Did I do this limit right? $\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1}$ This is how I did this limit:
$$\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1} =\lim_{x \to 0}\frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)} = \lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)(\sqrt{x^2+2}+\sqrt{2})}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} = \lim_{x \to 0}\frac{({x^2+2}-2)(\sqrt{x^2+1}+1)}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} =\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{1}{2\sqrt{2}}$$
But Wolframalpha gave me another answer! 
So did I do it right?

As a backup check, you could use $\sqrt{x+a} = \sqrt{a}+\frac{1}{2\sqrt{a}} x +o(x)$, or $\sqrt{x+a} - \sqrt{a} =\frac{1}{2\sqrt{a}} x +o(x)$.
Then $\frac{\sqrt{x^2+2}-\sqrt{2}}{ \sqrt{x^2+1}-\sqrt{1} } = \frac{ \frac{1}{2\sqrt{2}} x^2 +o(x^2)  }{ \frac{1}{2\sqrt{1}} x^2 +o(x^2)  } = 
\frac{ \frac{1}{2\sqrt{2}} +\frac{ o(x^2) }{x^2}  }{ \frac{1}{2\sqrt{1}} +\frac{ o(x^2) }{x^2}  }$. Taking limits gives the desired result.
