<s>[INST] Determine limit: $\lim_{x \to -\infty} (\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) $ Determine the limit of:
$$\lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I've tried a few times, most notably the following two versions. I'm looking for a comment on both, since both amount to a wrong answer.
First attempt
$$L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I thought it'd be nice to get rid of as much ugliness as possible by moving $x$ out of the roots:
$$
\begin{split}
L &= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}}\right)
                              - x\left(\sqrt{1 - \frac{2}{x}}\right)\right)\\
  &= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}} - \sqrt{1 - \frac{2}{x}}\right)\right)
\end{split}
$$
Seems simple enough, since $\frac{1}{x}$ should be an (infinitely) small number I thought it could be discarded with regard to $\sqrt1$, yielding:
$$L = \lim_{x \to -\infty} \left(x\left(\sqrt{1} - \sqrt{1}\right)\right) = 0$$
Wrong, Second attempt
Multply both sides with the conjugate:
$$
\begin{split}
L &= \lim_{x \to -\infty} \sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) \\
  &= \lim_{x \to -\infty} \frac{4x}{(\sqrt{x^2 +2x} + \sqrt{x^2 - 2x})} \\
  &= \lim_{x \to -\infty} \frac{4x}{x\sqrt{1 + \frac{2}{x}}
                                  + x\sqrt{1 - \frac{2}{x}}} \\
  &= \lim_{x \to -\infty} \frac{4}{\sqrt{1 + \frac{2}{x}}
                                 + \sqrt{1 - \frac{2}{x}}}\\
  &= \frac{4}{\sqrt{1} + \sqrt{1}} = \frac{4}{2} = 2
\end{split}
$$
Wrong as well... The answer should be $-2$, but I don't get how to 'get the negative' in...
 [/INST] The second is almost good, to mistake from the fact that if $x<0$, $\sqrt{x^2}=-x$.
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