How can I evaluate $\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)}{ x}$? I am having trouble with this practice problem on limits: 
$$\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)} {x}$$
The answer is $\sqrt{2} \over 4$, but I'm having trouble seeing how the answer was reached.  Any help would be appreciated.
 A: Please use LaTex next time. For now I assume your expression is 
$$
\lim_{x \to 0}\frac{\sqrt{2x}-\sqrt{2}}{x}=\sqrt{2} \lim_{x \to 0}\frac{\sqrt{x}-1}{x-1+1}=\sqrt{2}\lim_{x \to 0}\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)+1}=\sqrt{2}{\lim_{x \to 0}\frac{1}{\sqrt{x}+1+\frac{1}{\sqrt{x}-1}}}
$$
Two-sided limit doesn't exist.
A: I assume that you made typing error (probably there should write $\sqrt{2+x}$, not $\sqrt{2x}$). Then, the result you have stated is a limit of the following expression:
\begin{eqnarray*}
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{\sqrt{2+x}-\sqrt{2}}{x}\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}+\sqrt{2}},\\
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{x+2-2}{x(\sqrt{x+2}+\sqrt{2})},\\
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{x}{x(\sqrt{x+2}+\sqrt{2})},\\
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{1}{\sqrt{x+2}+\sqrt{2}},
\end{eqnarray*}
so
$$\lim_{x\rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0} \frac{1}{\sqrt{x+2}+\sqrt{2}}=\frac{\sqrt{2}}{4}.$$
A: The limit is just the derivative of $f(x)=\sqrt{x+2}$ at $x=0$. Differentiating wrt $x$ gives: $$f'(x)=\dfrac{1}{2\sqrt{x+2}}\Biggl|_{x=0}=\dfrac{1}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{4}$$
