$\lim_{x\to\infty}\frac{x-\sin x}{x^3}=?$

Question

My attempt:First I argue that $-1\leq \sin x\leq 1$,therefore $$\lim_{x\to\infty}\frac{x-\sin x}{x^3}=0$$

But from series expansion I think this limit is not zero.

Please help me to answer this problem.

Thanks.

@ViníciusNovelli what is problem with series expansion?May be because the terms are alternating in sign. — user163993, Oct 11 '14 at 03:54
You are probably using a series expansion around 0, but such an expansion is only valid when considering values close to 0. To analyze the behavior of a function as x goes to infinity, you should use an asymptotic series expansion, i.e. a series expansion "at infinity." — ajd, Oct 11 '14 at 04:14
Are you really considering the limit to $\infty$ (which is zero as you said). Or do you want the limit as $x \to 0$ in which case you could do worse than look here: http://math.stackexchange.com/questions/400541/find-the-limit-without-use-of-lh%C3%B4pital-or-taylor-series-lim-limits-x-right/579503#579503 — user164587, Oct 11 '14 at 04:22
...or here: http://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3/158134#158134 — Hans Lundmark, Oct 11 '14 at 09:30

score 2 · Answer 1 · answered Oct 11 '14 at 09:21

The easier formalization can be done, I think, using the squeeze theorem:

$$-1\le\sin x\le 1\iff1\ge-\sin x\ge -1\implies$$

$$\implies 0\xleftarrow[\infty\leftarrow x ]{} \frac{x+1}{x^3}\ge\frac{x-\sin x}{x^3}\ge\frac{x-1}{x^3}\xrightarrow[x\to\infty]{}0$$

1 Answers1