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

Question

I can't figure out how to mathematically solve $\lim_{x \to -\infty} (x +\sqrt{x^2+2x})$.

According to Desmos, this limit equals $-1$, and this answer was accepted by the software I'm using.

I have tried:

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

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

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

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

But this doesn't seem to point to $-1$ being the limit.

There are two separate questions, one in the title and one in the question body. — GNUSupporter 8964民主女神地下教會, Feb 26 '19 at 19:59
@GNUSupporter8964民主女神地下教會 My mistake, sorry! — LuminousNutria, Feb 26 '19 at 20:00

score 5 · Accepted Answer · answered Feb 26 '19 at 20:10

The problem happens when you factor $x$ out of the square root. Recall that $$\sqrt{x^2} = |x|,$$ so since $x \to -\infty$, we have $\sqrt{x^2} = -x$. Your last line should therefore be $$=\lim_{x \to -\infty} \frac{-2x}{x + x\sqrt{1+\frac{2}{x}}},$$ which you can see will approach $-1$.

score 2 · Answer 2 · answered Feb 26 '19 at 20:05

2

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

answered Feb 26 '19 at 20:05

Peter Foreman

20,095

I'm not sure what $o(\frac{1}{x^2})$ represents here. Is $o$ a variable? – LuminousNutria Feb 26 '19 at 20:13
1

That's the little-O. – GNUSupporter 8964民主女神地下教會 Feb 26 '19 at 20:13
@GNUSupporter8964民主女神地下教會 Sorry, I don't know what that is. – LuminousNutria Feb 26 '19 at 20:14
@LuminousNutria Little o notation describes asymptotic behaviour. – Théophile Feb 26 '19 at 20:14
1

@LuminousNutria We're here to learn, so feel free to ask. – GNUSupporter 8964民主女神地下教會 Feb 26 '19 at 20:15
@GNUSupporter8964民主女神地下教會 The OP did ask, and you answered that $o$ is a little O. That's not the most helpful reference... – Théophile Feb 26 '19 at 21:11
@Théophile I suppose that a key word would suffice for a Google search, and that we don't need to spoon-feed OP with a specific URL. Nonetheless, thanks for critic. I admit that I can make that clearer by saying "little-O notation". I just wanted to say there's no need to apologize for not knowing anything. – GNUSupporter 8964民主女神地下教會 Feb 26 '19 at 21:17

score 1 · Answer 3 · answered Feb 26 '19 at 20:18

$\lim_{x \rightarrow - \infty} (x + \sqrt{x^2 + 2x})$

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

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

Now use $x = -\sqrt{x^2}$ when $x < 0$ to get:

$=\lim_{x \rightarrow - \infty} (\frac{-2}{1 + {\sqrt{1 + 2/x}}}) = -1$

score 0 · Answer 4 · answered Feb 26 '19 at 20:07

0

it is $$\frac{-2x}{x+x\sqrt{1+\frac{2}{x}}}$$ and this tends to $-1$ for $x$ tends to $-\infty$

answered Feb 26 '19 at 20:07

Dr. Sonnhard Graubner

97,058

Thank you, is there some way I could tell that without plugging-in multiple values for x? – LuminousNutria Feb 26 '19 at 20:10

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

4 Answers4