Answer
$$\lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8} =\frac{1}{4} $$
Work Step by Step
Given $$\lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8}$$
let $$ f(x) = \lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8}$$
Since, we have
$$ f(8)= \frac{\sqrt{8-4}-2}{8-8}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}L&= \lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8} \\
&=\lim _{x \rightarrow 4} \frac{\sqrt{x-4}-2}{x-8} \times \frac{\sqrt{x-4}+2}{\sqrt{x-4}+2}\\
&=\lim _{x \rightarrow 4} \frac{(\sqrt{x-4})^{2}-2^{2}}{(x-8)(\sqrt{x-4}+2)}\\
&= \lim _{x \rightarrow 4} \frac{(x-4-4)}{(x-8)(\sqrt{x-4}+2)}\\
&=\lim _{x \rightarrow 4} \frac{(x-8)}{(x-8)(\sqrt{x-4}+2)} \\
&=\lim _{x \rightarrow 4} \frac{1}{\sqrt{x-4}+2} \\
&=\frac{1}{\sqrt{8-4}+2} \\ &=\frac{1}{\sqrt{4}+2} \\
&=\frac{1}{2+2}\\
&=\frac{1}{4} \end{aligned}
