)\)
2. \(\displaystyle \lim_{x\to0^-} f(x)\)
3. \(\displaystyle \lim_{x\to0^+} f(x)\)
4. \(\displaystyle \lim_{x\to1^+} f(x)\)
5. \(\displaystyle \lim_{x\to2^+} f(x)\)

Нажмите, чтобы увидеть решение

Ключевая идея: First factor both numerator and denominator to identify removable discontinuities and vertical asymptotes.

1. Factor the expressions:

   • Numerator: \(x^2 - 3x + 2 = (x-1)(x-2)\)
   • Denominator: \(x^3 - 4x = x(x^2-4) = x(x-2)(x+2)\)

2. Simplify (for \(x \neq 2\)): \[f(x) = \frac{(x-1)(x-2)}{x(x-2)(x+2)} = \frac{x-1}{x(x+2)} \quad (x \neq 2)\]

3. Evaluate each limit:

   a) \(\displaystyle \lim_{x\to-2^+} f(x)\): As \(x \to -2^+\): numerator \((x-1) \to -3\); denominator \(x(x+2) \to (-2)(0^+) = 0^-\) \[\lim_{x\to-2^+} \frac{x-1}{x(x+2)} = \frac{-3}{0^-} = +\infty\]

   b) \(\displaystyle \lim_{x\to0^-} f(x)\): As \(x \to 0^-\): numerator \((x-1) \to -1\); denominator \(x(x+2) \to (0^-)(2) = 0^-\) \[\lim_{x\to0^-} \frac{x-1}{x(x+2)} = \frac{-1}{0^-} = +\infty\]

   c) \(\displaystyle \lim_{x\to0^+} f(x)\): As \(x \to 0^+\): numerator \((x-1) \to -1\); denominator \(x(x+2) \to (0^+)(2) = 0^+\) \[\lim_{x\to0^+} \frac{x-1}{x(x+2)} = \frac{-1}{0^+} = -\infty\]

   d) \(\displaystyle \lim_{x\to1^+} f(x)\): Direct substitution works since \(x = 1\) is not a discontinuity of the simplified function: \[\lim_{x\to1^+} \frac{x-1}{x(x+2)} = \frac{1-1}{1(1+2)} = \frac{0}{3} = 0\]

   e) \(\displaystyle \lim_{x\to2^+} f(x)\): Although \(x = 2\) was cancelled, we use the simplified form: \[\lim_{x\to2^+} \frac{x-1}{x(x+2)} = \frac{2-1}{2(2+2)} = \frac{1}{8}\]

Answers:

• 1. \(+\infty\)
• 2. \(+\infty\)
• 3. \(-\infty\)
• 4. \(0\)
• 5. \(\frac{1}{8}\)

4.5. Find Asymptotes of Rational Functions (Лаба 12, Задание 3)

Find the equations of the asymptotes of the following functions:

1. \(f(x) = \dfrac{x^2 - 1}{2x + 4}\)

2. \(g(x) = \dfrac{x^3 + 1}{x^2}\)

Нажмите, чтобы увидеть решение

Ключевая идея: There are three types of asymptotes:

• Vertical asymptotes occur where the denominator equals zero (and numerator doesn’t)
• Horizontal asymptotes are found by evaluating \(\lim_{x\to\pm\infty} f(x)\)
• Oblique (slant) asymptotes exist when the degree of numerator exceeds denominator by exactly 1

(a) \(f(x) = \dfrac{x^2 - 1}{2x + 4}\):

1. Find vertical asymptotes: Set denominator \(= 0\): \(2x + 4 = 0 \Rightarrow x = -2\)

   Check numerator at \(x = -2\): \((-2)^2 - 1 = 3 \neq 0\)

   Vertical asymptote: \(x = -2\)

2. Check for horizontal asymptote: Degree of numerator (2) > degree of denominator (1), so no horizontal asymptote.

3. Find oblique asymptote: Since degree difference = 1, perform polynomial long division: \[\frac{x^2 - 1}{2x + 4} = \frac{x^2 - 1}{2(x + 2)}\]

   Dividing \(x^2 - 1\) by \(x + 2\): \[x^2 - 1 = (x + 2)(x - 2) + 3\]

   So: \[f(x) = \frac{(x-2)(x+2) + 3}{2(x+2)} = \frac{x-2}{2} + \frac{3}{2(x+2)} = \frac{x}{2} - 1 + \frac{3}{2(x+2)}\]

   As \(x \to \pm\infty\), the term \(\frac{3}{2(x+2)} \to 0\)

   Oblique asymptote: \(y = \dfrac{x}{2} - 1\) (or equivalently \(y = \dfrac{x-2}{2}\))

Asymptotes for (a): Vertical: \(x = -2\); Oblique: \(y = \dfrac{x}{2} - 1\)

(b) \(g(x) = \dfrac{x^3 + 1}{x^2}\):

1. Find vertical asymptotes: Set denominator \(= 0\): \(x^2 = 0 \Rightarrow x = 0\)

   Check numerator at \(x = 0\): \(0^3 + 1 = 1 \neq 0\)

   Vertical asymptote: \(x = 0\)

2. Check for horizontal asymptote: Degree of numerator (3) > degree of denominator (2), so no horizontal asymptote.

3. Find oblique asymptote: Perform polynomial division: \[g(x) = \frac{x^3 + 1}{x^2} = \frac{x^3}{x^2} + \frac{1}{x^2} = x + \frac{1}{x^2}\]

   As \(x \to \pm\infty\), the term \(\frac{1}{x^2} \to 0\)

   Oblique asymptote: \(y = x\)

Asymptotes for (b): Vertical: \(x = 0\); Oblique: \(y = x\)

Answers:

• 1. Vertical asymptote: \(x = -2\); Oblique asymptote: \(y = \dfrac{x}{2} - 1\)
• 2. Vertical asymptote: \(x = 0\); Oblique asymptote: \(y = x\)

4.6. Find Asymptotes with Square Root and Parameters (Лаба 13, Задание 1)

Assume that constants \(a\) and \(b\) are positive. Find equations for all horizontal and vertical asymptotes for the graph of \(y = \frac{\sqrt{ax^2 + 4}}{x - b}\).

Нажмите, чтобы увидеть решение

Ключевая идея: For functions involving square roots, we must carefully handle the sign of \(x\) when taking limits at \(\pm\infty\). Remember that \(\sqrt{x^2} = |x|\).

1. Find vertical asymptotes: Set denominator \(= 0\): \(x - b = 0 \Rightarrow x = b\)

   Check numerator at \(x = b\): \(\sqrt{ab^2 + 4} > 0\) (since \(a, b > 0\))

   Vertical asymptote: \(x = b\)

2. Find horizontal asymptotes as \(x \to \infty\): \[\lim_{x \to \infty} \frac{\sqrt{ax^2 + 4}}{x - b}\]

   For \(x > 0\), we have \(\sqrt{x^2} = x\). Factor out \(x\) from the square root: \[= \lim_{x \to \infty} \frac{\sqrt{x^2(a + \frac{4}{x^2})}}{x - b} = \lim_{x \to \infty} \frac{x\sqrt{a + \frac{4}{x^2}}}{x - b}\]

   Divide numerator and denominator by \(x\): \[= \lim_{x \to \infty} \frac{\sqrt{a + \frac{4}{x^2}}}{1 - \frac{b}{x}} = \frac{\sqrt{a}}{1} = \sqrt{a}\]

3. Find horizontal asymptotes as \(x \to -\infty\): \[\lim_{x \to -\infty} \frac{\sqrt{ax^2 + 4}}{x - b}\]

   For \(x < 0\), we have \(\sqrt{x^2} = |x| = -x\). Thus: \[= \lim_{x \to -\infty} \frac{-x\sqrt{a + \frac{4}{x^2}}}{x - b}\]

   Divide numerator and denominator by \(x\) (note: \(x < 0\)): \[= \lim_{x \to -\infty} \frac{-\sqrt{a + \frac{4}{x^2}}}{1 - \frac{b}{x}} = \frac{-\sqrt{a}}{1} = -\sqrt{a}\]

Answer:

• Vertical asymptote: \(x = b\)
• Horizontal asymptotes: \(y = \sqrt{a}\) (as \(x \to \infty\)) and \(y = -\sqrt{a}\) (as \(x \to -\infty\))

4.7. Indeterminate Form \(1^{\infty}\) with Squared Exponent (Лаба 13, Задание 2a)

Evaluate \(\displaystyle \lim_{x \to \infty} \left(\frac{x}{2x + 1}\right)^{x^2}\).

