Comprehensive Study Guide: Limits, Continuity, Circular, and Hyperbolic Functions

Set Notation and Fundamental Limit Concepts

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• Set Notation

  • The notation $n \in \mathbb{N}$ indicates that $n$ is an element of the set of natural numbers $\mathbb{N}$.
  • Standard Number Sets:
  • Natural numbers: $\mathbb{N}$
  • Integers: $\mathbb{Z}$
  • Rational numbers: $\mathbb{Q}$
  • Real numbers: $\mathbb{R}$
  • Examples of set membership:
  • $4 \in \mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$
  • $-1 \in \mathbb{Z}, \mathbb{Q}, \mathbb{R}$
  • $\frac{1}{2} \in \mathbb{Q}, \mathbb{R}$
  • $\pi \in \mathbb{R}$

• Core Concepts of Limits

  • In determining $\lim_{x \to a} f(x)$, only values of $x$ arbitrarily close to $a$ are considered; the value of the function at $x = a$ itself is irrelevant.
  • The value of $\lim_{x \to a} f(x)$ is not necessarily equal to $f(a)$.
  • Evaluating a limit requires investigating the behavior of $f(x)$ as $x$ approaches $a$ simultaneously from both the left and right sides.
  • Piecewise Example:
  • Let $f(x) = \begin{cases} x^2, & \text{if } x \neq 10 \\ 99, & \text{if } x = 10 \end{cases}$
  • Determine $\lim_{x \to 10} f(x)$.
  • Solution: As $x \to 10$, $x \neq 10$, which means $f(x) = x^2$. Thus, $\lim_{x \to 10} f(x) = \lim_{x \to 10} x^2 = 100$. Notice that $\lim_{x \to 10} f(x) = 100 \neq f(10) = 99$.

• Basic Limit Theorems

  • Uniqueness Theorem: If $\lim_{x \to a} f(x)$ exists, then the limit value is unique.
  • Constant Rule: If $c \in \mathbb{R}$, then $\lim_{x \to a} c = c$.
  • Identity Rule: $\lim_{x \to a} x = a$.

• Algebraic Limit Laws

  • Suppose $\lim_{x \to a} f(x) = L_1$ and $\lim_{x \to a} g(x) = L_2$:
  1. Sum and Difference Law: $\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) = L_1 \pm L_2$
  2. Product Law: $\lim_{x \to a} [f(x)g(x)] = \left(\lim_{x \to a} f(x)\right)\left(\lim_{x \to a} g(x)\right) = L_1 L_2$
  3. Constant Multiple Law: $\lim_{x \to a} [c \cdot f(x)] = c \lim_{x \to a} f(x) = c L_1$
  4. Quotient Law: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{L_1}{L_2}$, provided $L_2 \neq 0$
  5. Power Law: $\lim_{x \to a} [f(x)]^n = \left(\lim_{x \to a} f(x)\right)^n = (L_1)^n$

• Direct Substitution Rules

  • Polynomial Functions: If $f$ is a polynomial function, then $\lim_{x \to a} f(x) = f(a)$.
  • Rational Functions: If $f$ is a rational function and $f(a)$ is defined, then $\lim_{x \to a} f(x) = f(a)$.
  • Radical Functions: If $\lim_{x \to a} f(x)$ exists and $n \in \mathbb{N}$, then $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}$, provided $\lim_{x \to a} f(x) > 0$ when $n$ is an even integer.

Indeterminate Forms of Type 0/0 and Algebraic Limit Techniques

• Definition of Indeterminate Form \frac{0}{0}

  • If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, then $\lim_{x \to a} \frac{f(x)}{g(x)}$ is called an indeterminate form of type $\frac{0}{0}$.
  • Algebraic techniques used to evaluate type $\frac{0}{0}$ limits include:
  • Factoring
  • Rationalization

• Essential Algebraic Identities

  1. Difference of Two Squares: $a^2 - b^2 = (a - b)(a + b)$
  2. Sum of Two Cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
  3. Difference of Two Cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

• Detailed Worked Examples

  • Example 1: Evaluate $\lim_{x \to -2} \frac{x^3 - x^2 - x + 10}{x^2 + 3x + 2}$
  • Direct substitution at $x = -2$ yields $\frac{(-2)^3 - (-2)^2 - (-2) + 10}{(-2)^2 + 3(-2) + 2} = \frac{-8 - 4 + 2 + 10}{4 - 6 + 2} = \frac{0}{0}$.
  • Factor numerator by division/grouping: $x^3 - x^2 - x + 10 = (x + 2)(x^2 - 3x + 5)$.
  • Factor denominator: $x^2 + 3x + 2 = (x + 2)(x + 1)$.
  • Cancel common factor $(x + 2)$, valid since $x \neq -2$:       $\lim_{x \to -2} \frac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)} = \lim_{x \to -2} \frac{x^2 - 3x + 5}{x + 1} = \frac{(-2)^2 - 3(-2) + 5}{-2 + 1} = \frac{4 + 6 + 5}{-1} = -15$
  • Example 2: Evaluate $\lim_{x \to 2} \frac{\sqrt{2x} - \sqrt{6 - x}}{4 - x^2}$
  • Direct substitution yields $\frac{\sqrt{4} - \sqrt{4}}{4 - 4} = \frac{0}{0}$.
  • Rationalize the numerator by multiplying by the conjugate $\sqrt{2x} + \sqrt{6 - x}$:       $\frac{(\sqrt{2x} - \sqrt{6 - x})(\sqrt{2x} + \sqrt{6 - x})}{(4 - x^2)(\sqrt{2x} + \sqrt{6 - x})} = \frac{2x - (6 - x)}{(2 - x)(2 + x)(\sqrt{2x} + \sqrt{6 - x})} = \frac{3x - 6}{(2 - x)(2 + x)(\sqrt{2x} + \sqrt{6 - x})}$
  • Rewrite numerator as $-3(2 - x)$ and cancel $(2 - x)$:       $\lim_{x \to 2} \frac{-3(2 - x)}{(2 - x)(2 + x)(\sqrt{2x} + \sqrt{6 - x})} = \lim_{x \to 2} \frac{-3}{(2 + x)(\sqrt{2x} + \sqrt{6 - x})} = \frac{-3}{(2 + 2)(\sqrt{4} + \sqrt{4})} = \frac{-3}{4(4)} = -\frac{3}{16}$
  • Example 3: Evaluate $\lim_{t \to 0} \frac{(t + a)^3 - a^3}{t}$
  • Direct substitution at $t = 0$ yields $\frac{a^3 - a^3}{0} = \frac{0}{0}$.
  • Expand numerator: $(t + a)^3 - a^3 = t^3 + 3t^2a + 3ta^2 + a^3 - a^3 = t(t^2 + 3ta + 3a^2)$.
  • Cancel $t$:       $\lim_{t \to 0} \frac{t(t^2 + 3ta + 3a^2)}{t} = \lim_{t \to 0} (t^2 + 3ta + 3a^2) = 0 + 0 + 3a^2 = 3a^2$

