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? 

Let $x=2+h$ and let $h \rightarrow 0$ hence
$$
\frac{\sqrt{4+h}-2}{h}=\frac{2\sqrt{1+\frac{h}{4}}-2}{h}=\frac{2\left(1+\frac{h}{8}-1+o\left(h\right)\right)}{h} \underset{h \rightarrow 0}{\rightarrow}\frac{1}{4}
$$
Hence

$$
\frac{\sqrt{x+2}-2}{x-2} \underset{x \rightarrow 2}{\rightarrow}\frac{1}{4}]
$$

