Solve $\lim_{x\to 0} \frac{\sin x-x}{x^3}$

Question

I'm trying to solve this limit

$$\lim_{x\to 0} \frac{\sin x-x}{x^3}$$

Solving using L'hopital rule, we have:

$$\lim_{x\to 0} \frac{\sin x-x}{x^3}= \lim_{x\to 0} \frac{\cos x-1}{3x^2}=\lim_{x\to 0} \frac{-\sin x}{6x}=\lim_{x\to 0} \frac{-\cos x}{6}=-\frac{1}{6}.$$

Am I right?

I'm trying to solve this using change of variables, I need help.

Thanks

EDIT

I didn't understand the answer and the commentaries, I'm looking for an answer using change of variables.

Is it right using L'hôpital rule several times? In page 109 of Rudin's mathematical analysis book he says in order to use this rule we have to have $f'(x)/g'(x)\to A$. — user42912, Jul 04 '14 at 04:55
@Fabien really interesting, but can we take the limit in an infinite sum? — user42912, Jul 04 '14 at 04:59
@user42912, more than taking the limit, you can evaluate in $0$. — Fabien, Jul 04 '14 at 05:47
@Fabien yes, I know this is true if this sum is finite, but why this is true in this infinite sum. — user42912, Jul 04 '14 at 06:41
Related: http://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3/ — Hans Lundmark, Jul 04 '14 at 09:29

score 10 · Accepted Answer · answered Jul 04 '14 at 07:16

I suppose the below counts as a change of variable.

Assuming that the limit exists, then you can compute the limit as follows:

Replace $x$ by $3x$, then the limit (say $L$) is

$$L = \lim_{x\to 0}\frac{\sin 3x - 3x}{27x^3} = \lim_{x\to 0}\frac{3\sin x - 3x - 4\sin^3 x}{27x^3} = $$ $$\lim_{x\to 0}\frac{1}{9}\left(\frac{\sin x - x}{x^3}\right) - \lim_{x\to 0}\frac{4}{27}\left(\frac{\sin^3 x}{x^3}\right)$$

(we used the formula $\sin 3x = 3\sin x - 4 \sin^3 x$).

Thus we get

$$L = \frac{L}{9} - \frac{4}{27} \implies L = -\frac{1}{6}$$

Of course, we still need to prove that the limit exists.

It's EXACTLY I was looking for, thank you very much – user42912 Jul 04 '14 at 07:30 — user42912, Jul 04 '14 at 07:30

Solve $\lim_{x\to 0} \frac{\sin x-x}{x^3}$

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