• Supplementary Exercises and Solutions

  • Problem: Determine all values of $a$ and $b$ such that $\lim_{x \to 0} \frac{x}{\sqrt{ax + b} - 2} = 1$.
  • Solution: Since numerator approaches $0$ as $x \to 0$, for the limit to equal $1$, the denominator must also approach $0$. Thus, $\sqrt{b} - 2 = 0 \implies b = 4$.
  • Rationalize denominator:       $\frac{x(\sqrt{ax + 4} + 2)}{(\sqrt{ax + 4} - 2)(\sqrt{ax + 4} + 2)} = \frac{x(\sqrt{ax + 4} + 2)}{ax + 4 - 4} = \frac{x(\sqrt{ax + 4} + 2)}{ax} = \frac{\sqrt{ax + 4} + 2}{a}$
  • Taking the limit as $x \to 0$:       $\frac{\sqrt{4} + 2}{a} = \frac{4}{a} = 1 \implies a = 4$

• Seatwork 1 Problems

  1. Evaluate $\lim_{x \to 1} \frac{x^3 - 1}{\sqrt{2x - 1} - 1}$:
  • Indeterminate form $\frac{0}{0}$. Factor numerator as $(x - 1)(x^2 + x + 1)$ and multiply conjugate of denominator:        $\frac{(x - 1)(x^2 + x + 1)(\sqrt{2x - 1} + 1)}{2x - 1 - 1} = \frac{(x - 1)(x^2 + x + 1)(\sqrt{2x - 1} + 1)}{2(x - 1)} = \frac{(x^2 + x + 1)(\sqrt{2x - 1} + 1)}{2}$
  • Evaluate at $x = 1$: $\frac{(1 + 1 + 1)(\sqrt{1} + 1)}{2} = \frac{3(2)}{2} = 3$.
  1. Evaluate $\lim_{s \to 3} \frac{2s^2 - 7s + 3}{s^2 - 4s + 3}$:
  • Indeterminate form $\frac{0}{0}$. Factor numerator and denominator:        $\frac{(2s - 1)(s - 3)}{(s - 1)(s - 3)} = \frac{2s - 1}{s - 1}$
  • Evaluate at $s = 3$: $\frac{2(3) - 1}{3 - 1} = \frac{5}{2}$.

Infinite Limits, Limits at Infinity, and Radical Behavior

• Definitions of Infinite Limits

  • If $f(x)$ increases without bound as $x \to a^+$, then $\lim_{x \to a^+} f(x) = +\infty$.
  • If $f(x)$ decreases without bound as $x \to a^+$, then $\lim_{x \to a^+} f(x) = -\infty$.
  • Analogous definitions hold for left-hand limits ($x \to a^-$).
  • Crucial Distinction: The expression $\lim_{x \to a} f(x) = \pm\infty$ signifies that the limit does not exist (DNE).
  • A limit that does not exist does NOT always imply $\lim_{x \to a} f(x) = \pm\infty$. For example, $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ does not exist due to infinite oscillation.

• Indeterminate Forms Involving Infinity

  • Indeterminate forms include: $\infty - \infty$, $\frac{\infty}{\infty}$, $-\infty + \infty$, $\frac{\infty}{-\infty}$, $\frac{-\infty}{\infty}$, $\frac{-\infty}{-\infty}$.
  • Example illustrating varied limit outcomes for $\infty - \infty$:
  • Let $f(x) = x$ and $g(x) = \sqrt{x}$. As $x \to \infty$, $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to \infty} g(x) = \infty$
  • Outcome 1: $\lim_{x \to \infty} [f(x) - g(x)] = \lim_{x \to \infty} (x - \sqrt{x}) = +\infty$
  • Outcome 2: $\lim_{x \to \infty} [[f(x) - g(x) + 1] - [f(x) - g(x)]] = 1$

