<s>[INST] Limit to Infinity question? $$\lim_{x\to\infty}\left(-\sqrt{-2x+x^2}+\sqrt{2x+x^2}\right)=2$$
I'm not sure how to go about solving this problem. 
 [/INST] $lim_{x \rightarrow \infty} \sqrt{2x + x^2} - \sqrt{-2x +x^2} =$ $lim_{x \rightarrow \infty} \sqrt{2x + x^2} - \sqrt{-2x +x^2}  . \displaystyle\frac{\sqrt{2x + x^2} + \sqrt{-2x +x^2}}{\sqrt{2x + x^2} +\sqrt{-2x +x^2}} = $
$=lim_{x \rightarrow \infty} \displaystyle\frac{2x + x^2 - (-2x + x^2)}{\sqrt{2x + x^2} +\sqrt{-2x +x^2}}$
$= lim_{x \rightarrow \infty} \displaystyle\frac{4x}{ \sqrt{x^2(1 + 2/x)}+ \sqrt{x^2(1 - 2/x)}}$
$ = lim_{x \rightarrow \infty} \displaystyle\frac{4x}{ x \sqrt{(1 + 2/x)}+   x \sqrt{(1 - 2/x)}} $
$ = lim_{x \rightarrow \infty} \displaystyle\frac{4}{  \sqrt{(1 + 2/x)}+    \sqrt{(1 - 2/x)}} = 2 $
because
$= lim_{x \rightarrow \infty} \displaystyle\frac{1}{  \sqrt{(1 + 2/x)}+    \sqrt{(1 - 2/x)}} = 1/2$
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