How can I evaluate $\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)}{ x}$?

Question

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.

Answer 1

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.

Answer 2

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 $$\lim_{x\rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0} \frac{1}{\sqrt{x+2}+\sqrt{2}}=\frac{\sqrt{2}}{4}.$$