• Quotient Limit Behaviors near Zero and Infinity

  • If $\lim_{x \to a} f(x) = L > 0$ and $g(x) \to 0^+$ as $x \to a$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = +\infty$.
  • If $\lim_{x \to a} f(x) = L > 0$ and $g(x) \to 0^-$ as $x \to a$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = -\infty$.
  • Signs flip when $\lim_{x \to a} f(x) = L < 0$.
  • If $\lim_{x \to a} f(x)$ exists and $\lim_{x \to a} g(x) = \pm\infty$, then \lim_{x \to a} \frac{f(x)}{g(x)} = 0$.\n\n- **Algebraic Identity for Radicals of Even Indices**\n - \sqrt{x^2} = |x| = \begin{cases} x, & \text{if } x \ge 0 \ -x, & \text{if } x < 0 \end{cases}\n - In general, for any positive even integer n:\n    \sqrt[n]{x^n} = |x| = \begin{cases} x, & \text{if } x \ge 0 \ -x, & \text{if } x < 0 \end{cases}\n\n- **Worked Examples**\n - **Example 1**: Evaluate \lim_{x \to -2^-} \left( \frac{x}{x + 2} + \frac{x - 2}{x^2 + x - 2} \right)\n - Combine expressions over a common denominator: x^2 + x - 2 = (x + 2)(x - 1).\n      \frac{x(x - 1) + (x - 2)}{(x + 2)(x - 1)} = \frac{x^2 - x + x - 2}{(x + 2)(x - 1)} = \frac{x^2 - 2}{(x + 2)(x - 1)}\n - Evaluate sign behavior as x \to -2^-$($x < -2):\n - Numerator: (-2)^2 - 2 = 2 > 0\n - (x + 2) \to 0^-\n - (x - 1) \to -3 < 0\n - Denominator: 0^- \cdot (-3) = 0^+\n - Result: \lim_{x \to -2^-} \frac{x^2 - 2}{(x + 2)(x - 1)} = \frac{2}{0^+} = +\infty\n - **Example 2**: Evaluate \lim_{x \to -\infty} \left(\sqrt{x^2 + 3x} + x\right)\n - Form +\infty - \infty$. Rationalize by multiplying by$\frac{\sqrt{x^2 + 3x} - x}{\sqrt{x^2 + 3x} - x}:\n      \frac{x^2 + 3x - x^2}{\sqrt{x^2 + 3x} - x} = \frac{3x}{\sqrt{x^2\left(1 + \frac{3}{x}\right)} - x}\n - Since x \to -\infty$,$x < 0$, so$\sqrt{x^2} = |x| = -x.\n      \frac{3x}{-x\sqrt{1 + \frac{3}{x}} - x} = \frac{3x}{-x\left(\sqrt{1 + \frac{3}{x}} + 1\right)} = \frac{3}{-\left(\sqrt{1 + \frac{3}{x}} + 1\right)}\n - Taking limit as x \to -\infty$,$\frac{3}{x} \to 0:\n      \frac{3}{-(1 + 1)} = -\frac{3}{2}\n - **Example 3**: Evaluate \lim_{x \to +\infty} \frac{2x + \sqrt{x^2 - 1}}{x + 3}\n - Since x \to +\infty$,$x > 0$, so$\sqrt{x^2} = x.\n - Factor x from numerator and denominator:\n      \frac{x\left(2 + \sqrt{1 - \frac{1}{x^2}}\right)}{x\left(1 + \frac{3}{x}\right)} = \frac{2 + \sqrt{1 - \frac{1}{x^2}}}{1 + \frac{3}{x}}\n - Taking limit as x \to +\infty$,$\frac{1}{x^2} \to 0$and$\frac{3}{x} \to 0:\n      \frac{2 + \sqrt{1 - 0}}{1 + 0} = \frac{3}{1} = 3\n\n- **Seatwork 3 Problems**\n 1. Evaluate \lim_{x \to -2^+} \left( \frac{4}{x + 2} + \frac{8}{x^2 + 2x} \right):\n - Combine fractions: \frac{4x + 8}{x(x + 2)} = \frac{4(x + 2)}{x(x + 2)} = \frac{4}{x}.\n - Taking limit as x \to -2^+$, result is$\frac{4}{-2} = -2$.
  1. Evaluate $\lim_{x \to 2^-} \frac{x - 2}{2 - \sqrt{4x - x^2}}$:
  • Rationalize denominator: multiply by $\frac{2 + \sqrt{4x - x^2}}{2 + \sqrt{4x - x^2}}$.
  • Denominator becomes 4 - (4x - x^2) = x^2 - 4x + 4 = (x - 2)^2$.\n - Fraction: \frac{(x - 2)(2 + \sqrt{4x - x^2})}{(x - 2)^2} = \frac{2 + \sqrt{4x - x^2}}{x - 2}.\n - As x \to 2^-$: Numerator$\to 2 + \sqrt{4} = 4 > 0$, Denominator$\to 0^-.\n - Limit is -\infty$.
  1. Evaluate $\lim_{x \to -\infty} \frac{2x + \sqrt{x^2 - 1}}{x + 3}$:
  • As $x \to -\infty$, $x < 0$, so \sqrt{x^2} = -x$.\n - Fraction: \frac{2x - x\sqrt{1 - \frac{1}{x^2}}}{x\left(1 + \frac{3}{x}\right)} = \frac{2 - \sqrt{1 - \frac{1}{x^2}}}{1 + \frac{3}{x}}.\n - Limit as x \to -\infty$:$\frac{2 - 1}{1 + 0} = 1$.

Inequality Properties and the Intermediate Value Theorem

• Polynomial Inequalities and Sign Analysis

  • Inequalities involving polynomials of degree 2 or higher are non-trivial and require table-of-signs analysis or complete factorization.
  • Example: Factor decomposition of $\left|2x - x^2\right|$:     $\left|2x - x^2\right| = |x(2 - x)| = |x| \cdot |2 - x|$
  • Explicit piecewise structure:     $\left|2x - x^2\right| = \begin{cases} 2x - x^2, & \text{if } 2x - x^2 \ge 0 \quad (x \in [0, 2]) \\ -(2x - x^2), & \text{if } 2x - x^2 < 0 \quad (x \in (-\infty, 0) \cup (2, +\infty)) \end{cases}$
  • As $x \to 0^-$, $x < 0$, which implies $|x| = -x$ and |2 - x| = 2 - x$.\n\n- **Inequality Direction Rules**\n - Inequality direction flips when:\n - Multiplying or dividing both sides by a negative real number.\n - Taking reciprocals of quantities of the same sign.\n - **Decreasing Function Rule**: If f(x)$is strictly decreasing on$(a, b)$and$m < n$, then$f(m) > f(n).\n - Example: Take f(x) = -3x$, which is strictly decreasing everywhere. Since$3 \le 5$, applying$f$yields$f(3) \ge f(5) \implies -9 \ge -15$.