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Ключевая идея: First determine if this is actually a \(1^{\infty}\) form by checking the base limit.

1. Check the base limit: \[\lim_{x \to \infty} \frac{x}{2x + 1} = \lim_{x \to \infty} \frac{1}{2 + \frac{1}{x}} = \frac{1}{2}\]

   The base approaches \(\frac{1}{2}\), not \(1\). This is NOT a \(1^{\infty}\) form.

2. Analyze the limit: Since the base approaches \(\frac{1}{2} < 1\) and the exponent approaches \(\infty\): \[\left(\frac{1}{2}\right)^{\infty} = 0\]

3. Verify rigorously: For large \(x\), \(\frac{x}{2x+1} < \frac{1}{2} + \varepsilon\) for any \(\varepsilon > 0\). Thus: \[\left(\frac{x}{2x+1}\right)^{x^2} \to 0 \text{ as } x \to \infty\]

Answer: \(0\)

4.8. Indeterminate Form \(1^{\infty}\) - Standard Type (Лаба 13, Задание 2b)

Evaluate \(\displaystyle \lim_{x \to \infty} \left(\frac{x + 1}{2x + 1}\right)^{x}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Check the base limit to determine the form type.

1. Check the base limit: \[\lim_{x \to \infty} \frac{x + 1}{2x + 1} = \lim_{x \to \infty} \frac{1 + \frac{1}{x}}{2 + \frac{1}{x}} = \frac{1}{2}\]

   The base approaches \(\frac{1}{2}\), not \(1\). This is NOT a \(1^{\infty}\) form.

2. Analyze the limit: Since \(\frac{1}{2} < 1\) and the exponent goes to \(\infty\): \[\left(\frac{1}{2}\right)^{\infty} = 0\]

3. Alternative verification using logarithms: \[L = \lim_{x \to \infty} \left(\frac{x + 1}{2x + 1}\right)^{x}\] \[\ln L = \lim_{x \to \infty} x \ln\left(\frac{x + 1}{2x + 1}\right) = \lim_{x \to \infty} x \ln\left(\frac{1 + \frac{1}{x}}{2 + \frac{1}{x}}\right)\]

   As \(x \to \infty\): \(\ln\left(\frac{1}{2}\right) = -\ln 2 < 0\)

   So \(\ln L = \lim_{x \to \infty} x \cdot (-\ln 2 + \text{terms} \to 0) = -\infty\)

   Therefore \(L = e^{-\infty} = 0\)

Answer: \(0\)

4.9. Indeterminate Form \(1^{\infty}\) with Trigonometric Functions (Лаба 13, Задание 2c)

Evaluate \(\displaystyle \lim_{x \to 0} (1 + x^2)^{\cot^2 x}\).

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Ключевая идея: This IS a \(1^{\infty}\) form since the base approaches \(1\) and \(\cot^2 x \to \infty\) as \(x \to 0\). Use the corollary: if \(\lim g(x) = 1\) and \(\lim h(x) = \infty\), then \(\lim (g(x))^{h(x)} = e^{\lim h(x)(g(x)-1)}\).

1. Identify the components:

   • \(g(x) = 1 + x^2\), so \(g(x) - 1 = x^2\)
   • \(h(x) = \cot^2 x = \frac{\cos^2 x}{\sin^2 x}\)

2. Apply the formula: \[L = e^{\lim_{x \to 0} \cot^2 x \cdot x^2} = e^{\lim_{x \to 0} \frac{x^2 \cos^2 x}{\sin^2 x}}\]

3. Evaluate the exponent: \[\lim_{x \to 0} \frac{x^2 \cos^2 x}{\sin^2 x} = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^2 \cdot \cos^2 x\]

   Using \(\lim_{x \to 0} \frac{x}{\sin x} = 1\) and \(\lim_{x \to 0} \cos x = 1\): \[= 1^2 \cdot 1^2 = 1\]

4. Compute the final answer: \[L = e^1 = e\]

Answer: \(e\)

4.10. Continuous Extension - Removable Discontinuity (Лаба 13, Задание 3)

Define \(g(3)\) in a way that extends \(g(x) = \frac{x^2 - 9}{x - 3}\) to be continuous at \(x = 3\).

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Ключевая идея: A function has a removable discontinuity at \(x = c\) if \(\lim_{x \to c} f(x)\) exists but differs from \(f(c)\) (or \(f(c)\) is undefined). To make it continuous, we define \(f(c)\) to equal the limit.

1. Simplify the function: \[g(x) = \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3 \quad (x \neq 3)\]
2. Find the limit at \(x = 3\): \[\lim_{x \to 3} g(x) = \lim_{x \to 3} (x + 3) = 3 + 3 = 6\]
3. Define \(g(3)\) for continuity: For \(g\) to be continuous at \(x = 3\), we need: \[g(3) = \lim_{x \to 3} g(x) = 6\]

Answer: \(g(3) = 6\)

4.11. Continuous Extension - Another Example (Лаба 13, Задание 4)

Define \(h(2)\) in a way that extends \(h(t) = \frac{t^2 + 3t - 10}{t - 2}\) to be continuous at \(t = 2\).

Нажмите, чтобы увидеть решение

Ключевая идея: Factor the numerator to cancel the problematic factor in the denominator.

1. Factor the numerator: We need to factor \(t^2 + 3t - 10\). Looking for two numbers that multiply to \(-10\) and add to \(3\): these are \(5\) and \(-2\). \[t^2 + 3t - 10 = (t + 5)(t - 2)\]
2. Simplify the function: \[h(t) = \frac{(t + 5)(t - 2)}{t - 2} = t + 5 \quad (t \neq 2)\]
3. Find the limit at \(t = 2\): \[\lim_{t \to 2} h(t) = \lim_{t \to 2} (t + 5) = 2 + 5 = 7\]
4. Define \(h(2)\) for continuity: \[h(2) = 7\]

Answer: \(h(2) = 7\)

4.12. Continuity of Piecewise Function - One Parameter (Лаба 13, Задание 5)

For what values of \(a\) is the function \[f(x) = \begin{cases} a^2x - 2a, & x \geq 2 \\ 12, & x < 2 \end{cases}\] continuous at every \(x\)?

Нажмите, чтобы увидеть решение

Ключевая идея: A piecewise function is continuous everywhere if: (1) each piece is continuous on its domain, and (2) the pieces “connect” at the boundary points.

1. Analyze each piece:
   • For \(x < 2\): \(f(x) = 12\) (constant, always continuous)
   • For \(x \geq 2\): \(f(x) = a^2x - 2a\) (linear, always continuous)
2. Check continuity at the boundary \(x = 2\): For continuity at \(x = 2\), we need: \[\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\]
3. Calculate the limits:
   • Left-hand limit: \(\lim_{x \to 2^-} f(x) = 12\)
   • Right-hand limit: \(\lim_{x \to 2^+} f(x) = a^2(2) - 2a = 2a^2 - 2a\)
   • Function value: \(f(2) = a^2(2) - 2a = 2a^2 - 2a\)
4. Set the limits equal: \[12 = 2a^2 - 2a\] \[2a^2 - 2a - 12 = 0\] \[a^2 - a - 6 = 0\] \[(a - 3)(a + 2) = 0\] \[a = 3 \quad \text{or} \quad a = -2\]

Answer: \(a = 3\) or \(a = -2\)

4.13. Continuity of Piecewise Function - Two Parameters (Лаба 13, Задание 6)

For what values of \(a\) and \(b\) is the function \[g(x) = \begin{cases} -2, & x \leq -1 \\ ax - b, & -1 < x < 1 \\ 3, & x \geq 1 \end{cases}\] continuous at every \(x\)?

