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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.

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  • $\begingroup$ It would be helpful if you explain what you have already tried so people can specifically address the points you are having trouble with. $\endgroup$
    – Sak
    Aug 13, 2013 at 15:27
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    $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Aug 13, 2013 at 15:27
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    $\begingroup$ And if the limit is what now appears after the edition then the limit is not what you say it is. $\endgroup$
    – DonAntonio
    Aug 13, 2013 at 15:28
  • $\begingroup$ It looks like the problem was supposed to have a $\sqrt{2+x}$ instead of $\sqrt{2x}$. $\endgroup$ Aug 13, 2013 at 15:32
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    $\begingroup$ 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}? $\endgroup$
    – alans
    Aug 13, 2013 at 15:33

3 Answers 3

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

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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.

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    $\begingroup$ I would rather use "DNE" than $\emptyset$, which has already a standard meaning. $\endgroup$
    – Pedro
    Aug 13, 2013 at 16:18
  • $\begingroup$ It is probably good to mention explicitly that this is an answer to the original version of the question - it has been edited since then, probably it contained a typo. $\endgroup$ Feb 5, 2019 at 8:07
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The limit is just the derivative of $f(x)=\sqrt{x+2}$ at $x=0$. Differentiating wrt $x$ gives: $$f'(x)=\dfrac{1}{2\sqrt{x+2}}\Biggl|_{x=0}=\dfrac{1}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{4}$$

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    $\begingroup$ The fact that this is exactly the derivative is a useful observation and it is useful for the OP to know this. But it should be also said that problems like this are often assigned before derivatives are taught and it is good also to know a solution without derivatives. $\endgroup$ Feb 5, 2019 at 8:08

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