• Intermediate Value Theorem (IVT)

  • Theorem: Let $f$ be continuous on a closed interval $[a, b]$ with $f(a) \neq f(b)$. For every $k$ between $f(a)$ and $f(b)$, there exists at least one $c \in (a, b)$ such that f(c) = k$.\n  \n  \n\n - **Remark**: The IVT asserts that a continuous function cannot skip any intermediate value between its initial point f(a)$and terminal point$f(b).\n\n- **Worked Examples using IVT**\n - **Example 1**: Show that f(x) = x^4 - 7x^2 + x + 4 has at least two real zeroes.\n - Solution:\n - f(x)$is a polynomial, so it is continuous on all of$\mathbb{R}.\n - Evaluate f(x) at selected test points:\n - f(0) = 0^4 - 7(0)^2 + 0 + 4 = 4 > 0\n - f(1) = 1^4 - 7(1)^2 + 1 + 4 = -1 < 0\n - f(3) = 3^4 - 7(3^2) + 3 + 4 = 81 - 63 + 3 + 4 = 25 > 0\n - On interval [0, 1]$:$f$is continuous,$f(1) = -1 < 0 < 4 = f(0)$. By IVT with$k = 0$, there exists$c_1 \in (0, 1)$such that$f(c_1) = 0$.
    • On interval $[1, 3]$: $f$ is continuous, $f(1) = -1 < 0 < 25 = f(3)$. By IVT with $k = 0$, there exists $c_2 \in (1, 3)$ such that f(c_2) = 0$.\n - Therefore, f(x) has at least two real zeroes.\n\n- **Supplementary Exercise on IVT**\n - Problem: Show that the equation x^5 + 4x^3 + 14 = 7x has at least one real solution.\n - Solution: Define g(x) = x^5 + 4x^3 - 7x + 14$. Function$g$is continuous on$\mathbb{R}.\n - Evaluate at test values:\n - g(0) = 14 > 0\n - g(-2) = (-2)^5 + 4(-2)^3 - 7(-2) + 14 = -32 - 32 + 14 + 14 = -36 < 0\n - Since g(-2) < 0 < g(0)$, by IVT applied to$g$on$[-2, 0]$with$k = 0$, there exists$c \in (-2, 0)$such that$g(c) = 0$, which means$c^5 + 4c^3 + 14 = 7c$.

The Squeeze Theorem and Limits of Trigonometric Functions

• The Squeeze Theorem

  • Theorem: Let $f(x)$, $g(x)$, and $h(x)$ be defined on an open interval $I$ containing $a$ (except possibly at $x = a$) such that:     $f(x) \le g(x) \le h(x) \quad \text{for all } x \in I \setminus \{a\}$     If $\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x)$, then \lim_{x \to a} g(x) = L$.\n - **Notation Note**: Symbol \setminus$represents "set minus", so$x \in I \setminus {a}$means$x \in I$but$x \neq a$.
  • The theorem holds analogously for one-sided limits ($x \to a^-$, $x \to a^+$) and limits at infinity ($x \to \pm\infty$).      

• Continuity and Domains of Trigonometric Functions

  • All circular trigonometric functions are continuous on their respective domains.
  • Continuous properties and domains:
  • $\text{dom}(\sin) = \mathbb{R} \implies \lim_{x \to a} \sin(f(x)) = \sin\left(\lim_{x \to a} f(x)\right)$ for any $a \in \mathbb{R}$.
  • $\text{dom}(\cos) = \mathbb{R}$.
  • $\text{dom}(\tan) = \text{dom}(\sec) = \left\{x \in \mathbb{R} : x \neq \frac{(2k+1)\pi}{2}, k \in \mathbb{Z}\right\}$.
  • $\text{dom}(\cot) = \text{dom}(\csc) = \left\{x \in \mathbb{R} : x \neq k\pi, k \in \mathbb{Z}\right\}$.

• Fundamental Trigonometric Limit

  • Fundamental Limit: $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$
  • Generalized Form: If $g(x) \to 0$ as $x \to a$ (i.e., $\lim_{x \to a} g(x) = 0$), then:     $\lim_{x \to a} \frac{\sin(g(x))}{g(x)} = 1$

• Worked Examples

  • Example 1: Evaluate $\lim_{x \to -\infty} \frac{2x^3 + 1 + 5\cos(x)}{3x^3}$
  • Split the fraction: $\lim_{x \to -\infty} \left( \frac{2x^3 + 1}{3x^3} + \frac{5\cos(x)}{3x^3} \right) = \lim_{x \to -\infty} \frac{2x^3 + 1}{3x^3} + \lim_{x \to -\infty} \frac{5\cos(x)}{3x^3}$.
  • First part: $\lim_{x \to -\infty} \left( \frac{2}{3} + \frac{1}{3x^3} \right) = \frac{2}{3}$.
  • Second part via Squeeze Theorem: Since $-1 \le \cos(x) \le 1$ for all $x \in \mathbb{R}$, for $x < 0$:       $\frac{5}{3x^3} \le \frac{5\cos(x)}{3x^3} \le -\frac{5}{3x^3}$
  • Since $\lim_{x \to -\infty} \frac{5}{3x^3} = 0$ and $\lim_{x \to -\infty} -\frac{5}{3x^3} = 0$, \lim_{x \to -\infty} \frac{5\cos(x)}{3x^3} = 0$.\n - Total limit: \frac{2}{3} + 0 = \frac{2}{3}.\n - **Example 2**: Evaluate \lim_{x \to \infty} x \sin\left(\frac{1}{x^2}\right)\n - Rewrite as: \lim_{x \to \infty} \frac{\sin\left(\frac{1}{x^2}\right)}{\frac{1}{x}} = \lim_{x \to \infty} \left[ \frac{1}{x} \cdot \frac{\sin\left(\frac{1}{x^2}\right)}{\frac{1}{x^2}} \right].\n - Let g(x) = \frac{1}{x^2}$. As$x \to \infty$,$g(x) \to 0$, so$\lim_{x \to \infty} \frac{\sin\left(\frac{1}{x^2}\right)}{\frac{1}{x^2}} = 1$.
  • Since $\lim_{x \to \infty} \frac{1}{x} = 0$, the total limit is 0 \cdot 1 = 0$.\n - **Example 3**: Evaluate \lim_{x \to 2} \frac{\sin(4x - 8)}{4x^2 - 5x - 6}\n - Factor denominator: 4x^2 - 5x - 6 = (4x + 3)(x - 2).\n - Factor argument: 4x - 8 = 4(x - 2).\n - Rewrite expression:\n      \lim_{x \to 2} \left[ \frac{\sin(4(x - 2))}{4(x - 2)} \cdot \frac{4}{4x + 3} \right]\n - As x \to 2$,$4(x - 2) \to 0$, so$\lim_{x \to 2} \frac{\sin(4(x - 2))}{4(x - 2)} = 1$.
  • Limit is $1 \cdot \frac{4}{4(2) + 3} = \frac{4}{11}$.

