The problem consists of three parts: a) Evaluate the limit $\lim_{x\to 2} \frac{|x-2| + x - 2}{x^2 - 4}$ by computing the one-sided limits. b) Express $\cosh^{-1} x$ in logarithmic form for $x \ge 1$. c) Given that $\sinh x = -\frac{3}{4}$, find $\coth x$.

1. Problem Description

The problem consists of three parts:

a) Evaluate the limit $\lim_{x\to 2} \frac{|x-2| + x - 2}{x^2 - 4}$ by computing the one-sided limits.

b) Express $\cosh^{-1} x$ in logarithmic form for $x \ge 1$.

c) Given that $\sinh x = -\frac{3}{4}$, find $\coth x$.

2. Solution Steps

a) We need to find $\lim_{x\to 2} \frac{|x-2| + x - 2}{x^2 - 4}$. We analyze the one-sided limits.

For $x \to 2^+$, we have $x > 2$, so $|x-2| = x-2$. Then

$$\lim_{x\to 2^+} \frac{|x-2| + x - 2}{x^2 - 4} = \lim_{x\to 2^+} \frac{(x-2) + (x-2)}{x^2 - 4} = \lim_{x\to 2^+} \frac{2(x-2)}{(x-2)(x+2)} = \lim_{x\to 2^+} \frac{2}{x+2} = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}.$$

For $x \to 2^-$, we have $x < 2$, so $|x-2| = -(x-2) = 2-x$. Then

$$\lim_{x\to 2^-} \frac{|x-2| + x - 2}{x^2 - 4} = \lim_{x\to 2^-} \frac{-(x-2) + (x-2)}{x^2 - 4} = \lim_{x\to 2^-} \frac{0}{(x-2)(x+2)} = \lim_{x\to 2^-} 0 = 0.$$

Since the one-sided limits are not equal, the limit does not exist.

b) Let $y = \cosh^{-1} x$, so $x = \cosh y$. We are given that $\cosh^2 y - \sinh^2 y = 1$ and $\cosh y + \sinh y = e^y$.

From $\cosh^2 y - \sinh^2 y = 1$, we have $\cosh^2 y - 1 = \sinh^2 y$, so $\sinh y = \pm \sqrt{\cosh^2 y - 1} = \pm \sqrt{x^2 - 1}$.

Since $y = \cosh^{-1} x$, $y \ge 0$, thus $\cosh y + \sinh y = e^y > 0$. Also, $\cosh y = x \ge 1$. Therefore, $\sinh y$ must have the same sign as $e^y - \cosh y = e^y - x$, which depends on the value of x.

Substituting $\cosh y = x$ and $\sinh y = \pm \sqrt{x^2 - 1}$ into $\cosh y + \sinh y = e^y$, we get $x \pm \sqrt{x^2 - 1} = e^y$. Thus, $y = \ln(x \pm \sqrt{x^2 - 1})$.

Since $y \ge 0$, and $\cosh y = x \ge 1$, $\sinh y = \sqrt{x^2 - 1} \ge 0$.

So $y = \ln(x + \sqrt{x^2 - 1})$.

c) Given that $\sinh x = -\frac{3}{4}$, we want to find $\coth x = \frac{\cosh x}{\sinh x}$.

We know that $\cosh^2 x - \sinh^2 x = 1$, so $\cosh^2 x = 1 + \sinh^2 x = 1 + \left(-\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}$.

Thus, $\cosh x = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}$. Since $\cosh x$ is always positive, $\cosh x = \frac{5}{4}$.

Then, $\coth x = \frac{\cosh x}{\sinh x} = \frac{5/4}{-3/4} = \frac{5}{-3} = -\frac{5}{3}$.

3. Final Answer

a) The limit does not exist because the one-sided limits are not equal.

b) $\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1})$

c) $\coth x = -\frac{5}{3}$