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

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.

• This is the correct value. – Adam Hughes Jul 4 '14 at 4:52
• You can checked it with Taylor series. – Fabien Jul 4 '14 at 4:54
• 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 4 '14 at 4:55
• @Fabien really interesting, but can we take the limit in an infinite sum? – user42912 Jul 4 '14 at 4:59
• @user42912, more than taking the limit, you can evaluate in $0$. – Fabien Jul 4 '14 at 5:47

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 4 '14 at 7:30

You can also try the Taylor expansion of $\sin x$. $$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots$$ Hence $\sin x-x=-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots$

• Very interesting, but it's not what I'm looking for. Thank you very much. – user42912 Jul 4 '14 at 4:56
• @user42912, then what are you looking for? – IAmNoOne Jul 4 '14 at 4:57
• @Nameless I'm trying to prove this using change of variables as I said in the question – user42912 Jul 4 '14 at 4:58
• Can we take the limit in this infinite sum? – user42912 Jul 4 '14 at 4:58