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

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

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.

Thanks.

• Your argument works, the limit is $0$. – Vinícius Novelli Oct 11 '14 at 3:47
• @ViníciusNovelli what is problem with series expansion?May be because the terms are alternating in sign. – user163993 Oct 11 '14 at 3:54
• 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 4:14
• 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: math.stackexchange.com/questions/400541/… – user164587 Oct 11 '14 at 4:22
• ...or here: math.stackexchange.com/questions/157903/… – Hans Lundmark Oct 11 '14 at 9:30

$$-1\le\sin x\le 1\iff1\ge-\sin x\ge -1\implies$$
$$\implies 0\xleftarrow[\infty\leftarrow x ]{} \frac{x+1}{x^3}\ge\frac{x-\sin x}{x^3}\ge\frac{x-1}{x^3}\xrightarrow[x\to\infty]{}0$$