<s>[INST] 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?
 [/INST] If you correctly simplify the very last line in your solution, you'll see you get the same answer: $\frac 2{2\sqrt 2}$
$$\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{2}{2\sqrt{2}} = \frac 1{\sqrt 2}$$
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