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 $0$?
 A: 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 $\lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0$.
A: We already have answers showing correct ways to solve the problem.
What I found interesting was the implied question:
where is the error in the steps taken in the body of the question,
which appeared to show that the limit is $1$?
The steps in question are
\begin{align}
\newcommand{\?}{\stackrel{?}{=}}
\lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} 
  &\? \lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}
                          \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} \tag1\\
  &\? \lim_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x} \tag2\\
  &\? \lim_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} \tag3\\
  &\? 1 \tag4
\end{align}
The error is in the third step (from line $2$ to line $3$).
In fact, if $x < 0,$ then $\sqrt{x^2} = -x,$ so
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x} 
 &= \lim_{x\to -\infty} \frac{-x\sqrt{1+\frac{2}{x}}+x}{2x} \\
  &= \lim_{x\to -\infty} \frac{-\sqrt{1+\frac{2}{x}}+1}{2} \\
  &\neq \lim_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2}
\end{align}
One way to avoid errors of this sort (other than constant vigilance
when manipulating square roots of expressions over negative-valued variables)
is to begin by observing that
$$
\lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x} - x} 
 = \lim_{y\to +\infty} \frac{1}{\sqrt{y^2-2y} + y}.
$$
One can then work on the limit over $y,$ confident in the knowledge that
$\sqrt{y^2}/y = 1.$
A: Let  $-1/x=h$
$x^2+2x=\dfrac{1-2h}{h^2}\implies\sqrt{x^2+2x}=\dfrac{\sqrt{1-2h}}{\sqrt{h^2}}=\dfrac{\sqrt{1-2h}}{|h|}$
As $h\to0^+, h>0,|h|=+h$
$$\lim_{x\to-\infty}\dfrac1{\sqrt{x^2+2x}-x}=\lim_{h\to0^+}\dfrac h{\sqrt{1-2h}+1}=?$$
