Answer
$$\lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16} =\frac{1}{8}$$
Work Step by Step
Given $$\lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16}$$
let $$ f(x) = \frac{\sqrt{x}-4}{x-16}$$
Since, we have
$$ f(4)=\frac{4-4}{16-16}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16}\\
&= \lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16} \ \ \frac{\sqrt{x}+4}{\sqrt{x}+4}\\
&= \lim _{x \rightarrow 16} \frac{( x-16) }{( x-16)(\sqrt{x}+4)} \ \\
&=\lim _{x \rightarrow 16} \frac{1}{\sqrt{x}+4 }\\
&=\frac{1}{4+4}\\
&=\frac{1}{8}\\
\end{aligned}
