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So I'm trying to evaluate $\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}$

These are my steps:

I first rationalize the expression (square root trick) - $$\lim \limits_{x\to -\infty} \frac{-2x}{x - \sqrt{x^2+2x}}$$ Then I simply divide by $x$ so $$\lim \limits_{x\to -\infty} \frac{-2}{1 - \sqrt{1+\frac{1}{2x}}}$$

Then I get the following by evaluating the limit $$\frac{-2}{1 - \sqrt{1}}$$ which then evaluates to $0$ in the denominator. Would really appreciate some help in understanding what I'm doing wrong here.

Thanks in advance!

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  • $\begingroup$ Haven't figured out what you did wrong, but I'd find the limit of $(x+1)+\sqrt{x^2+2x}$ and then subtract one. Then the same approach becomes clearer. $\endgroup$ Commented Nov 21, 2019 at 20:28
  • $\begingroup$ For negative $x, \sqrt{x^2}=-x$ $\endgroup$ Commented Nov 21, 2019 at 20:30
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    $\begingroup$ $\frac{1}{x}\sqrt{x^2+2x}=-\sqrt{1+2/x}$ when $x<0.$ $\endgroup$ Commented Nov 21, 2019 at 20:32
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    $\begingroup$ The trap is the minus. $\endgroup$
    – user65203
    Commented Nov 21, 2019 at 20:38

4 Answers 4

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What you did wrong:

for $x<0, \sqrt{x^2}=-x$

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  • $\begingroup$ so if I take the square root of 1 at the end, It should instead be 1 + 1 in the denominator? $\endgroup$
    – Eli S.
    Commented Nov 21, 2019 at 20:33
  • $\begingroup$ Yes, that’s correct $\endgroup$ Commented Nov 21, 2019 at 20:34
  • $\begingroup$ Thank you! That helps. $\endgroup$
    – Eli S.
    Commented Nov 21, 2019 at 20:35
  • $\begingroup$ @EliS. But clearly 1 is greater than 0 . Please tell why you would take out $-1$ out of square root if 1 is greater than zero. $\endgroup$ Commented Sep 19, 2022 at 10:57
  • $\begingroup$ @J.W.Tanner But clearly 1 is greater than 0 . Please tell why you would take out $−1$ out of square root if 1 is greater than zero. $\endgroup$ Commented Sep 19, 2022 at 10:58
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What you got wrong is in the step "I simply divide by $x$", where you should get $$ \frac{-2x}{x - \sqrt{x^2+2x}}= \frac{-2}{1-\frac{1}{x}\sqrt{x^2+2x}} =\frac{-2}{1{\color{red}+}\sqrt{\frac{x^2+2x}{x^2}}} =\frac{-2}{1{\color{red}+}\sqrt{1+\frac{2}{x}}} $$

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To avoid the trap,

$$\lim_{x\to-\infty}(\sqrt{x^2+2x}+x)=\lim_{x\to\infty}(\sqrt{x^2-2x}-x).$$

Then

$$\lim_{x\to\infty}(\sqrt{x^2-2x}-x)=\lim_{x\to\infty}\frac{-2x}{\sqrt{x^2-2x}+x}=-1.$$

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Things become clear if we set $x=-\dfrac1h\implies h\to0^+$

$$\sqrt{x^2+2x}=\sqrt{\dfrac{1-2h}{h^2}}=\dfrac{\sqrt{1-2h}}{\sqrt{h^2}}=\dfrac{\sqrt{1-2h}}h$$

as $\sqrt{h^2}=|h|$ which is $+h$ as $h>0$

$$F=\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}=\lim_{ h\to0^+}\dfrac{\sqrt{1-2h}-1}h$$

Now set $\sqrt{1-2h}-1=2u\implies-2h=(2u+1)^2-1=4u(u+1)$

$$F=-2\lim_{u\to0^+}\dfrac{2u}{4u(u+1)}=?$$

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