Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$

Question

Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):

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

Answer 1

Letting $y = x^3$, note that your limit is the same as $$ \lim_{y \rightarrow 0} \frac{\sin(y^{1 \over 3}) - y^{1 \over 3}\cos(y^{1 \over 3})}{y} $$ If you let the mean value theorem be one of your standard rules, then you can say that ${\displaystyle \frac{\sin(y^{1 \over 3}) - y^{1 \over 3}\cos(y^{1 \over 3})}{y} = f'(z)}$ for some $z$ between $0$ and $y$. Calculating the derivative reveals that $$ f'(z) = {1 \over 3}{\sin(z^{1 \over 3}) \over z^{1 \over 3}} $$ Observe that as $y$ goes to zero, so does $z$, and therefore so does $z^{1 \over 3}$. Using that ${\displaystyle \lim_{x \rightarrow 0} {\sin(x) \over x} = 1}$ one sees therefore that $$ \lim_{y \rightarrow 0} f'(z) = {1 \over 3} $$ So ${1 \over 3}$ is the limit.

Answer 2

Use that $$\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots$$ and that $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots$$ Hence $$\frac{\sin(x)-x\cos(x)}{x^3}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}- x+\frac{x^3}{2}- \frac {x^5}{4!} \pm \dots}{x^3}=\frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}\pm \dots$$ Hence the Limit of $$\lim_{x\to 0} \frac{\sin(x)-x\cos(x)}{x^3} =\lim_{x\to 0} \frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}=\frac{1}{3}$$

Answer 3

Even though I like the approach of Dominic Michaelis, here is another one using infinite products.

First note that \begin{align} \sin(x) &= x \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2}) \\ &= x(1-\sum_{n=1}^\infty \frac{x^2}{\pi^2 n^2} + O(x^4)) \end{align} and \begin{align} \cos(x) &= \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2(n-1/2)^2}) \\ \Leftrightarrow x\cos(x) &= x\prod_{n=1}^\infty (1-\frac{x^2}{\pi^2(n-1/2)^2}) \\ &= x(1-\sum_{n=1}^\infty \frac{x^2}{\pi^2 (n-1/2)^2} + O(x^4)) \end{align} Therefore \begin{align} \frac{\sin(x)-x\cos(x)}{x^3}&= \frac{1}{x^2}\Bigl[\sum_{n=1}^\infty \frac{x^2}{\pi^2 (n-1/2)^2} - \sum_{n=1}^\infty \frac{x^2}{\pi^2 n^2} +O(x^4) \Bigr] \\ &=\sum_{n=1}^\infty \frac{1}{\pi^2 (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{\pi^2 n^2} +O(x^2) \\ &=\frac{1}{\pi^2} \Bigl[\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{n^2} \Bigr]+O(x^2) \end{align} Together with the fact that \begin{align} &\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} =\frac{\pi^2}{2} \quad \text{and} \\ &\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \end{align} we get \begin{align} \frac{\sin(x)-x\cos(x)}{x^3} &=\frac{1}{\pi^2} \Bigl[\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{n^2} \Bigr]+O(x^2) \\ &= \frac{1}{\pi^2} (\frac{\pi^2}{2}-\frac{\pi^2}{6}) +O(x^2) \\ &= \frac{1}{3} + O(x^2) = 0 \quad \text{for } x\rightarrow 0 \end{align} Note: I am aware, that this is not as simple as requested by Maths'problem and even more complex than the answer of Dominic Michaelis. We also need to know the infinite products and the identities involving $\sum 1/n^2$ and $\sum 1/(n-1/2)^2$. However, I just wanted to provide another method to solve the problem.