Нажмите, чтобы увидеть решение

Ключевая идея: We need to ensure continuity at both boundary points \(x = -1\) and \(x = 1\).

1. Analyze each piece:

   • For \(x \leq -1\): \(f(x) = -2\) (constant, continuous)
   • For \(-1 < x < 1\): \(f(x) = ax - b\) (linear, continuous)
   • For \(x \geq 1\): \(f(x) = 3\) (constant, continuous)

2. Check continuity at \(x = -1\):

   • Left-hand limit: \(\lim_{x \to -1^-} g(x) = -2\)
   • Right-hand limit: \(\lim_{x \to -1^+} g(x) = a(-1) - b = -a - b\)
   • Function value: \(g(-1) = -2\)

   For continuity: \(-a - b = -2\), which gives us: \[a + b = 2 \quad \text{...(Equation 1)}\]

3. Check continuity at \(x = 1\):

   • Left-hand limit: \(\lim_{x \to 1^-} g(x) = a(1) - b = a - b\)
   • Right-hand limit: \(\lim_{x \to 1^+} g(x) = 3\)
   • Function value: \(g(1) = 3\)

   For continuity: \(a - b = 3 \quad \text{...(Equation 2)}\)

4. Solve the system of equations: From Equation 1: \(a + b = 2\) From Equation 2: \(a - b = 3\)

   Adding these equations: \(2a = 5 \Rightarrow a = \frac{5}{2}\)

   Substituting back: \(\frac{5}{2} + b = 2 \Rightarrow b = 2 - \frac{5}{2} = -\frac{1}{2}\)

Answer: \(a = \frac{5}{2}\) and \(b = -\frac{1}{2}\)

4.14. Continuity of Piecewise Function - Three Pieces with Two Parameters (Лаба 13, Задание 7)

For what values of \(a\) and \(b\) is the function \[h(x) = \begin{cases} ax + 2b, & x \leq 0 \\ x^2 + 3a - b, & 0 < x \leq 2 \\ 3x - 5, & x > 2 \end{cases}\] continuous at every \(x\)?

Нажмите, чтобы увидеть решение

Ключевая идея: We need continuity at both boundary points \(x = 0\) and \(x = 2\).

1. Analyze each piece:

   • For \(x \leq 0\): \(h(x) = ax + 2b\) (linear, continuous)
   • For \(0 < x \leq 2\): \(h(x) = x^2 + 3a - b\) (polynomial, continuous)
   • For \(x > 2\): \(h(x) = 3x - 5\) (linear, continuous)

2. Check continuity at \(x = 0\):

   • Left-hand limit and function value: \(\lim_{x \to 0^-} h(x) = h(0) = a(0) + 2b = 2b\)
   • Right-hand limit: \(\lim_{x \to 0^+} h(x) = 0^2 + 3a - b = 3a - b\)

   For continuity: \(2b = 3a - b\), which gives us: \[3b = 3a \Rightarrow a = b \quad \text{...(Equation 1)}\]

3. Check continuity at \(x = 2\):

   • Left-hand limit and function value: \(\lim_{x \to 2^-} h(x) = h(2) = 2^2 + 3a - b = 4 + 3a - b\)
   • Right-hand limit: \(\lim_{x \to 2^+} h(x) = 3(2) - 5 = 1\)

   For continuity: \(4 + 3a - b = 1\), which gives us: \[3a - b = -3 \quad \text{...(Equation 2)}\]

4. Solve the system of equations: Substituting Equation 1 (\(a = b\)) into Equation 2: \[3a - a = -3\] \[2a = -3\] \[a = -\frac{3}{2}\]

   From Equation 1: \(b = a = -\frac{3}{2}\)

Answer: \(a = -\frac{3}{2}\) and \(b = -\frac{3}{2}\)

4.15. Evaluating a Limit Using Factorization (Глава 5, Пример 1)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{5x^3 + 8x^2}{3x^4 - 16x^2}\).

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Ключевая идея: Factor out common terms from numerator and denominator to cancel and remove the \(\frac{0}{0}\) indeterminate form.

1. Factor the numerator and denominator: \[\frac{5x^3 + 8x^2}{3x^4 - 16x^2} = \frac{x^2(5x + 8)}{x^2(3x^2 - 16)}\]
2. Cancel common factors (valid since \(x \neq 0\)): \[= \frac{5x + 8}{3x^2 - 16}\]
3. Evaluate the limit by direct substitution: \[\lim_{x \to 0} \frac{5x + 8}{3x^2 - 16} = \frac{5(0) + 8}{3(0)^2 - 16} = \frac{8}{-16} = -\frac{1}{2}\]

Answer: \(-\frac{1}{2}\)

4.16. Simplifying Before Taking the Limit (Глава 5, Пример 2)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\frac{1}{x-1} + \frac{1}{x+1}}{x}\).

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Ключевая идея: Combine fractions in the numerator, then simplify.

1. Combine the fractions in the numerator: \[\frac{1}{x-1} + \frac{1}{x+1} = \frac{(x+1) + (x-1)}{(x-1)(x+1)} = \frac{2x}{x^2-1}\]
2. Rewrite the original limit: \[\lim_{x \to 0} \frac{\frac{2x}{x^2-1}}{x} = \lim_{x \to 0} \frac{2x}{x(x^2-1)} = \lim_{x \to 0} \frac{2}{x^2-1}\]
3. Evaluate by direct substitution: \[= \frac{2}{0^2-1} = \frac{2}{-1} = -2\]

Answer: \(-2\)

4.17. Factoring and Canceling (Глава 5, Пример 3)

Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x^4 - 16}\).

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1. Factor using difference formulas:
   • Numerator: \(x^3 - 8 = (x-2)(x^2 + 2x + 4)\)
   • Denominator: \(x^4 - 16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4)\)
2. Cancel the common factor \((x-2)\): \[\frac{x^3 - 8}{x^4 - 16} = \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)(x^2+4)} = \frac{x^2 + 2x + 4}{(x+2)(x^2+4)}\]
3. Evaluate by substitution: \[\lim_{x \to 2} \frac{x^2 + 2x + 4}{(x+2)(x^2+4)} = \frac{4 + 4 + 4}{(4)(8)} = \frac{12}{32} = \frac{3}{8}\]

Answer: \(\frac{3}{8}\)

4.18. Polynomial Division (Глава 5, Пример 4)

Evaluate \(\displaystyle \lim_{x \to -1} \frac{x^3 - x^2 - 5x - 3}{(x+1)^2}\).

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Ключевая идея: Since \(x = -1\) is a root of the numerator, factor out \((x+1)\).

1. Check if \(x = -1\) is a root of the numerator: \[(-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0\]

   Yes, so \((x+1)\) is a factor.

2. Factor the numerator by polynomial division or synthetic division: \[x^3 - x^2 - 5x - 3 = (x+1)(x^2 - 2x - 3) = (x+1)(x-3)(x+1) = (x+1)^2(x-3)\]

3. Simplify and evaluate: \[\lim_{x \to -1} \frac{(x+1)^2(x-3)}{(x+1)^2} = \lim_{x \to -1} (x-3) = -1 - 3 = -4\]

Answer: \(-4\)

4.19. Rationalizing the Denominator (Глава 5, Пример 5)

Evaluate \(\displaystyle \lim_{x \to 1} \frac{x-1}{\sqrt{x+3} - 2}\).

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Ключевая идея: Multiply by the conjugate of the denominator.