• Supplementary Exercises and Seatwork 5

  • Evaluate $\lim_{x \to +\infty} \frac{\sin(x) + 3\cos(x)}{x}$:
  • Bound numerator: $-4 \le \sin(x) + 3\cos(x) \le 4$. For $x > 0$:       $-\frac{4}{x} \le \frac{\sin(x) + 3\cos(x)}{x} \le \frac{4}{x}$
  • Since $\lim_{x \to +\infty} \pm \frac{4}{x} = 0$, by Squeeze Theorem the limit is 0$.\n - Evaluate \lim_{x \to 0} \frac{\sin(\sin(x))}{x}:\n - Rewrite as \frac{\sin(\sin(x))}{\sin(x)} \cdot \frac{\sin(x)}{x}$. As$x \to 0$,$\sin(x) \to 0$, so both factors approach$1$. Limit is$1 \cdot 1 = 1$.
  • Evaluate $\lim_{x \to 0} \frac{x \tan(5x)}{\sin^2(2x)}$:
  • Convert to sines/cosines:       $\frac{x \sin(5x)}{\cos(5x) \sin^2(2x)} = \frac{1}{\cos(5x)} \cdot \left(\frac{\sin(5x)}{5x} \cdot 5x\right) \cdot \frac{x}{\left(\frac{\sin(2x)}{2x} \cdot 2x\right)^2} = \frac{1}{\cos(5x)} \cdot \frac{\sin(5x)}{5x} \cdot \frac{1}{\left(\frac{\sin(2x)}{2x}\right)^2} \cdot \frac{5x^2}{4x^2}$
  • Taking limit as $x \to 0$ yields 1 \cdot 1 \cdot 1 \cdot \frac{5}{4} = \frac{5}{4}$.\n\n# Exponential, Logarithmic, and Inverse Circular Functions\n\n- **Limits of Exponential Functions**\n - Case a > 1:\n - \lim_{x \to +\infty} a^x = +\infty\n - \lim_{x \to -\infty} a^x = 0\n - Case a \in (0, 1):\n - \lim_{x \to +\infty} a^x = 0\n - \lim_{x \to -\infty} a^x = +\infty\n - **Remark**: If a \in (0, 1)$, then$a^{-1} > 1$, reducing to the$a > 1 case:\n    \lim_{x \to +\infty} \left(\frac{1}{2}\right)^x = \lim_{x \to +\infty} 2^{-x} = \lim_{y \to -\infty} 2^y = 0\n\n- **Limits of Logarithmic Functions**\n - Case a > 1:\n - \lim_{x \to +\infty} \log_a(x) = +\infty\n - \lim_{x \to 0^+} \log_a(x) = -\infty\n - Case a \in (0, 1):\n - \lim_{x \to +\infty} \log_a(x) = -\infty\n - \lim_{x \to 0^+} \log_a(x) = +\infty\n - **Remark**: If a \in (0, 1)$, then$\log_a(x) = \frac{\log(x)}{\log(a)} = \frac{\log(x)}{-\log(a^{-1})} = -\log_{a^{-1}}(x).\n\n- **Inverse Circular Functions - Notation and Asymptotic Limits**\n - **Notation**: \sin^{-1}(x)$denotes the inverse sine function; the superscript$-1$denotes functional inverse, NOT an exponent. An equivalent notation is$\arcsin(x).\n - **Inverse Sine**: f(x) = \sin^{-1}(x)$, domain$[-1, 1]$, range$[-\frac{\pi}{2}, \frac{\pi}{2}].\n - \lim_{x \to -1^+} \sin^{-1}(x) = -\frac{\pi}{2}\n    \n    \n\n - **Inverse Cosine**: f(x) = \cos^{-1}(x)$, domain$[-1, 1]$, range$[0, \pi].\n - \lim_{x \to 1^-} \cos^{-1}(x) = 0\n    \n    \n\n - **Inverse Tangent**: f(x) = \tan^{-1}(x)$, domain$\mathbb{R}$, range$(-\frac{\pi}{2}, \frac{\pi}{2}).\n - \lim_{x \to +\infty} \tan^{-1}(x) = \frac{\pi}{2}\n - \lim_{x \to -\infty} \tan^{-1}(x) = -\frac{\pi}{2}\n    \n    \n\n - **Inverse Cotangent**: f(x) = \cot^{-1}(x)$, domain$\mathbb{R}$, range$(0, \pi).\n - \lim_{x \to +\infty} \cot^{-1}(x) = 0\n - \lim_{x \to -\infty} \cot^{-1}(x) = \pi\n    \n    \n\n - **Inverse Secant**: f(x) = \sec^{-1}(x)$, domain$(-\infty, -1] \cup [1, +\infty).\n - \lim_{x \to +\infty} \sec^{-1}(x) = \frac{\pi}{2}\n - \lim_{x \to -\infty} \sec^{-1}(x) = \frac{3\pi}{2}\n    \n    \n\n - **Inverse Cosecant**: f(x) = \csc^{-1}(x)$, domain$(-\infty, -1] \cup [1, +\infty).\n - \lim_{x \to +\infty} \csc^{-1}(x) = 0\n - \lim_{x \to -\infty} \csc^{-1}(x) = -\pi\n    \n    \n\n- **Worked Limit Examples**\n - **Example 1**: Evaluate \lim_{x \to -\infty} \frac{4e^{4x} - e^{-2x}}{6e^{4x} - e^{2x} + 3e^{-x}}\n - Divide numerator and denominator by dominant term e^{-x}$as$x \to -\infty:\n      \frac{4e^{5x} - e^{-x}}{6e^{5x} - e^{3x} + 3}\n - Alternatively, factor e^{-2x} from numerator and denominator:\n      \frac{e^{-2x}(4e^{6x} - 1)}{e^{-2x}(6e^{6x} - e^{4x} + 3e^x)} = \frac{4e^{6x} - 1}{6e^{6x} - e^{4x} + 3e^x}\n - Taking limit as x \to -\infty$($e^{6x} \to 0, e^{4x} \to 0, e^x \to 0^+):\n      \frac{0 - 1}{0 - 0 + 0^+} = -\infty\n - **Example 2**: Evaluate \lim_{x \to 0^+} \cot^{-1}(\ln(x))\n - As x \to 0^+$,$\ln(x) \to -\infty.\n - Let u = \ln(x) \to -\infty$. Then$\lim_{u \to -\infty} \cot^{-1}(u) = \pi$.
  • Example 3: Evaluate $\lim_{x \to -\infty} \csc^{-1}(e^x + x^2)$
  • As $x \to -\infty$, $e^x \to 0$ and $x^2 \to +\infty$, so e^x + x^2 \to +\infty$.\n - Let u = e^x + x^2 \to +\infty$. Then$\lim_{u \to +\infty} \csc^{-1}(u) = 0$.
  • Example 4: Evaluate $\lim_{x \to +\infty} \sec\left(2\tan^{-1}\left(\frac{2x}{3}\right)\right)$
  • As $x \to +\infty$, $\frac{2x}{3} \to +\infty$, so $\tan^{-1}\left(\frac{2x}{3}\right) \to \frac{\pi}{2}^-$.
  • Argument approaches 2\left(\frac{\pi}{2}\right) = \pi$.\n - Limit is \sec(\pi) = -1$.

