What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there? 

Hint
$$\frac{\sqrt{x+2}-2}{x-2}\cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}= \frac{\sqrt{x+2}^2-2^2}{(x-2)\sqrt{x+2}+2}$$
$$ \frac{x-2}{(x-2)\sqrt{x+2}+2}= \frac{1}{\sqrt{x+2}+2}$$
