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

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. Jul 4, 2014 at 4:52
• You can checked it with Taylor series. Jul 4, 2014 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 $f'(x)/g'(x)\to A$. Jul 4, 2014 at 4:55
• @Fabien really interesting, but can we take the limit in an infinite sum? Jul 4, 2014 at 4:59
• @user42912, more than taking the limit, you can evaluate in $0$. Jul 4, 2014 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 Jul 4, 2014 at 7:30
• Excellent answer. Aug 14, 2021 at 23:11