What is the limit?

The initial limit is of the indeterminate form $\frac{0}{0}$ , so we can apply L'Hôpital's Rule:

$\begin{aligned} \lim_{x \to 0} \dfrac{\sin x - x \cos x}{x^3} &= \lim_{x \to 0} \dfrac{x \sin x}{3 x^2} \\ \\ &= \lim_{x \to 0} \dfrac{\sin x}{3 x} \\ \\ &= \dfrac{1}{3} \ \lim_{x \to 0} \dfrac{\sin x}{x} & \small \color{#3D99F6}{\text{[ Now use the fact that } \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \ ]} \\ \\ &= \boxed{\dfrac{1}{3}} \end{aligned}$

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin x - x\cos x}{x^3} & \small \color{#3D99F6} \text{Using Maclaurin series} \\ & = \lim_{x \to 0} \frac {\left(x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \cdots \right) - x \left(1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!} + \cdots \right)}{x^3} \\ & = \lim_{x \to 0} \frac {\frac {x^3}3 - \frac {x^5}{30} + \frac {x^7}{840} - \cdots}{x^3} & \small \color{#3D99F6} \text{Divide up and down by }x^3 \\ & = \lim_{x \to 0} \frac 13 - \frac {x^2}{30} + \frac {x^5}{840} - \cdots \\ & = \boxed{\dfrac 13} \end{aligned}$

Given that $L = \lim_{x\to 0} \dfrac{\sin x - x\cos x }{x^3}$ Also we have $\sin x =x-\dfrac{x^3}{3!} +\dfrac{x^5}{5!} +\cdots\phantom{cccc} \cos x = 1- \dfrac{x^2}{2!} +\dfrac{x^4}{4!} +\cdots \\ \therefore \sin x - x\cos x =x-\dfrac{x^3}{3!} +\dfrac{x^5}{5!} +\cdots - x\left(1- \dfrac{x^2}{2!} +\dfrac{x^4}{4!} +\cdots \right)\\ \sin x - x\cos x = \dfrac{x^3}{3!} +\dfrac{x^5}{5!} +\cdots +\dfrac{x^3}{2!} -\dfrac{x^4}{3!} +\cdots = x^3\left(\dfrac{1}{3!} + \dfrac{x^2}{5!} +\cdots+ \dfrac{1!}{2!} -\dfrac{x}{3!} +\cdots \right) \implies \\ \lim_{x\to 0} \dfrac{ x^3\left(\dfrac{1}{3!} + \dfrac{x^2}{5!} +\cdots+ \dfrac{1!}{2!} -\dfrac{x}{3!} +\cdots \right) }{x^3} = \dfrac{1}{6} +\dfrac{1}{2} =\dfrac{1}{3}$

Both denominator and numerator of the fraction go to zero for $x \to 0$ . Therefore, the rule of l'Hospital can be applied $\begin{aligned} \lim_{x \to 0} \frac{\sin x - x \cos x}{x^3} &= \lim_{x \to 0} \frac{(\sin x - x \cos x)'}{(x^3)'} & |& \text{l'Hospital}\\ &= \lim_{x \to 0} \frac{x \sin x}{3 x^2} & |& \text{still } \tfrac{0}{0} \\ &= \lim_{x \to 0} \frac{(x \sin x)'}{(3 x^2)'} & |& \text{l'Hospital}\\ &= \lim_{x \to 0} \frac{\sin x + x \cos x}{6 x} & |& \text{still } \tfrac{0}{0} \\ &= \lim_{x \to 0} \frac{(\sin x + x \cos x)'}{(6 x)'} & |& \text{l'Hospital}\\ &= \lim_{x \to 0} \frac{2 \cos x - x \sin x}{6} \\ &= \frac{1}{3} \end{aligned}$