1. Multiply numerator and denominator by the conjugate: \[\lim_{x \to 1} \frac{x-1}{\sqrt{x+3} - 2} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2}\]
2. Simplify: \[= \lim_{x \to 1} \frac{(x-1)(\sqrt{x+3} + 2)}{(x+3) - 4} = \lim_{x \to 1} \frac{(x-1)(\sqrt{x+3} + 2)}{x-1}\]
3. Cancel \((x-1)\) and evaluate: \[= \lim_{x \to 1} (\sqrt{x+3} + 2) = \sqrt{4} + 2 = 2 + 2 = 4\]

Answer: \(4\)

4.20. Rationalizing the Numerator (Глава 5, Пример 6)

Evaluate \(\displaystyle \lim_{x \to -3} \frac{2 - \sqrt{x^2-5}}{x+3}\).

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1. Multiply by the conjugate of the numerator: \[\lim_{x \to -3} \frac{2 - \sqrt{x^2-5}}{x+3} \cdot \frac{2 + \sqrt{x^2-5}}{2 + \sqrt{x^2-5}}\]
2. Simplify: \[= \lim_{x \to -3} \frac{4 - (x^2-5)}{(x+3)(2 + \sqrt{x^2-5})} = \lim_{x \to -3} \frac{9 - x^2}{(x+3)(2 + \sqrt{x^2-5})}\]
3. Factor the numerator: \[= \lim_{x \to -3} \frac{(3-x)(3+x)}{(x+3)(2 + \sqrt{x^2-5})} = \lim_{x \to -3} \frac{-(x-3)(x+3)}{(x+3)(2 + \sqrt{x^2-5})}\]
4. Cancel \((x+3)\) and evaluate: \[= \lim_{x \to -3} \frac{-(x-3)}{2 + \sqrt{x^2-5}} = \frac{-(-3-3)}{2 + \sqrt{9-5}} = \frac{6}{2 + 2} = \frac{6}{4} = \frac{3}{2}\]

Answer: \(\frac{3}{2}\)

4.21. Another Rationalization Problem (Глава 5, Пример 7)

Evaluate \(\displaystyle \lim_{x \to 3} \frac{x^2-9}{\sqrt{x^2+7} - 4}\).

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1. Multiply by the conjugate of the denominator: \[\lim_{x \to 3} \frac{x^2-9}{\sqrt{x^2+7} - 4} \cdot \frac{\sqrt{x^2+7} + 4}{\sqrt{x^2+7} + 4}\]
2. Simplify: \[= \lim_{x \to 3} \frac{(x^2-9)(\sqrt{x^2+7} + 4)}{(x^2+7) - 16} = \lim_{x \to 3} \frac{(x^2-9)(\sqrt{x^2+7} + 4)}{x^2-9}\]
3. Cancel \((x^2-9)\) and evaluate: \[= \lim_{x \to 3} (\sqrt{x^2+7} + 4) = \sqrt{9+7} + 4 = 4 + 4 = 8\]

Answer: \(8\)

4.22. Cube Root Rationalization (Глава 5, Пример 8)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\sqrt[3]{x+1} - 1}{x}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Use the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Here, let \(a = \sqrt[3]{x+1}\) and \(b = 1\).

1. Multiply by the conjugate factor: \[\lim_{x \to 0} \frac{\sqrt[3]{x+1} - 1}{x} \cdot \frac{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\]
2. Use the difference of cubes in the numerator: \[= \lim_{x \to 0} \frac{(\sqrt[3]{x+1})^3 - 1^3}{x((\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1)} = \lim_{x \to 0} \frac{(x+1) - 1}{x((\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1)}\]
3. Simplify and evaluate: \[= \lim_{x \to 0} \frac{x}{x((\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1)} = \lim_{x \to 0} \frac{1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\] \[= \frac{1}{1 + 1 + 1} = \frac{1}{3}\]

Answer: \(\frac{1}{3}\)

4.23. One-Sided Limit from the Right (Глава 5, Пример 9)

Evaluate \(\displaystyle \lim_{x \to -2^+} \frac{(x+3)|x+2|}{x+2}\).

Нажмите, чтобы увидеть решение

Ключевая идея: As \(x \to -2^+\), we have \(x > -2\), so \(x + 2 > 0\), which means \(|x+2| = x+2\).

1. Simplify using \(|x+2| = x+2\) when \(x > -2\): \[\frac{(x+3)|x+2|}{x+2} = \frac{(x+3)(x+2)}{x+2} = x+3\]
2. Evaluate the limit: \[\lim_{x \to -2^+} (x+3) = -2 + 3 = 1\]

Answer: \(1\)

4.24. One-Sided Limit from the Left (Глава 5, Пример 10)

Evaluate \(\displaystyle \lim_{x \to -2^-} \frac{(x+3)|x+2|}{x+2}\).

Нажмите, чтобы увидеть решение

Ключевая идея: As \(x \to -2^-\), we have \(x < -2\), so \(x + 2 < 0\), which means \(|x+2| = -(x+2)\).

1. Simplify using \(|x+2| = -(x+2)\) when \(x < -2\): \[\frac{(x+3)|x+2|}{x+2} = \frac{(x+3)(-(x+2))}{x+2} = -(x+3)\]
2. Evaluate the limit: \[\lim_{x \to -2^-} (-(x+3)) = -(-2+3) = -1\]

Answer: \(-1\)

4.25. Left-Hand Limit with Nested Radicals (Глава 5, Пример 11)

Evaluate \(\displaystyle \lim_{h \to 0^-} \frac{\sqrt{6} - \sqrt{5h^2+11h+6}}{h}\).

Нажмите, чтобы увидеть решение

1. Multiply by the conjugate: \[\lim_{h \to 0^-} \frac{\sqrt{6} - \sqrt{5h^2+11h+6}}{h} \cdot \frac{\sqrt{6} + \sqrt{5h^2+11h+6}}{\sqrt{6} + \sqrt{5h^2+11h+6}}\]
2. Simplify: \[= \lim_{h \to 0^-} \frac{6 - (5h^2+11h+6)}{h(\sqrt{6} + \sqrt{5h^2+11h+6})} = \lim_{h \to 0^-} \frac{-5h^2-11h}{h(\sqrt{6} + \sqrt{5h^2+11h+6})}\]
3. Factor and cancel \(h\): \[= \lim_{h \to 0^-} \frac{h(-5h-11)}{h(\sqrt{6} + \sqrt{5h^2+11h+6})} = \lim_{h \to 0^-} \frac{-5h-11}{\sqrt{6} + \sqrt{5h^2+11h+6}}\]
4. Evaluate: \[= \frac{-11}{\sqrt{6} + \sqrt{6}} = \frac{-11}{2\sqrt{6}} = -\frac{11\sqrt{6}}{12}\]

Answer: \(-\frac{11\sqrt{6}}{12}\)

4.26. Tangent Limit (Глава 5, Пример 12)

Evaluate \(\displaystyle \lim_{\theta \to 0} \frac{\tan \theta}{\theta}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and known limits.

1. Rewrite: \[\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = \lim_{\theta \to 0} \frac{\sin \theta}{\theta \cos \theta} = \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta}\]
2. Apply limit laws: \[= \left(\lim_{\theta \to 0} \frac{\sin \theta}{\theta}\right) \cdot \left(\lim_{\theta \to 0} \frac{1}{\cos \theta}\right) = 1 \cdot \frac{1}{1} = 1\]

Answer: \(1\)

4.27. Ratio of Sine Functions (Глава 5, Пример 13)

Evaluate \(\displaystyle \lim_{y \to 0} \frac{\sin 3y}{\sin 5y}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Multiply and divide to create the standard limit form.

1. Rewrite: \[\lim_{y \to 0} \frac{\sin 3y}{\sin 5y} = \lim_{y \to 0} \frac{\sin 3y}{3y} \cdot \frac{5y}{\sin 5y} \cdot \frac{3y}{5y}\]
2. Use substitutions: Let \(u = 3y\) and \(v = 5y\). As \(y \to 0\), both \(u \to 0\) and \(v \to 0\). \[= \lim_{u \to 0} \frac{\sin u}{u} \cdot \lim_{v \to 0} \frac{v}{\sin v} \cdot \frac{3}{5} = 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}\]

Answer: \(\frac{3}{5}\)

4.28. Tangent and Sine Combined (Глава 5, Пример 14)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\tan 3x}{\sin 8x}\).

