Calculus: Indeterminate Forms & L'Hopital's Rule

$\lim_{x \to 0} \left( \frac{\sin x - x}{x \sin x} \right)$ As $x \to 0$, numerator $\to 0 - 0 = 0$ and denominator $\to 0 \times 0 = 0$, giving an indeterminate form of $\frac{0}{0}$.

Differentiating numerator and denominator with respect to $x$: $= \lim_{x \to 0} \frac{\cos x - 1}{\sin x + x \cos x}$ At $x = 0$, numerator is $1 - 1 = 0$ and denominator is $0 + 0 = 0$ (still $\frac{0}{0}$).

Differentiating again: $= \lim_{x \to 0} \frac{-\sin x}{\cos x + (\cos x - x \sin x)} = \lim_{x \to 0} \frac{-\sin x}{2\cos x - x \sin x}$ Substituting $x = 0$: $= \frac{-\sin 0}{2\cos 0 - 0 \sin 0} = \frac{0}{2(1) - 0} = \frac{0}{2} = 0$