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This problem is from Calculus: A complete course 8 ed.(Adams/Essex). Problem no. 35 chapt 1.2.

Evaluate the limit or explain why it does not exist.

$\displaystyle\lim_{x\to 0}\frac{\sqrt{2+x^2}-\sqrt{2+x^2}}{x^2}$

So I multiplied the fraction with the conjugate of the numerator and ended up with this. I see that $x^2$ can be divided from both numerator and denominator. But after that I am not sure how I should continue.

$\displaystyle\lim_{x\to 0}\frac {2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$

The answer states that next step is

$\displaystyle\frac {2}{\sqrt{2} + \sqrt{2}}$

But I do not see how it got there. I tried looking up square root laws and how simplify but could not see anything that was useful for me.

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  • $\begingroup$ I may be missing something but the numerator appears to be $0$. $\endgroup$ – badjohn Sep 1 '19 at 8:58
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$\displaystyle\quad\lim_{x\to 0}\frac {2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$

$\displaystyle=\lim_{x\to 0}\frac {2}{(\sqrt{2+x^2} + \sqrt{2-x^2})}$

$\displaystyle=\frac {2}{(\sqrt{2+0^2} + \sqrt{2-0^2})}$

$\displaystyle=\frac {2}{(\sqrt{2} + \sqrt{2})}$

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