Нажмите, чтобы увидеть решение

1. Rewrite tangent: \[\lim_{x \to 0} \frac{\tan 3x}{\sin 8x} = \lim_{x \to 0} \frac{\sin 3x}{\cos 3x \cdot \sin 8x}\]
2. Rearrange: \[= \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{8x}{\sin 8x} \cdot \frac{3x}{8x} \cdot \frac{1}{\cos 3x}\]
3. Evaluate: \[= 1 \cdot 1 \cdot \frac{3}{8} \cdot \frac{1}{1} = \frac{3}{8}\]

Answer: \(\frac{3}{8}\)

4.29. Nested Trigonometric Functions (Глава 5, Пример 15)

Evaluate \(\displaystyle \lim_{t \to 0} \frac{\sin(1-\cos t)}{2 - 2\cos t}\).

Нажмите, чтобы увидеть решение

1. Factor the denominator: \[\lim_{t \to 0} \frac{\sin(1-\cos t)}{2(1 - \cos t)}\]
2. Let \(u = 1 - \cos t\). As \(t \to 0\), \(u \to 0\): \[= \lim_{u \to 0} \frac{\sin u}{2u} = \frac{1}{2} \lim_{u \to 0} \frac{\sin u}{u} = \frac{1}{2} \cdot 1 = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.30. Sine of Hyperbolic Sine (Глава 5, Пример 16)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\sin(\sinh x)}{2\sinh x}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Use the fact that \(\lim_{x \to 0} \frac{\sinh x}{x} = 1\).

1. Let \(u = \sinh x\). As \(x \to 0\), \(u \to 0\): \[\lim_{x \to 0} \frac{\sin(\sinh x)}{2\sinh x} = \lim_{u \to 0} \frac{\sin u}{2u} = \frac{1}{2} \lim_{u \to 0} \frac{\sin u}{u} = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.31. Product of Trigonometric Functions (Глава 5, Пример 17)

Evaluate \(\displaystyle \lim_{x \to 0} 8x^2 (\cot x)(\csc x)\).

Нажмите, чтобы увидеть решение

1. Rewrite in terms of sine and cosine: \[\lim_{x \to 0} 8x^2 \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \lim_{x \to 0} \frac{8x^2 \cos x}{\sin^2 x}\]
2. Rearrange: \[= \lim_{x \to 0} 8 \left(\frac{x}{\sin x}\right)^2 \cos x = 8 \cdot 1^2 \cdot 1 = 8\]

Answer: \(8\)

4.32. Sine Squared in Denominator (Глава 5, Пример 18)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{x - x\cos x}{\sin^2 3x}\).

Нажмите, чтобы увидеть решение

1. Factor the numerator: \[\lim_{x \to 0} \frac{x(1 - \cos x)}{\sin^2 3x}\]
2. Use the identity \(1 - \cos x = 2\sin^2(x/2)\): \[= \lim_{x \to 0} \frac{2x\sin^2(x/2)}{\sin^2 3x} = \lim_{x \to 0} 2x \cdot \frac{\sin^2(x/2)}{\sin^2 3x}\]
3. Rearrange: \[= \lim_{x \to 0} 2x \cdot \left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \left(\frac{3x}{\sin 3x}\right)^2 \cdot \frac{(x/2)^2}{(3x)^2}\] \[= \lim_{x \to 0} 2x \cdot 1^2 \cdot 1^2 \cdot \frac{1/4}{9} = \lim_{x \to 0} 2x \cdot \frac{1}{36} = \lim_{x \to 0} \frac{x}{18} = 0\]

Answer: \(0\)

4.33. Product of Trigonometric Ratios (Глава 5, Пример 19)

Evaluate \(\displaystyle \lim_{z \to 0} \frac{\sin(3z) \cot(5z)}{z \cot(4z)}\).

Нажмите, чтобы увидеть решение

1. Rewrite cotangents: \[\lim_{z \to 0} \frac{\sin(3z) \cdot \frac{\cos 5z}{\sin 5z}}{z \cdot \frac{\cos 4z}{\sin 4z}} = \lim_{z \to 0} \frac{\sin(3z) \cos(5z) \sin(4z)}{z \sin(5z) \cos(4z)}\]
2. Rearrange: \[= \lim_{z \to 0} \frac{\sin 3z}{3z} \cdot \frac{4z}{\sin 4z} \cdot \frac{5z}{\sin 5z} \cdot \frac{3z \cdot 4z}{z \cdot 5z} \cdot \frac{\cos 5z}{\cos 4z}\] \[= 1 \cdot 1 \cdot 1 \cdot \frac{12}{5} \cdot \frac{1}{1} = \frac{12}{5}\]

Answer: \(\frac{12}{5}\)

4.34. Limit of Rational Function at Infinity (Глава 5, Пример 20)

Find \(\displaystyle \lim_{x \to \infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4}\) and \(\displaystyle \lim_{x \to -\infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4}\).

Нажмите, чтобы увидеть решение

Ключевая идея: Divide numerator and denominator by the highest power of \(x\).

1. Divide by \(x^4\): \[\lim_{x \to \infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4} = \lim_{x \to \infty} \frac{9 + \frac{1}{x^3}}{\frac{6}{x^4} - \frac{1}{x^3} + \frac{5}{x^2} + 2}\]
2. Evaluate as \(x \to \infty\): \[= \frac{9 + 0}{0 - 0 + 0 + 2} = \frac{9}{2}\]
3. For \(x \to -\infty\), the same calculation applies: \[\lim_{x \to -\infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4} = \frac{9}{2}\]

Answer: Both limits equal \(\frac{9}{2}\)

4.35. Limit at Infinity (Higher Degree in Denominator) (Глава 5, Пример 21)

Find \(\displaystyle \lim_{x \to \infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5}\) and \(\displaystyle \lim_{x \to -\infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5}\).

Нажмите, чтобы увидеть решение

1. Divide by \(x^6\): \[\lim_{x \to \infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5} = \lim_{x \to \infty} \frac{\frac{10}{x} + \frac{1}{x^2} + \frac{3}{x^6}}{1 - \frac{2}{x^4} + \frac{5}{x^6}} = \frac{0}{1} = 0\]
2. For \(x \to -\infty\): \[\lim_{x \to -\infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5} = 0\]

Answer: Both limits equal \(0\)

4.36. Limit at Infinity (Higher Degree in Numerator) (Глава 5, Пример 22)

Find \(\displaystyle \lim_{x \to \infty} \frac{5x^8 - 2x^3 + 9}{3 + x - 4x^5}\) and \(\displaystyle \lim_{x \to -\infty} \frac{5x^8 - 2x^3 + 9}{3 + x - 4x^5}\).

Нажмите, чтобы увидеть решение

1. Divide by \(x^8\): \[\lim_{x \to \infty} \frac{5x^8 - 2x^3 + 9}{3 + x - 4x^5} = \lim_{x \to \infty} \frac{5 - \frac{2}{x^5} + \frac{9}{x^8}}{\frac{3}{x^8} + \frac{1}{x^7} - \frac{4}{x^3}}\]
2. As \(x \to \infty\), the numerator approaches \(5\) and denominator approaches \(0^-\): \[= -\infty\]
3. For \(x \to -\infty\), numerator approaches \(5\) and denominator approaches \(0^+\): \[= \infty\]

Answer: \(\lim_{x \to \infty} = -\infty\), \(\lim_{x \to -\infty} = \infty\)

4.37. Square Root Limit at Infinity (Глава 5, Пример 23)

Evaluate \(\displaystyle \lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}}\).

Нажмите, чтобы увидеть решение

1. Move the limit inside (since square root is continuous): \[\lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}} = \sqrt{\lim_{x \to \infty} \frac{8x^2 - 3}{2x^2 + x}}\]
2. Divide by \(x^2\): \[= \sqrt{\lim_{x \to \infty} \frac{8 - \frac{3}{x^2}}{2 + \frac{1}{x}}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\]

Answer: \(2\)

4.38. Cube Root Limit as \(x \to -\infty\) (Глава 5, Пример 24)

Evaluate \(\displaystyle \lim_{x \to -\infty} \left(\frac{x^2 + x - 1}{8x^2 - 3}\right)^{1/3}\).

