Calculus: L'Hopital's Rule Indeterminate Limit

$\lim_{x \to 0} \frac{\sin x - x}{x \sin x}$ As $x \to 0$, numerator $\sin 0 - 0 = 0$ and denominator $0 \cdot 0 = 0$. This is in $\frac{0}{0}$ indeterminate form.

Differentiate 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 $1 - 1 = 0$, denominator is $0 + 0 = 0$. Still in $\frac{0}{0}$ form.

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