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. 
 A: Hints:
$\lim_{x\to \infty}(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})
=\lim_{x\to \infty}\frac{(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})\times (\sqrt{-2x+x^2}+\sqrt{2x+x^2})}{(\sqrt{-2x+x^2}+\sqrt{2x+x^2})}=?$
Then Do you know what to do next step?
In fact, $$=\lim_{x\to \infty}\frac{4x}{\sqrt{-2x+x^2}+\sqrt{2x+x^2}}=
\lim_{x\to \infty}\frac{4}{\sqrt{-2\frac1x+1}+\sqrt{2\frac1x+1}}=2$$
A: Setting the limit to $0^+$ has often eased of my calculation 
Setting $h=-\dfrac1x,$
$$F=\lim_{x\to \infty}(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})$$
$$=\lim_{h\to0^+}\frac{-\sqrt{1-2h}+\sqrt{1+2h}}h$$
$$=\lim_{h\to0^+}\frac{1+2h-(1-2h)}{h(\sqrt{1-2h}+\sqrt{1+2h})}$$
Now cancelling $h$ as $h\ne0$ as $h\to0^+$
$$F=\lim_{h\to0^+}\frac4{\sqrt{1-2h}+\sqrt{1+2h}}$$
$$F=\lim_{h\to0^+}\frac4{\sqrt{1-2\cdot0}+\sqrt{1+2\cdot0}}=?$$
A: Another way using Taylor expansions : rewrite $$\sqrt{x^2+2x}-\sqrt{x^2-2x}=x\Big(\sqrt{1+\frac{2}{x}}-\sqrt{1+\frac{2}{x}}\Big)$$ and use the fact that, for small $y$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Replace $y$ by $\frac{2}{x}$ for the first radical and by  $-\frac{2}{x}$ for the second radical.You so obtain $$\sqrt{x^2+2x}-\sqrt{x^2-2x}\approx x \times \frac{2}{x}=2$$ If you had used one extra term for the series expansion of $\sqrt{1+y}$, you would have arrived to $$\sqrt{x^2+2x}-\sqrt{x^2-2x}\approx2+\frac{1}{x^2}$$ which shows both the limit as well as the manner the limit is approached.
A: $lim_{x \rightarrow \infty} \sqrt{2x + x^2} - \sqrt{-2x +x^2} =$ $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
$= lim_{x \rightarrow \infty} \displaystyle\frac{1}{  \sqrt{(1 + 2/x)}+    \sqrt{(1 - 2/x)}} = 1/2$
A: $\quad \lim_{x\to \infty}(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})$
$=\lim_{x\to \infty}(-\sqrt{-2x+x^2+1}+\sqrt{2x+x^2+1})\\$ (because $\lim_{x\to\infty}\sqrt{N+1}-\sqrt{N}=0$)
$=\lim_{x\to \infty}(-\sqrt{(x-1)^2}+\sqrt{(x+1)^2})$
$=\lim_{x\to \infty}(-{(x-1)}+{(x+1)})=2$