• Seatwork 6 Problems

  1. Evaluate $\lim_{x \to +\infty} \frac{4e^{4x} - e^{-2x}}{6e^{4x} - e^{2x} + 3e^{-x}}$:
  • Multiply numerator and denominator by $e^{-4x}$:        $\frac{4 - e^{-6x}}{6 - e^{-2x} + 3e^{-5x}}$
  • As $x \to +\infty$, negative powers of $e$ approach 0$. Limit is \frac{4 - 0}{6 - 0 + 0} = \frac{2}{3}.\n 2. Evaluate \lim_{x \to 0^-} \sec^{-1}\left(\frac{1}{x}\right):\n - As x \to 0^-$,$\frac{1}{x} \to -\infty.\n - \lim_{u \to -\infty} \sec^{-1}(u) = \frac{3\pi}{2}.\n 3. Evaluate \lim_{x \to -\infty} \frac{1}{\pi + 2\tan^{-1}(x)}:\n - As x \to -\infty$,$\tan^{-1}(x) \to -\frac{\pi}{2}^+$, so$\pi + 2\tan^{-1}(x) \to 0^+.\n - Limit is \frac{1}{0^+} = +\infty$.

Hyperbolic Functions and Identity Theorems

• Definitions of Hyperbolic Functions

  1. Hyperbolic Sine: $\sinh(x) = \frac{e^x - e^{-x}}{2}$, $x \in \mathbb{R}$
  2. Hyperbolic Cosine: $\cosh(x) = \frac{e^x + e^{-x}}{2}$, $x \in \mathbb{R}$
  3. Hyperbolic Tangent: $\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$, $x \in \mathbb{R}$
  4. Hyperbolic Cotangent: $\coth(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}}$, $x \in \mathbb{R} \setminus \{0\}$
  5. Hyperbolic Secant: $\text{sech}(x) = \frac{2}{e^x + e^{-x}}$, $x \in \mathbb{R}$
  6. Hyperbolic Cosecant: $\text{csch}(x) = \frac{2}{e^x - e^{-x}}$, $x \in \mathbb{R} \setminus \{0\}$

• Initial Numerical Evaluations

  • Evaluate \sinh(0) = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0$.\n - Evaluate \cosh(\ln 2) = \frac{e^{\ln 2} + e^{-\ln 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4}$.

• Graphs and Boundary Limits of Hyperbolic Functions

  • Hyperbolic Sine: $f(x) = \sinh(x)$

  • $\lim_{x \to +\infty} \sinh(x) = +\infty$

  • $\lim_{x \to -\infty} \sinh(x) = -\infty$

  • Hyperbolic Cosine: $f(x) = \cosh(x)$, y-intercept $(0, 1)$

  • $\lim_{x \to \pm\infty} \cosh(x) = +\infty$

  • Hyperbolic Tangent: $f(x) = \tanh(x)$, horizontal asymptotes $y = 1, y = -1$

  • $\lim_{x \to +\infty} \tanh(x) = 1$

  • $\lim_{x \to -\infty} \tanh(x) = -1$

  • Hyperbolic Cotangent: $f(x) = \coth(x)$, horizontal asymptotes $y = 1, y = -1$, vertical asymptote $x = 0$

  • $\lim_{x \to +\infty} \coth(x) = 1$, $\lim_{x \to -\infty} \coth(x) = -1$

  • $\lim_{x \to 0^+} \coth(x) = +\infty$, $\lim_{x \to 0^-} \coth(x) = -\infty$

  • Hyperbolic Secant: $f(x) = \text{sech}(x)$, peak $(0, 1)$, horizontal asymptote $y = 0$

  • $\lim_{x \to \pm\infty} \text{sech}(x) = 0$

  • Hyperbolic Cosecant: $f(x) = \text{csch}(x)$, horizontal asymptote $y = 0$, vertical asymptote $x = 0$

  • $\lim_{x \to 0^+} \text{csch}(x) = +\infty$, $\lim_{x \to 0^-} \text{csch}(x) = -\infty$

  • $\lim_{x \to \pm\infty} \text{csch}(x) = 0$

• Hyperbolic Identity Theorems

  • Reciprocal Identities:
  1. $\text{sech}(x) = \frac{1}{\cosh(x)}$
  2. $\text{csch}(x) = \frac{1}{\sinh(x)}$
  3. $\coth(x) = \frac{1}{\tanh(x)}$
  • Quotient Identities:
  1. $\tanh(x) = \frac{\sinh(x)}{\cosh(x)}$
  2. $\coth(x) = \frac{\cosh(x)}{\sinh(x)}$
  • Exponential Combination Theorems:
  1. $\cosh(x) + \sinh(x) = e^x$
  2. $\cosh(x) - \sinh(x) = e^{-x}$
  • Hyperbolic Pythagorean Identities:
  1. $\cosh^2(x) - \sinh^2(x) = 1$
  2. $1 - \tanh^2(x) = \text{sech}^2(x)$
  3. $1 - \coth^2(x) = -\text{csch}^2(x)$
  • Sum and Difference Formulas:
  1. $\sinh(x \pm y) = \sinh(x)\cosh(y) \pm \cosh(x)\sinh(y)$
  2. $\cosh(x \pm y) = \cosh(x)\cosh(y) \pm \sinh(x)\sinh(y)$
  • Double Measure Formulas:
  1. $\sinh(2x) = 2\sinh(x)\cosh(x)$
  2. $\cosh(2x) = \cosh^2(x) + \sinh^2(x) = 1 + 2\sinh^2(x) = 2\cosh^2(x) - 1$

• Geometric Foundation of Hyperbolic Functions

  • Trigonometric circular functions parameterize the Unit Circle $x^2 + y^2 = 1$ via points $(\cos(t), \sin(t))$.      

