Evaluating $\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}$ which step is wrong?

Question

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!

Answer 1

What you did wrong:

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

Answer 2

What you got wrong is in the step "I simply divide by $x$", where you should get $$ \frac{-2x}{x - \sqrt{x^2+2x}}= \frac{-2}{1-\frac{1}{x}\sqrt{x^2+2x}} =\frac{-2}{1{\color{red}+}\sqrt{\frac{x^2+2x}{x^2}}} =\frac{-2}{1{\color{red}+}\sqrt{1+\frac{2}{x}}} $$

Answer 3

To avoid the trap,

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

Then

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

Answer 4

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

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