A calculus problem by Shubhamkar Ayare

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 0} \frac 1{x^2} \left(\frac {\sin x}x - \cos x \right) \\ & = \lim_{x \to 0} \frac {\sin x - x \cos x}{x^3} & \small {\color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0} \frac {\cos x - \cos x + x \sin x}{3x^2} & \small {\color{#3D99F6} \text{Differentiate up and down.}} \\ & = \lim_{x \to 0} \frac {\sin x}{3x} & \small {\color{#3D99F6}\text{A 0/0 case, differentiate again.}} \\ & = \lim_{x \to 0} \frac {\cos x}3 \\ & = \boxed{\dfrac 13} \end{aligned}$

$sinx = x - \frac{x^3}{3!} + \frac{x^5}{5!} + ...$

Therefore, $\frac{sinx}{x} = 1 - \frac{x^2}{6} + ...$

Also, $cosx = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + ...$

Therefore, $\lim_{x\to0} \frac{1}{x^2}(\frac{sinx}{x} - cosx)$

$= \lim_{x\to0} \frac{1}{x^2}((1 - \frac{x^2}{6}) - (1 - \frac{x^2}{2}))$ .......... (Higher order terms may be neglected with respect to $x^2$ term, which has a non-zero coefficient.)

$= \lim_{x\to0} \frac{1}{x^2} \frac{x^2}{3}$

$= \frac{1}{3}$

Edit: I had made this problem to illustrate that ' $\lim_{x\to0} \frac{sinx}{x}$ is not always 1'. In this regard, please see the comment by Calvin Lin. (I still need to grok it.)