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