Question: Evaluate $\lim\_{x\to45} \frac{\sqrt{x+4}-7}{x-45}$

So far I multiply both sides by $x+45$, but I don't know how to finish it.

Please Help!!

Answer: 
Hint:
$$\frac{(\sqrt{x+4}-7)(\sqrt{x+4}+7)}{(x-45)(\sqrt{x+4}+7)}=\frac{1}{\sqrt{x+4}+7}.$$
You want
$\lim\_{x\to45} \frac{\sqrt{x+4}-7}{x-45}
$.

Since it's best to expand around zero,
let
$y = x-45$,
so
$x = y+45$.

You then want

$\begin{align}
\frac{\sqrt{y+49}-7}{y}
&=\frac{\sqrt{y+49}-7}{y}\frac{\sqrt{y+49}+7}{\sqrt{y+49}+7}\\
&=\frac{(\sqrt{y+49}-7)(\sqrt{y+49}+7)}{y(\sqrt{y+49}+7)}\\
&=\frac{(y+49)-49}{y(\sqrt{y+49}+7)}\\
&=\frac{y}{y(\sqrt{y+49}+7)}\\
&=\frac{1}{\sqrt{y+49}+7}\\
&=\frac{1}{\sqrt{y+49}+7}\\
\end{align}
$

so
$
\lim\_{y\to 0} 
\frac{\sqrt{y+49}-7}{y}
=\lim\_{y\to 0} 
=\frac{1}{\sqrt{y+49}+7}
=\frac1{14}
$.