TITLE: How do i evaluate $\displaystyle \lim_{x \to -2}\frac {\sqrt{x+2}}{(\sqrt{x}+\sqrt{2} )}$?
QUESTION [5 upvotes]: How do i evaluate $\displaystyle \lim_{x \to -2}\frac {\sqrt{x+2}}{(\sqrt{x}+\sqrt{2} )}$.
I'm confused in wolfram alpha the result is $0$ however $\sqrt{x}$ at $x=-2$ 
is undefined in $\mathbb{R}$ then how it is $0$ ?

REPLY [3 votes]: Notice, $$\lim_{x\to -2}\frac{\sqrt{x+2}}{\sqrt x+\sqrt 2}$$
$$=\lim_{x\to -2}\sqrt{x+2}\times \lim_{x\to -2} \frac{1}{\sqrt x+\sqrt 2}$$
$$=\sqrt{-2+2}\times \frac{1}{\sqrt {-2}+\sqrt 2}$$
$$=(0)\frac{1}{\sqrt {2}+i\sqrt 2}$$
$$=\color{red}{0}$$