When is $\sqrt{x^2+x}-x$=o(1) (little-o notation)?

Question

This is the question: determine when $\sqrt{x^2+x}-x$ is equal to little-$o$ of 1. The options are:

a) $x \to +\infty$

b) $x \to -\infty$

If I understand correctly, $\sqrt{x^2+x}-x = o(1)$ if $$\lim_{x \to x_0}\frac{\sqrt{x^2+x}-x}{1}=0$$ Mutliply top and bottom by $\sqrt{x^2+x}+x$ and then divide top and bottom by x, we get: $$\lim_{x \to x_0}\frac{1}{\frac{x}{|x|}\sqrt{1+\frac{1}{x}}+1}$$

If $x_0=+\infty$ then this limit is equal to $1/2$, and if $x_0=-\infty$, then the limit is equal to $+\infty$.

The answer in the book is supposed to be a) $x\to+\infty$, but I don't see how this is the case. I'm stuck, I'd really appreciate if you guys could help me out!!

Important note: when $x>0,$ $\sqrt{x^2}=x$; when $x<0, $ $\sqrt{x^2}=-x$ — J. W. Tanner, Nov 13 '19 at 02:10
it does; are you sure it's little o notation and not big O notation? — J. W. Tanner, Nov 13 '19 at 02:21
maybe, because I agree with you that $\lim_\limits{x\to+\infty}(\sqrt{x^2+x}-x)=\frac12$ and $\lim_\limits{x\to-\infty}(\sqrt{x^2+x}-x)=\infty$, and $\frac12=O(1)$ though not $o(1)$ — J. W. Tanner, Nov 13 '19 at 02:28

score 0 · Accepted Answer · answered Nov 13 '19 at 02:53

0

It seems there was a typo in the textbook.

$$\frac{1}{2} = O(1)$$ as $x\to+\infty$

(Big-O notation and not little-o)

answered Nov 13 '19 at 02:53

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When is $\sqrt{x^2+x}-x$=o(1) (little-o notation)?

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