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: 
Try multiplying top and bottom by the quantity $\sqrt{x + 4} + 7$. This leads to

$$\frac{\sqrt{x + 4} - 7}{x - 45} = \frac{(\sqrt{x + 4} - 7)(\sqrt{x + 4} + 7)}{(x - 45)(\sqrt{x + 4} + 7)}$$

The numerator can be simplified to get $x - 45$.
Hint: think of $f(x) = \sqrt{x+4}$. What does the limit you've described remind you of?