[Math] How to find $\lim_{x \to 7} \frac{x-7}{\sqrt{x+2}-3}$

Question

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…

Answer 1

Hint:- Rationalize.

Solution:-

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