$ \lim_{x \to -\infty}{\sqrt{x^2+2x}+x} $

Question

I have to solve the following limit $$ \lim_{x \to -\infty}{\sqrt{x^2+2x}+x} $$ My solution is:

$ \lim\limits_{x \to -\infty}{\sqrt{x^2+2x}+x}= \lim\limits_{x \to -\infty}{x \cdot\left(\sqrt{1+\frac{2}{x}}+1\right)}=- \infty$

while the correct result is $-1$,

but I can't understand where I'm making mistakes.

Hint - try rationalizing the expression first, right now you have $\infty +/- \infty $ in your calculations. — ljaniec, Sep 12 '20 at 11:16

Ralph Clausen · Accepted Answer · 2020-09-12T11:46:14.627

4

See this: $$ \lim_{x \to -\infty} \frac{(\sqrt{x^2 + 2x} + x)(\sqrt{x^2 + 2x} - x)}{(\sqrt{x^2 + 2x} - x)} = \lim_{x \to -\infty} \frac{2x}{\sqrt{x^2 + 2x} - x}$$ $$ = \lim_{x \to -\infty} \frac{2}{-\sqrt{1 + \frac{2}{x}}-1}$$ $$ = \frac{2}{-1-1} = -1 \quad [\text{as }x< 0] $$

edited Sep 12 '20 at 11:46

answered Sep 12 '20 at 11:20

Ralph Clausen

2,341

how did you conclude $\sqrt {1+\frac{2}{x}}$ is $-1$ – Not a Salmon Fish Sep 12 '20 at 11:26
hmm i don't see the connection between $x<0$ and $\sqrt{1} = -1$. – Jamāl Sep 12 '20 at 11:31
if we're working with reals, then square root function should always give a positive result. – Jamāl Sep 12 '20 at 11:32
@ViktorGlombik i think thats because it should be $\lim_{x \to -\infty} \frac{2}{-\sqrt{1 + \frac{2}{x}}-1}$ – Jamāl Sep 12 '20 at 11:44
1

Yeah, thanks guys. It's changed. – Ralph Clausen Sep 12 '20 at 11:47

score 2 · Answer 2 · answered Sep 12 '20 at 11:18

2

$\sqrt {x^{2}+2x} =-x (\sqrt {1+\frac 2 x})$ for $x <0$. Now use the fact that $\sqrt {1+\frac 2 x} \sim 1+\frac 1 x+o(\frac 1 x)$ to see that the limit is $-1$.

answered Sep 12 '20 at 11:18

Kavi Rama Murthy

311,013

$ \lim_{x \to -\infty}{\sqrt{x^2+2x}+x} $

2 Answers2