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

Question

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

How do I go about doing this? I can see no simple way of doing this. Application of l'Hopital's rule would be very laborious. A Taylor expansion seems feasible but is that the best way? It seems like it may be also very laborious.

Neither of those approaches is laborious at all. – May 31 '15 at 21:34 — , May 31 '15 at 21:34

score 3 · Accepted Answer · answered May 31 '15 at 21:33

Taylor expanding we find that: \begin{equation} \begin{aligned} \lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}&=\lim_{x\to 0}\frac{x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)-x\left[1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right]}{x^3}\\ &=\lim_{x\to 0}\frac{x^3\left(\frac{1}{2}-\frac{1}{6}\right)+\mathcal{O}\left(x^4\right)}{x^3}\\ &=\frac{1}{3} \end{aligned} \end{equation}

score 3 · Answer 2 · edited May 31 '15 at 21:42

3

Use l'Hopital, not laborious at all..

$$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3} = \lim_{x\to 0}\frac{\cos x-(\cos x - xsinx)}{3x^2} = \lim_{x\to 0}\frac{sinx}{3x} = \frac{1}{3} $$

edited May 31 '15 at 21:42

user26486

11,331

answered May 31 '15 at 21:35

alonkol

299

$\sin(0)=0$ though? How have you got $\frac13$? – RobChem May 31 '15 at 21:51
@RobChem You should be aware of $\lim_{x\to 0}\frac{\sin x}{x}=1$, from which $\lim_{x\to 0}\frac{\frac{\sin x}{x}}{3}=\frac{1}{3}$. For a proof see here (do not use L'Hopital to prove it, since to prove $\sin' x=\cos x$ you need the limit and it is circular reasoning). – user26486 May 31 '15 at 22:03

score 1 · Answer 3 · answered May 31 '15 at 21:36

I may use classic limits (which can easily be computed by hospital's rule (twice) \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2} \end{eqnarray*} The idea is to write the original expression using those above, \begin{equation*} \frac{\sin x-x\cos x}{x^{3}}=\frac{\sin x-x+x-x\cos x}{x^{3}}=\frac{\sin x-x% }{x^{3}}+\frac{1-\cos x}{x^{2}} \end{equation*} It follows that \begin{equation*} \lim_{x\rightarrow 0}\frac{\sin x-x\cos x}{x^{3}}=\lim_{x\rightarrow 0}\frac{% \sin x-x}{x^{3}}+\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=-\frac{1}{6}+% \frac{1}{2}=\frac{1}{3}. \end{equation*}

score 1 · Answer 4 · edited May 31 '15 at 22:43

1

Taylor series

$$ \sin x = x - \frac 16 x^3 + O(x^5)$$ $$ x \cos x = x - \frac 12 x^3 + O(x^5)$$

so

$$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3} = \lim_{x\to 0} \left(\frac 13 + O(x^2)\right) = \frac 13$$

edited May 31 '15 at 22:43

user26486

11,331

answered May 31 '15 at 21:36

WW1

10,497

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

4 Answers4