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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...

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    How did you go from $\sqrt{x^2\pm2x}\to x\sqrt{1\pm\frac{1}{x}}$? – 2017-01-18
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    Using the fact that $a \sqrt b = \sqrt{a^2} \sqrt{b} = \sqrt{a^2 b}$ – 2017-01-18
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    I think you misunderstand me - as you should be getting $$x\sqrt{1\pm\frac{2}{x}}$$ – 2017-01-18
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    @Apeiron It's a bit wrong. $\sqrt{a^2}=|a|$, not just $a$. If $a\ge 0$, then $|a|=a$, if $a<0$, then $|a|=-a$. – 2017-01-18
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    @Chinny84 Ah, that's a typo, that got copied subsequently. Will fix this, thanks :) – 2017-01-18
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    @Chinny84 $-x\sqrt{1\pm\frac{2}{x}}$, because $x<0$, so $\sqrt{x^2}=|x|=-x$. – 2017-01-18
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    @user236182 I was just highlighting the typo of the factor $2$ – 2017-01-18

2 Answers 2

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Your Second attempt is true, but when you bring $x$ from radicals, take a minus to $x$, because $x<0$.

For your first attempt use this: $$L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$ $$ \begin{split} &= \lim_{x \to -\infty} \left(\left(\sqrt{(x+1)^2-1}\right) - \left(\sqrt{(x-1)^2-1}\right)\right)\\ &= \lim_{x \to -\infty} |x+1| - |x-1| \end{split} $$ $$=\lim_{x \to -\infty}\,-x-1+x-1=-2$$

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    And this is because the limit approaches negative infitnity? – 2017-01-18
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    Yes. $\sqrt{x^2+2x}=|x|\sqrt{1+\dfrac1x}=-x\sqrt{1+\dfrac1x}$ since $x<0$ (or $x\to-\infty$) – 2017-01-18
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    where is the mistake in the first one? – 2017-01-18
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    @MyGlasses Ok, clear. What about the first attempt? Is that strategy just doomed? And if so, how do I recognize or see that for future reference? Or is my algebra just wrong? Since it doesn't have a root in the denominator, using the conjugate was just a random strategy, since I didn't know what else to do with it... – 2017-01-18
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    Especially the clarification of the second part was really helpful! – 2017-01-19
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The second is almost good, to mistake from the fact that if $x<0$, $\sqrt{x^2}=-x$.

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    $\sqrt{x^2}=|x|$ for all $x\in\mathbb R$. If $x<0$, then $|x|=-x$. – 2017-01-18
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    :-P tricky little things. – 2017-01-18