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: 
The denominator is $(x+4)-49$, which is of the form $a^2-b^2=(a-b)(a+b)$. Then
$$
\frac{\sqrt{x+4}-7}{x-45}=\frac{\sqrt{x+4}-7}{(x+4)-49}=\frac{\sqrt{x+4}-7}{(\sqrt{x+4}-7)(\sqrt{x+4}+7)}=\frac{1}{\sqrt{x+4}+7}.
$$
Now we can obtain the limit by evaluating at $x=45$, to get $1/14$.
This is of indeterminate form $\frac{0}{0}$, so we may apply [L'Hôpital's Rule](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule). In this specific case, if $f(x)=\sqrt{x+4}-7$ and $g(x)=x-45$, then $$\lim\_{x \rightarrow 45} \frac{f(x)}{g(x)} = \lim\_{x \rightarrow 45} \frac{f'(x)}{g'(x)}$$ where $f'$ and $g'$ are the derivatives of $f$ and $g$, respectively.