<s>[INST] 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? 
 [/INST] For fun:
$\dfrac{\sqrt{x+2} -2}{(\sqrt{x+2})^2-4}=$
$\dfrac{\sqrt{x+2}-2}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}=$
$\dfrac{1}{\sqrt{x+2}+2}.$
The limit $x \rightarrow 2$ is?
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