# 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.

• It would be helpful if you explain what you have already tried so people can specifically address the points you are having trouble with.
– Sak
Aug 13, 2013 at 15:27
• If you meant the limit of $\,\frac{\sqrt{2x}-\sqrt2}x\;$ then this is not well defined as $\,\sqrt{2x}\;$ is defined only for $\,x\ge 0\;$ as a real function. So either your expression is different or the limit is from the right: $\,x\to 0^+\;$ . Please do use LaTeX to write mathematics in this site. Aug 13, 2013 at 15:27
• And if the limit is what now appears after the edition then the limit is not what you say it is. Aug 13, 2013 at 15:28
• It looks like the problem was supposed to have a $\sqrt{2+x}$ instead of $\sqrt{2x}$. Aug 13, 2013 at 15:32
• Is it a typing error - the result you stated is a limit of $\lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}$, since $\lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}+\sqrt{2}}=\lim_{x\rightarrow 0}\frac{1}{\sqrt{x+2}+\sqrt{2}}$=\frac{\sqrt{2}}{4}? Aug 13, 2013 at 15:33

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 $$\lim_{x\rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0} \frac{1}{\sqrt{x+2}+\sqrt{2}}=\frac{\sqrt{2}}{4}.$$
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.
• I would rather use "DNE" than $\emptyset$, which has already a standard meaning.
The limit is just the derivative of $$f(x)=\sqrt{x+2}$$ at $$x=0$$. Differentiating wrt $$x$$ gives: $$f'(x)=\dfrac{1}{2\sqrt{x+2}}\Biggl|_{x=0}=\dfrac{1}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{4}$$