How to find $\lim_{x \to 7} \frac{x-7}{\sqrt{x+2}-3}$? I'm totally at a loss for this... someone please help!
$$\lim_{x \to 7} \frac{x-7}{\sqrt{x+2}-3}$$
I've tried multiplying by the conjugate, but after doing that I couldn't find a way to cancel out the $x-7$ still (the only thing making the limit not work).
Edit: Thank you all so much! I made a simple math error when I multiplied by the conjugate...
 A: Hint:- Rationalize.
Solution:-

$\displaystyle\lim_{x\to 7}\dfrac{x-7}{\sqrt{x+2}-3}=\displaystyle\lim_{x\to 7}\dfrac{(x-7)(\sqrt{x+2}+3)}{x+2-9}=\displaystyle\lim_{x\to 7}\dfrac{(x-7)(\sqrt{x+2}+3)}{x-7}=\displaystyle\lim_{x\to 7}(\sqrt{x+2}+3)=?$

A: Note that 
$$\lim_{x \to 7}\frac{x-7}{\sqrt{x+2}-3} = \lim_{h \to 0}\frac{h}{\sqrt{9+h}-3};$$
but
$$
\frac{h}{\sqrt{9+h}-3} = \frac{h(\sqrt{9+h})+3h}{h} = \sqrt{9+h}+3 \to 6
$$
as $h \to 0$.
A: Let $f(x) = \sqrt {x+2}.$ The expression equals
$$\frac{1}{(f(x) - f(7))/(x-7)} \to 1/f'(7)$$
by the definition of the derivative. 
A: $$\lim_{x -> 7} \frac{x-7}{\sqrt{x+2}-3} = \lim_{x -> 7} \frac{(x-7)(\sqrt{x+2}+3)}{x+2-9} = 6$$
A: Multiply by (conjugate/conjugate) as you suggested:  
$$\lim\limits_{x \to 7} (\frac{x-7}{\sqrt{x+2}-3}\times \frac {\sqrt{x+2}+3}{\sqrt{x+2}+3})=\lim\limits_{x \to 7} \frac{(x-7)(\sqrt{x+2}+3)}{{x+2}-9}=\lim\limits_{x \to 7} \frac{(x-7)(\sqrt{x+2}+3)}{x-7}
  =\lim\limits_{x \to 7} \sqrt{x+2}+3=6$$
Do you see how $x-7$ cancelled out?