Нажмите, чтобы увидеть решение

1. Move limit inside: \[\lim_{x \to -\infty} \left(\frac{x^2 + x - 1}{8x^2 - 3}\right)^{1/3} = \left(\lim_{x \to -\infty} \frac{x^2 + x - 1}{8x^2 - 3}\right)^{1/3}\]
2. Divide by \(x^2\): \[= \left(\lim_{x \to -\infty} \frac{1 + \frac{1}{x} - \frac{1}{x^2}}{8 - \frac{3}{x^2}}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.39. Fifth Power Limit (Глава 5, Пример 25)

Evaluate \(\displaystyle \lim_{x \to -\infty} \left(\frac{1 - x^3}{x^2 + 7x}\right)^5\).

Нажмите, чтобы увидеть решение

1. Move limit inside: \[\left(\lim_{x \to -\infty} \frac{1 - x^3}{x^2 + 7x}\right)^5\]

2. Divide by \(x^3\): \[= \left(\lim_{x \to -\infty} \frac{\frac{1}{x^3} - 1}{\frac{1}{x} + \frac{7}{x^2}}\right)^5 = \left(\frac{0 - 1}{0 + 0}\right)^5\]

   The denominator approaches 0 while numerator approaches \(-1\), so the fraction goes to \(-\infty\).

Answer: \(-\infty\) (or more precisely, the limit does not exist as a finite number)

4.40. Mixed Root and Polynomial (Глава 5, Пример 26)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{2\sqrt{x} + x^{-1}}{3x - 7}\).

Нажмите, чтобы увидеть решение

1. Divide by \(x\): \[\lim_{x \to \infty} \frac{2\sqrt{x} + x^{-1}}{3x - 7} = \lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}} + \frac{1}{x^2}}{3 - \frac{7}{x}} = \frac{0 + 0}{3 - 0} = 0\]

Answer: \(0\)

4.41. Square Root Ratio (Глава 5, Пример 27)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{2 + \sqrt{x}}{2 - \sqrt{x}}\).

Нажмите, чтобы увидеть решение

1. Divide numerator and denominator by \(\sqrt{x}\): \[\lim_{x \to \infty} \frac{2 + \sqrt{x}}{2 - \sqrt{x}} = \lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}} + 1}{\frac{2}{\sqrt{x}} - 1} = \frac{0 + 1}{0 - 1} = -1\]

Answer: \(-1\)

4.42. Cube and Fifth Root (Глава 5, Пример 28)

Evaluate \(\displaystyle \lim_{x \to -\infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}\).

Нажмите, чтобы увидеть решение

1. For \(x < 0\), write \(x = -|x|\). As \(x \to -\infty\), \(|x| \to \infty\): \[\sqrt[3]{x} = -\sqrt[3]{|x|}, \quad \sqrt[5]{x} = -\sqrt[5]{|x|}\]
2. Substitute: \[\lim_{x \to -\infty} \frac{-\sqrt[3]{|x|} - (-\sqrt[5]{|x|})}{-\sqrt[3]{|x|} + (-\sqrt[5]{|x|})} = \lim_{|x| \to \infty} \frac{-\sqrt[3]{|x|} + \sqrt[5]{|x|]}{-\sqrt[3]{|x|} - \sqrt[5]{|x|}}\]
3. Divide by \(\sqrt[3]{|x|}\) (higher order): \[= \lim_{|x| \to \infty} \frac{-1 + \frac{1}{|x|^{3/5 - 1/3}}}{-1 - \frac{1}{|x|^{1/3 - 1/5}}} = \lim_{|x| \to \infty} \frac{-1 + \frac{1}{|x|^{2/15}}}{-1 - \frac{1}{|x|^{2/15}}} = \frac{-1 + 0}{-1 - 0} = 1\]

Answer: \(1\)

4.43. Negative Powers (Глава 5, Пример 29)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{x^{-1} + x^{-4}}{x^{-2} - x^{-3}}\).

Нажмите, чтобы увидеть решение

1. Multiply numerator and denominator by \(x^4\): \[\lim_{x \to \infty} \frac{x^{-1} + x^{-4}}{x^{-2} - x^{-3}} = \lim_{x \to \infty} \frac{x^3 + 1}{x^2 - x}\]
2. Divide by \(x^3\): \[= \lim_{x \to \infty} \frac{1 + \frac{1}{x^3}}{\frac{1}{x} - \frac{1}{x^2}} = \frac{1}{0} = \infty\]

Answer: \(\infty\)

4.44. Fractional Powers (Глава 5, Пример 30)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{2x^{5/3} - x^{1/3} + 7}{x^{8/5} + 3x + \sqrt{x}}\).

Нажмите, чтобы увидеть решение

1. Identify the highest power in each:

   • Numerator: \(x^{5/3} \approx x^{1.667}\) is highest
   • Denominator: \(x^{8/5} = x^{1.6}\) is highest (since \(x^{8/5} > x^1 = x\) and \(x^{8/5} > x^{1/2} = \sqrt{x}\))

2. Divide by \(x^{5/3}\): \[= \lim_{x \to \infty} \frac{2 - x^{-4/3} + 7x^{-5/3}}{x^{8/5 - 5/3} + 3x^{1 - 5/3} + x^{1/2 - 5/3}}\] \[= \lim_{x \to \infty} \frac{2 - x^{-4/3} + 7x^{-5/3}}{x^{-1/15} + 3x^{-2/3} + x^{-7/6}}\]

   Since numerator \(\to 2\) and denominator \(\to 0^+\):

Answer: \(\infty\)

4.45. Polynomial Under Square Root (Глава 5, Пример 31)

Evaluate \(\displaystyle \lim_{x \to -\infty} \frac{4 - 3x^3}{\sqrt{x^6 + 9}}\).

Нажмите, чтобы увидеть решение

1. For \(x < 0\), \(\sqrt{x^6} = |x^3| = -x^3\): \[\lim_{x \to -\infty} \frac{4 - 3x^3}{\sqrt{x^6 + 9}} = \lim_{x \to -\infty} \frac{4 - 3x^3}{-x^3\sqrt{1 + \frac{9}{x^6}}}\]
2. Divide numerator by \(x^3\): \[= \lim_{x \to -\infty} \frac{\frac{4}{x^3} - 3}{-\sqrt{1 + \frac{9}{x^6}}} = \frac{0 - 3}{-\sqrt{1 + 0}} = \frac{-3}{-1} = 3\]

Answer: \(3\)

4.46. One-Sided Limits at a Point (Глава 5, Пример 32)

Calculate \(\displaystyle \lim_{x \to ?} \frac{x^2 - 3x + 2}{x^3 - 2x^2}\) as \(x \to 2^+\), \(x \to 2^-\), \(x \to 2\), \(x \to 0^+\), \(x \to 0\).

Нажмите, чтобы увидеть решение

1. Factor: \[\frac{x^2 - 3x + 2}{x^3 - 2x^2} = \frac{(x-1)(x-2)}{x^2(x-2)} = \frac{x-1}{x^2} \quad (x \neq 2)\]
2. For \(x \to 2^+\), \(x \to 2^-\), and \(x \to 2\): \[\lim_{x \to 2} \frac{x-1}{x^2} = \frac{2-1}{4} = \frac{1}{4}\]
3. For \(x \to 0^+\): \[\lim_{x \to 0^+} \frac{x-1}{x^2} = \frac{-1}{0^+} = -\infty\]
4. For \(x \to 0\): Since both one-sided limits are \(-\infty\): \[\lim_{x \to 0} \frac{x-1}{x^2} = -\infty\]

Answer:

• \(x \to 2^+\), \(x \to 2^-\), \(x \to 2\): all equal \(\frac{1}{4}\)
• \(x \to 0^+\): \(-\infty\)
• \(x \to 0\): \(-\infty\)

4.47. Difference of Square Roots (Глава 5, Пример 33)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{x+9} - \sqrt{x+4})\).

