Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^2}\sqrt{1-\frac{2}{x}}+x}{-2x}$
$\lim \limits_{x\to +\infty} \frac{\sqrt{1-\frac{2}{x}}+1}{-2} = -1 $
The next exercise is $\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$
I thought that I could solve this one in the same way by:
$\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} $
$\lim \limits_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}$
$\lim \limits_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} = 1 $
But apparently, the answer is $0$...
Why can't the second one be solved in the same way as the first?
And why is the answer $0$?

When $x\rightarrow-\infty$, you don't have $0$ as denominator, but $+\infty$, because when $x \rightarrow-\infty, \sqrt{x^2 -2x} \sim |x|$ and, as $x < 0$, $|x| = -x$, so you have $\lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0$.
