Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 26

Determine the $\displaystyle \lim_{x \to 0} \frac{\sin x - x}{x^3}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.
$\displaystyle \lim_{x \to 0 } \frac{\sin x - x}{x^3} = \frac{\sin 0 - 0}{0^3} = \frac{0}{0} \text{ Indeterminate}$

Thus, by applying L'Hospitals rule,
$\displaystyle \lim_{x \to 0} \frac{\sin x - x}{x^3} = \lim_{x \to 0} \frac{\cos x - 1}{3x^2}$
If we evaluate the limit, we will still get an indeterminate form, hence, we need to apply L'Hospitals Rule once more, so...

$\displaystyle \lim_{x \to 0} \frac{\cos x - 1}{3x^2} = \lim_{x \to 0} \frac{(-\sin x)-0}{6x} = \lim_{x \to 0} \frac{-\sin x}{6x}$

Again, by applying L'Hospital's Rule for the third time, since we still get indeterminate form.

$\begin{equation} \begin{aligned} \lim_{x \to 0} - \frac{\sin x}{6x} &= \lim_{x \to 0} - \frac{\cos x}{6}\\ \\ &= \frac{-\cos(0)}{6}\\ \\ &= \frac{-(1)}{6}\\ \\ &= \frac{-1}{6} \end{aligned} \end{equation}$