  • Hyperbolic functions parameterize the Unit Hyperbola $x^2 - y^2 = 1$ via points $(\cosh(t), \sinh(t))$.      

• Worked Identity Example

  • Problem: Given $\tanh(x) = \frac{12}{13}$, find the values of the remaining five hyperbolic functions of $x$.
  • Solution:
  • Use $1 - \tanh^2(x) = \text{sech}^2(x)$: $\text{sech}^2(x) = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}$. Since $\text{sech}(x) > 0$ for all $x$, \text{sech}(x) = \frac{5}{13}$.\n - \cosh(x) = \frac{1}{\text{sech}(x)} = \frac{13}{5}.\n - \sinh(x) = \tanh(x)\cosh(x) = \frac{12}{13} \cdot \frac{13}{5} = \frac{12}{5}.\n - \coth(x) = \frac{1}{\tanh(x)} = \frac{13}{12}.\n - \text{csch}(x) = \frac{1}{\sinh(x)} = \frac{5}{12}.\n\n# Inverse Hyperbolic Functions and Advanced Limit Applications\n\n- **One-to-One Nature and Domain Restriction**\n - Functions \sinh$,$\text{csch}$,$\tanh$, and$\coth are strictly one-to-one across their entire domains.\n - Functions \cosh$and$\text{sech}$fail the horizontal line test on$\mathbb{R} and are NOT one-to-one.\n - Domain restriction solution: Restrict the domains of \cosh(x)$and$\text{sech}(x)$to$[0, +\infty), where both are strictly monotonic and one-to-one, allowing inverses to be defined.\n\n- **Explicit Logarithmic Representations of Inverse Hyperbolic Functions**\n 1. \sinh^{-1}(x) = \ln\left(x + \sqrt{x^2 + 1}\right)$, for all$x \in \mathbb{R}\n 2. \cosh^{-1}(x) = \ln\left(x + \sqrt{x^2 - 1}\right)$, for all$x \in [1, +\infty)\n 3. \tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)$, for all$x \in (-1, 1)\n 4. \coth^{-1}(x) = \frac{1}{2} \ln\left(\frac{x + 1}{x - 1}\right)$, for all$x \in (-\infty, -1) \cup (1, +\infty)\n 5. \text{sech}^{-1}(x) = \ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)$, for all$x \in (0, 1]\n 6. \text{csch}^{-1}(x) = \ln\left(\frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|}\right)$, for all$x \in \mathbb{R} \setminus {0}\n\n- **Numerical Inverse Evaluation Example**\n - Evaluate \coth^{-1}\left(\frac{5}{4}\right):\n    \coth^{-1}\left(\frac{5}{4}\right) = \frac{1}{2} \ln\left( \frac{\frac{5}{4} + 1}{\frac{5}{4} - 1} \right) = \frac{1}{2} \ln\left( \frac{\frac{9}{4}}{\frac{1}{4}} \right) = \frac{1}{2} \ln(9) = \ln(3)\n\n- **Graphs and Boundary Limits of Inverse Hyperbolic Functions**\n - **Inverse Hyperbolic Sine**: f(x) = \sinh^{-1}(x)\n - \lim_{x \to -\infty} \sinh^{-1}(x) = -\infty\n - \lim_{x \to +\infty} \sinh^{-1}(x) = +\infty\n - **Inverse Hyperbolic Cosine**: f(x) = \cosh^{-1}(x)$, point$(1, 0)\n - \lim_{x \to 1^+} \cosh^{-1}(x) = 0\n - \lim_{x \to +\infty} \cosh^{-1}(x) = +\infty\n - **Inverse Hyperbolic Tangent**: f(x) = \tanh^{-1}(x)$, vertical asymptotes$x = -1, x = 1\n - \lim_{x \to -1^+} \tanh^{-1}(x) = -\infty\n - \lim_{x \to 1^-} \tanh^{-1}(x) = +\infty\n - **Inverse Hyperbolic Cotangent**: f(x) = \coth^{-1}(x)$, vertical asymptotes$x = -1, x = 1\n - \lim_{x \to -1^-} \coth^{-1}(x) = -\infty$,$\lim_{x \to 1^+} \coth^{-1}(x) = +\infty\n - \lim_{x \to \pm\infty} \coth^{-1}(x) = 0\n - **Inverse Hyperbolic Secant**: f(x) = \text{sech}^{-1}(x)$, point$(1, 0)\n - \lim_{x \to 0^+} \text{sech}^{-1}(x) = +\infty\n - \lim_{x \to 1^-} \text{sech}^{-1}(x) = 0\n - **Inverse Hyperbolic Cosecant**: f(x) = \text{csch}^{-1}(x)\n - \lim_{x \to 0^-} \text{csch}^{-1}(x) = -\infty$,$\lim_{x \to 0^+} \text{csch}^{-1}(x) = +\infty\n - \lim_{x \to \pm\infty} \text{csch}^{-1}(x) = 0\n\n- **Comprehensive Limit Examples**\n - **Example 1**: Evaluate \lim_{x \to +\infty} \frac{\cosh(x)}{e^x}\n - Solution: Recall \lim_{x \to -\infty} e^x = 0$. Substitute explicit formula for$\cosh(x):\n      \lim_{x \to +\infty} \frac{\frac{e^x + e^{-x}}{2}}{e^x} = \frac{1}{2} \lim_{x \to +\infty} \frac{e^x + e^{-x}}{e^x} = \frac{1}{2} \lim_{x \to +\infty} (1 + e^{-2x}) = \frac{1}{2}(1 + 0) = \frac{1}{2}\n - **Example 2**: Evaluate \lim_{x \to -\infty} \frac{3\text{csch}(3x)}{2\text{sech}(2x)}\n - Solution:\n      \lim_{x \to -\infty} \frac{3\text{csch}(3x)}{2\text{sech}(2x)} = \frac{3}{2} \lim_{x \to -\infty} \frac{\cosh(2x)}{\sinh(3x)} = \frac{3}{2} \lim_{x \to -\infty} \frac{\frac{e^{2x} + e^{-2x}}{2}}{\frac{e^{3x} - e^{-3x}}{2}} = \frac{3}{2} \lim_{x \to -\infty} \frac{e^{2x} + e^{-2x}}{e^{3x} - e^{-3x}}\n - Multiply numerator and denominator by e^{2x}$, yielding form$\frac{\infty}{-\infty}:\n      \frac{3}{2} \lim_{x \to -\infty} \frac{e^{4x} + 1}{e^{5x} - e^{-x}}\n - As x \to -\infty$,$e^{4x} \to 0$,$e^{5x} \to 0$,$e^{-x} \to +\infty$. Thus:       $\frac{3}{2} \cdot \frac{0 + 1}{0 - \infty} = 0$
  • Example 3: Evaluate $\lim_{x \to 0^-} \tanh\left(\text{csch}^{-1}\left(\sinh^{-1}(x)\right)\right)$
  • Solution: Trace composite inner limits as $x \to 0^-$:
    • $x \to 0^- \implies \sinh^{-1}(x) \to 0^-$
    • $u = \sinh^{-1}(x) \to 0^- \implies \text{csch}^{-1}(u) \to -\infty$
    • $v = \text{csch}^{-1}(u) \to -\infty \implies \tanh(v) \to -1^+$
  • Hence, \lim_{x \to 0^-} \tanh\left(\text{csch}^{-1}\left(\sinh^{-1}(x)\right)\right) = -1$.\n - **Example 4**: Evaluate \lim_{x \to 0^+} \ln\left(\text{sech}^{-1}(x)\right)\n - Solution:\n - x \to 0^+ \implies \text{sech}^{-1}(x) \to +\infty\n - u = \text{sech}^{-1}(x) \to +\infty \implies \ln(u) \to +\infty\n - Hence, \lim_{x \to 0^+} \ln\left(\text{sech}^{-1}(x)\right) = +\infty$.
  • Example 5: Evaluate $\lim_{x \to +\infty} \sin(x)\text{csch}(x)$
  • Solution: Note $-1 \le \sin(x) \le 1$ for all $x \in \mathbb{R}$. Since $\text{csch}(x) > 0$ for $x > 0$:       $-\text{csch}(x) \le \sin(x)\text{csch}(x) \le \text{csch}(x) \quad \text{for all } x > 0$
  • Evaluate bounding limits: $\lim_{x \to +\infty} -\text{csch}(x) = 0$ and \lim_{x \to +\infty} \text{csch}(x) = 0$.\n - By Squeeze Theorem, \lim_{x \to +\infty} \sin(x)\text{csch}(x) = 0$.