Нажмите, чтобы увидеть решение

Ключевая идея: Multiply by the conjugate to rationalize.

1. Multiply by conjugate: \[\lim_{x \to \infty} (\sqrt{x+9} - \sqrt{x+4}) \cdot \frac{\sqrt{x+9} + \sqrt{x+4}}{\sqrt{x+9} + \sqrt{x+4}}\]
2. Simplify: \[= \lim_{x \to \infty} \frac{(x+9) - (x+4)}{\sqrt{x+9} + \sqrt{x+4}} = \lim_{x \to \infty} \frac{5}{\sqrt{x+9} + \sqrt{x+4}}\]
3. As \(x \to \infty\), denominator \(\to \infty\): \[= 0\]

Answer: \(0\)

4.48. Difference of Square Roots with Same Leading Term (Глава 5, Пример 34)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{x^2+25} - \sqrt{x^2-1})\).

Нажмите, чтобы увидеть решение

1. Multiply by conjugate: \[\lim_{x \to \infty} \frac{(\sqrt{x^2+25} - \sqrt{x^2-1})(\sqrt{x^2+25} + \sqrt{x^2-1})}{\sqrt{x^2+25} + \sqrt{x^2-1}}\]
2. Simplify: \[= \lim_{x \to \infty} \frac{(x^2+25) - (x^2-1)}{\sqrt{x^2+25} + \sqrt{x^2-1}} = \lim_{x \to \infty} \frac{26}{\sqrt{x^2+25} + \sqrt{x^2-1}} = 0\]

Answer: \(0\)

4.49. Sum with Square Root (Negative Infinity) (Глава 5, Пример 35)

Calculate \(\displaystyle \lim_{x \to -\infty} (\sqrt{x^2+3} + x)\).

Нажмите, чтобы увидеть решение

1. For \(x < 0\), write \(\sqrt{x^2} = |x| = -x\). Multiply by conjugate: \[\lim_{x \to -\infty} (\sqrt{x^2+3} + x) \cdot \frac{\sqrt{x^2+3} - x}{\sqrt{x^2+3} - x}\]
2. Simplify: \[= \lim_{x \to -\infty} \frac{(x^2+3) - x^2}{\sqrt{x^2+3} - x} = \lim_{x \to -\infty} \frac{3}{\sqrt{x^2+3} - x}\]
3. For \(x \to -\infty\), write \(\sqrt{x^2+3} = -x\sqrt{1 + \frac{3}{x^2}}\) (since \(x < 0\)): \[= \lim_{x \to -\infty} \frac{3}{-x\sqrt{1 + \frac{3}{x^2}} - x} = \lim_{x \to -\infty} \frac{3}{-x(\sqrt{1 + \frac{3}{x^2}} + 1)}\]
4. As \(x \to -\infty\) (so \(x\) is large negative), \(-x\) is large positive: \[= \lim_{x \to -\infty} \frac{3}{-x \cdot 2} = 0\]

Answer: \(0\)

4.50. Linear Plus Square Root (Глава 5, Пример 36)

Calculate \(\displaystyle \lim_{x \to -\infty} (2x + \sqrt{4x^2+3x-2})\).

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1. Multiply by conjugate: \[\lim_{x \to -\infty} (2x + \sqrt{4x^2+3x-2}) \cdot \frac{2x - \sqrt{4x^2+3x-2}}{2x - \sqrt{4x^2+3x-2}}\]
2. Simplify: \[= \lim_{x \to -\infty} \frac{4x^2 - (4x^2+3x-2)}{2x - \sqrt{4x^2+3x-2}} = \lim_{x \to -\infty} \frac{-3x+2}{2x - \sqrt{4x^2+3x-2}}\]
3. For \(x \to -\infty\), \(\sqrt{4x^2} = 2|x| = -2x\): \[= \lim_{x \to -\infty} \frac{-3x+2}{2x - (-2x)\sqrt{1 + \frac{3}{4x} - \frac{2}{4x^2}}}\]
4. Divide by \(x\) (noting \(x < 0\)): \[= \lim_{x \to -\infty} \frac{-3 + \frac{2}{x}}{2 + 2\sqrt{1 + \frac{3}{4x} - \frac{1}{2x^2}}} = \frac{-3}{2 + 2} = -\frac{3}{4}\]

Answer: \(-\frac{3}{4}\)

4.51. Quadratic Minus Linear Under Root (Глава 5, Пример 37)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{9x^2-x} - 3x)\).

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1. Multiply by conjugate: \[\lim_{x \to \infty} \frac{(\sqrt{9x^2-x} - 3x)(\sqrt{9x^2-x} + 3x)}{\sqrt{9x^2-x} + 3x}\]
2. Simplify: \[= \lim_{x \to \infty} \frac{9x^2 - x - 9x^2}{\sqrt{9x^2-x} + 3x} = \lim_{x \to \infty} \frac{-x}{\sqrt{9x^2-x} + 3x}\]
3. Divide by \(x\) (for \(x > 0\), \(\sqrt{x^2} = x\)): \[= \lim_{x \to \infty} \frac{-1}{\sqrt{9 - \frac{1}{x}} + 3} = \frac{-1}{3 + 3} = -\frac{1}{6}\]

Answer: \(-\frac{1}{6}\)

4.52. Difference of Square Roots with Different Powers (Глава 5, Пример 38)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{x^2+3x} - \sqrt{x^2-2x})\).

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1. Multiply by conjugate: \[\lim_{x \to \infty} \frac{(x^2+3x) - (x^2-2x)}{\sqrt{x^2+3x} + \sqrt{x^2-2x}} = \lim_{x \to \infty} \frac{5x}{\sqrt{x^2+3x} + \sqrt{x^2-2x}}\]
2. Divide by \(x\): \[= \lim_{x \to \infty} \frac{5}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 - \frac{2}{x}}} = \frac{5}{1 + 1} = \frac{5}{2}\]

Answer: \(\frac{5}{2}\)

4.53. Find All Asymptotes (Глава 5, Пример 39)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2}{x-1}\).

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1. Vertical asymptote: Domain excludes \(x = 1\). Check limits: \[\lim_{x \to 1^+} \frac{x^2}{x-1} = \frac{1}{0^+} = \infty\] \[\lim_{x \to 1^-} \frac{x^2}{x-1} = \frac{1}{0^-} = -\infty\]

   Vertical asymptote: \(x = 1\)

2. Horizontal asymptote: Check limits at infinity: \[\lim_{x \to \pm\infty} \frac{x^2}{x-1} = \lim_{x \to \pm\infty} \frac{x}{1 - \frac{1}{x}} = \pm\infty\]

   No horizontal asymptote.

3. Oblique asymptote: Perform polynomial division: \[\frac{x^2}{x-1} = x + 1 + \frac{1}{x-1}\]

   As \(x \to \pm\infty\), \(\frac{1}{x-1} \to 0\), so the oblique asymptote is \(y = x + 1\).

Answer: Vertical: \(x = 1\); Oblique: \(y = x + 1\)

4.54. Asymptotes of Rational Function (Глава 5, Пример 40)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2+1}{x-1}\).

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1. Vertical asymptote: \(x = 1\) (same analysis as previous problem)

2. Oblique asymptote: Polynomial division: \[\frac{x^2+1}{x-1} = x + 1 + \frac{2}{x-1}\]

   Oblique asymptote: \(y = x + 1\)

Answer: Vertical: \(x = 1\); Oblique: \(y = x + 1\)

4.55. Another Asymptote Problem (Глава 5, Пример 41)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2-4}{x-1}\).

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1. Vertical asymptote: \(x = 1\)

2. Oblique asymptote: Polynomial division: \[\frac{x^2-4}{x-1} = x + 1 - \frac{3}{x-1}\]

   Oblique asymptote: \(y = x + 1\)

Answer: Vertical: \(x = 1\); Oblique: \(y = x + 1\)

4.56. Asymptotes with Factored Denominator (Глава 5, Пример 42)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2-1}{2x+4}\).

