a) We are given the function $f(x) = \sqrt{ax - b}$, where $a, b \in \mathbb{R}$ and $a > 0$. We need to verify that $\lim_{x \to \frac{b+a}{a}} f(x) = \sqrt{a}$ using the definition of a limit. b) We need to find the value of $x \in \mathbb{R}$ for which $\sinh^{-1} x + \cosh^{-1} (x+2) = 0$.

1. Problem Description

a) We are given the function $f(x) = \sqrt{ax - b}$, where $a, b \in \mathbb{R}$ and $a > 0$. We need to verify that $\lim_{x \to \frac{b+a}{a}} f(x) = \sqrt{a}$ using the definition of a limit.

b) We need to find the value of $x \in \mathbb{R}$ for which $\sinh^{-1} x + \cosh^{-1} (x+2) = 0$.

2. Solution Steps

a) To verify the limit, we need to show that for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - \frac{b+a}{a}| < \delta$, then $|f(x) - \sqrt{a}| < \epsilon$.

We have $|f(x) - \sqrt{a}| = |\sqrt{ax - b} - \sqrt{a}|$. We want to make this less than $\epsilon$.

Multiplying and dividing by the conjugate, we get:

$|\sqrt{ax - b} - \sqrt{a}| = |\frac{(\sqrt{ax - b} - \sqrt{a})(\sqrt{ax - b} + \sqrt{a})}{\sqrt{ax - b} + \sqrt{a}}| = |\frac{ax - b - a}{\sqrt{ax - b} + \sqrt{a}}| = |\frac{a(x - \frac{b+a}{a})}{\sqrt{ax - b} + \sqrt{a}}|$

So, $|f(x) - \sqrt{a}| = \frac{a|x - \frac{b+a}{a}|}{\sqrt{ax - b} + \sqrt{a}}$.

Since $x \to \frac{b+a}{a}$, we can assume $x > \frac{b}{a}$, so $ax - b > 0$.

Also, $ax - b$ is close to $a$ when $x$ is close to $\frac{b+a}{a}$, so $\sqrt{ax - b}$ is close to $\sqrt{a}$. Thus $\sqrt{ax - b} + \sqrt{a}$ is close to $2\sqrt{a}$.

$|f(x) - \sqrt{a}| = \frac{a|x - \frac{b+a}{a}|}{\sqrt{ax - b} + \sqrt{a}} < \frac{a|x - \frac{b+a}{a}|}{\sqrt{a}} = \sqrt{a} |x - \frac{b+a}{a}|$.

So, if we choose $\delta = \frac{\epsilon}{\sqrt{a}}$, then whenever $|x - \frac{b+a}{a}| < \delta$, we have $|f(x) - \sqrt{a}| < \sqrt{a} \delta = \sqrt{a} \frac{\epsilon}{\sqrt{a}} = \epsilon$.

Thus, we have shown that $\lim_{x \to \frac{b+a}{a}} f(x) = \sqrt{a}$.

b) We are given $\sinh^{-1} x + \cosh^{-1} (x+2) = 0$.

So, $\sinh^{-1} x = - \cosh^{-1} (x+2)$.

We know that $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$ and $\cosh^{-1} y = \ln(y + \sqrt{y^2 - 1})$ for $y \geq 1$.

Thus, $\ln(x + \sqrt{x^2 + 1}) = - \ln(x+2 + \sqrt{(x+2)^2 - 1}) = \ln(\frac{1}{x+2 + \sqrt{(x+2)^2 - 1}})$.

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

$(x + \sqrt{x^2 + 1}) (x+2 + \sqrt{x^2 + 4x + 3}) = 1$.

Also, since $\cosh^{-1}(x+2)$ is defined, we require $x+2 \geq 1$, which means $x \geq -1$.

$\sinh^{-1} x = - \cosh^{-1} (x+2)$

Taking sinh on both sides, $x = \sinh(-\cosh^{-1}(x+2)) = - \sinh(\cosh^{-1}(x+2))$.

We know that $\cosh^2 y - \sinh^2 y = 1$, so $\sinh y = \pm \sqrt{\cosh^2 y - 1}$.

Thus, $x = - \sqrt{\cosh^2(\cosh^{-1}(x+2)) - 1} = - \sqrt{(x+2)^2 - 1} = - \sqrt{x^2 + 4x + 3}$.

Squaring both sides gives $x^2 = x^2 + 4x + 3$, so $4x + 3 = 0$, and $x = -\frac{3}{4}$.

Now, we need to check if this value of $x$ works. Since $x \geq -1$, $x = -3/4$ is a candidate.

When $x = -\frac{3}{4}$, the original equation becomes:

$\sinh^{-1}(-\frac{3}{4}) + \cosh^{-1}(-\frac{3}{4} + 2) = \sinh^{-1}(-\frac{3}{4}) + \cosh^{-1}(\frac{5}{4})$.

$\sinh^{-1}(-\frac{3}{4}) = \ln(-\frac{3}{4} + \sqrt{\frac{9}{16} + 1}) = \ln(-\frac{3}{4} + \frac{5}{4}) = \ln(\frac{2}{4}) = \ln(\frac{1}{2}) = - \ln 2$.

$\cosh^{-1}(\frac{5}{4}) = \ln(\frac{5}{4} + \sqrt{\frac{25}{16} - 1}) = \ln(\frac{5}{4} + \frac{3}{4}) = \ln(\frac{8}{4}) = \ln 2$.

So, $\sinh^{-1}(-\frac{3}{4}) + \cosh^{-1}(\frac{5}{4}) = - \ln 2 + \ln 2 = 0$.

3. Final Answer

a) Verified.

b) $x = -\frac{3}{4}$