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