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1. Vertical asymptote: \(2x + 4 = 0 \implies x = -2\)

2. Oblique asymptote: Polynomial division: \[\frac{x^2-1}{2x+4} = \frac{x^2-1}{2(x+2)} = \frac{1}{2}x - 1 + \frac{1}{2(x+2)}\]

   Oblique asymptote: \(y = \frac{1}{2}x - 1\)

Answer: Vertical: \(x = -2\); Oblique: \(y = \frac{x}{2} - 1\)

4.57. Asymptotes of \(\frac{x^2-1}{x}\) (Глава 5, Пример 43)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2-1}{x}\).

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1. Vertical asymptote: \(x = 0\) \[\lim_{x \to 0^+} \frac{x^2-1}{x} = \frac{-1}{0^+} = -\infty\]

2. Oblique asymptote: Simplify: \[\frac{x^2-1}{x} = x - \frac{1}{x}\]

   As \(x \to \pm\infty\), \(\frac{1}{x} \to 0\), so oblique asymptote: \(y = x\)

Answer: Vertical: \(x = 0\); Oblique: \(y = x\)

4.58. Asymptotes with Cubic Numerator (Глава 5, Пример 44)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^3+1}{x^2}\).

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1. Vertical asymptote: \(x = 0\)

2. Oblique asymptote: Polynomial division: \[\frac{x^3+1}{x^2} = x + \frac{1}{x^2}\]

   As \(x \to \pm\infty\), \(\frac{1}{x^2} \to 0\), so oblique asymptote: \(y = x\)

Answer: Vertical: \(x = 0\); Oblique: \(y = x\)

4.59. Limit with Exponential Form \((1+\frac{1}{x})^x\) (Глава 5, Пример 45)

Find \(\displaystyle L_1 = \lim_{x \to \infty} \left(\frac{x+2}{x+1}\right)^x\).

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Ключевая идея: Rewrite in the form \((1 + \frac{a}{t})^t\).

1. Rewrite: \[L_1 = \lim_{x \to \infty} \left(\frac{x+2}{x+1}\right)^x = \lim_{x \to \infty} \left(1 + \frac{1}{x+1}\right)^x\]
2. Substitute \(t = x + 1\), so \(x = t - 1\). As \(x \to \infty\), \(t \to \infty\): \[L_1 = \lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^{t-1} = \lim_{t \to \infty} \frac{\left(1 + \frac{1}{t}\right)^t}{\left(1 + \frac{1}{t}\right)}\]
3. Evaluate: \[= \frac{\lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^t}{\lim_{t \to \infty} \left(1 + \frac{1}{t}\right)} = \frac{e}{1} = e\]

Answer: \(e\)

4.60. Limit with \((1 + \frac{1}{x^2})^x\) (Глава 5, Пример 46)

Find \(\displaystyle L_2 = \lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x\).

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Ключевая идея: Use the relationship \(a^b = e^{b \ln a}\).

1. Rewrite: \[L_2 = \lim_{x \to \infty} \left[\left(1 + \frac{1}{x^2}\right)^{x^2}\right]^{1/x}\]
2. Note that \(\lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^{x^2} = e\): \[L_2 = \lim_{x \to \infty} e^{1/x} = e^0 = 1\]

Answer: \(1\)

4.61. Complex Exponential Limit (Глава 5, Пример 47)

Find \(\displaystyle L_3 = \lim_{x \to \infty} \left(\frac{5x-2}{2x+1}\right)^{2x-1}\).

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1. Use the exponential form: \[L_3 = \lim_{x \to \infty} e^{(2x-1) \ln\left(\frac{5x-2}{2x+1}\right)}\]
2. Evaluate the limit in the exponent: \[\lim_{x \to \infty} (2x-1) \cdot \ln\left(\frac{5x-2}{2x+1}\right) = \lim_{x \to \infty} (2x-1) \cdot \ln\left(\frac{5 - \frac{2}{x}}{2 + \frac{1}{x}}\right)\] \[= \lim_{x \to \infty} (2x-1) \cdot \ln\left(\frac{5}{2}\right) = \infty \cdot \ln\left(\frac{5}{2}\right) = \infty\]
3. Since \(\ln(5/2) > 0\): \[L_3 = e^{\infty} = \infty\]

Answer: \(\infty\)

4.62. Indeterminate Form \(1^{\infty}\) Example 1 (Глава 5, Пример 48)

Find \(\displaystyle L_4 = \lim_{x \to \infty} \left(\frac{x^2+1}{x^2-2}\right)^{x^2}\).

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Ключевая идея: For \(1^{\infty}\) form, use \(\lim_{x \to c} (g(x))^{h(x)} = e^{\lim_{x \to c} h(x)(g(x)-1)}\).

1. Let \(g(x) = \frac{x^2+1}{x^2-2}\) and \(h(x) = x^2\): \[L_4 = e^{\lim_{x \to \infty} x^2 \left(\frac{x^2+1}{x^2-2} - 1\right)}\]
2. Simplify the expression: \[\lim_{x \to \infty} x^2 \left(\frac{x^2+1 - x^2+2}{x^2-2}\right) = \lim_{x \to \infty} x^2 \cdot \frac{3}{x^2-2} = \lim_{x \to \infty} \frac{3x^2}{x^2-2}\] \[= \lim_{x \to \infty} \frac{3}{1 - \frac{2}{x^2}} = 3\]
3. Therefore: \[L_4 = e^3\]

Answer: \(e^3\)

4.63. Indeterminate Form \(1^{\infty}\) Example 2 (Глава 5, Пример 49)

Find \(\displaystyle L_5 = \lim_{x \to 0} \left(\frac{1+\tan x}{1+\sin x}\right)^{1/\sin x}\).

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1. Let \(g(x) = \frac{1+\tan x}{1+\sin x}\) and \(h(x) = \frac{1}{\sin x}\): \[L_5 = e^{\lim_{x \to 0} \frac{1}{\sin x}\left(\frac{1+\tan x}{1+\sin x} - 1\right)}\]
2. Simplify: \[h(x)(g(x) - 1) = \frac{1}{\sin x} \cdot \frac{1+\tan x - 1 - \sin x}{1+\sin x} = \frac{\tan x - \sin x}{\sin x(1+\sin x)}\]
3. Rewrite tangent: \[= \frac{\frac{\sin x}{\cos x} - \sin x}{\sin x(1+\sin x)} = \frac{\sin x(1 - \cos x)}{\cos x \cdot \sin x(1+\sin x)} = \frac{1-\cos x}{\cos x(1+\sin x)}\]
4. Use \(1 - \cos x = 2\sin^2(x/2)\): \[\lim_{x \to 0} \frac{2\sin^2(x/2)}{\cos x(1+\sin x)} = \frac{0}{1 \cdot 1} = 0\]
5. Therefore: \[L_5 = e^0 = 1\]

Answer: \(1\)

4.64. Limit of \(\frac{a^x-1}{x}\) (Глава 5, Пример 50)

Find \(\displaystyle L_6 = \lim_{x \to 0} \frac{a^x-1}{x}\) for \(a > 0\).

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Ключевая идея: Use the fact that \(a^x = e^{x \ln a}\) and known limits.

1. Note that: \[\frac{x}{a^x-1} = \frac{x}{e^{x\ln a} - 1}\]
2. Let \(t = x \ln a\). As \(x \to 0\), \(t \to 0\): \[\frac{x}{a^x-1} = \frac{t/\ln a}{e^t - 1} = \frac{1}{\ln a} \cdot \frac{t}{e^t-1}\]
3. We know that \(\lim_{t \to 0} \frac{e^t-1}{t} = 1\), so: \[\lim_{x \to 0} \frac{x}{a^x-1} = \frac{1}{\ln a} \cdot 1 = \frac{1}{\ln a}\]
4. Therefore: \[L_6 = \lim_{x \to 0} \frac{a^x-1}{x} = \ln a\]

Answer: \(\ln a\)