• Supplementary Exercises and Seatwork 7 Solutions

  • Supplementary Exercise Limits:
  1. $\lim_{x \to -\infty} e^x \sinh(x) = \lim_{x \to -\infty} e^x \left(\frac{e^x - e^{-x}}{2}\right) = \lim_{x \to -\infty} \frac{e^{2x} - 1}{2} = -\frac{1}{2}$
  2. $\lim_{x \to 0^+} \sec^{-1}(\text{csch}(x)) = \frac{\pi}{2}$, since $\text{csch}(x) \to +\infty$
  3. $\lim_{x \to -\infty} \tan^{-1}(\cosh(x)) = \frac{\pi}{2}$, since $\cosh(x) \to +\infty$
  4. $\lim_{x \to 0^-} 2^{\text{csch}^{-1}(x)} = 0$, since $\text{csch}^{-1}(x) \to -\infty$
  5. $\lim_{x \to 0^+} \ln(\text{sech}^{-1}(x)) = +\infty$
  6. $\lim_{x \to 0^+} \left(\frac{1}{2}\right)^{\text{csch}^{-1}(x)} = 0$, since $\text{csch}^{-1}(x) \to +\infty$
  • Exact Values:
  1. $\text{sech}(\ln 2) = \frac{1}{\cosh(\ln 2)} = \frac{4}{5}$
  2. $\cosh(\ln 5 + \ln 6) = \cosh(\ln 30) = \frac{e^{\ln 30} + e^{-\ln 30}}{2} = \frac{30 + \frac{1}{30}}{2} = \frac{901}{60}$
  3. $\tanh^{-1}(0.5) = \frac{1}{2}\ln\left(\frac{1 + 0.5}{1 - 0.5}\right) = \frac{1}{2}\ln(3) = \ln(\sqrt{3})$
  • Seatwork 7 Problems:
  1. Evaluate $\lim_{x \to 0^+} 2\coth\left(\text{csch}^{-1}\left(\tanh^{-1}(x)\right)\right)$:
     • As $x \to 0^+$, $\tanh^{-1}(x) \to 0^+$.
     • As $u \to 0^+$, $\text{csch}^{-1}(u) \to +\infty$
     • As $v \to +\infty$, $\coth(v) \to 1$
     • Limit is 2(1) = 2$.\n 2. Evaluate \lim_{x \to -\infty} \cos(x) \coth^{-1}(x) using Squeeze Theorem:\n - For x < -1$,$-1 \le \cos(x) \le 1$. Since$\coth^{-1}(x) < 0$for$x < -1:\n         \coth^{-1}(x) \le \cos(x)\coth^{-1}(x) \le -\coth^{-1}(x)\n - Since \lim_{x \to -\infty} \coth^{-1}(x) = 0$and$\lim_{x \to -\infty} -\coth^{-1}(x) = 0, by the Squeeze Theorem:\n         \lim_{x \to -\infty} \cos(x) \coth^{-1}(x) = 0$$