Limit of a difference of reciprocals via L'Hopital's Rule

First combine the two terms into a single fraction: $\frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x\sin x}.$ Substituting $x=0$ into numerator and denominator gives $\sin(0)-0=0$ and $0\cdot\sin(0)=0$, so this is the indeterminate form $0/0$, and L'Hopital's Rule applies.

Differentiate numerator and denominator separately: the derivative of $\sin x - x$ is $\cos x - 1$, and the derivative of $x\sin x$ is $\sin x + x\cos x$ (product rule). Substituting $x=0$ gives $\cos(0)-1 = 0$ and $\sin(0)+0\cdot\cos(0)=0$, so it is again $0/0$, and L'Hopital's Rule must be applied a second time.

Differentiate again: the derivative of $\cos x - 1$ is $-\sin x$, and the derivative of $\sin x + x\cos x$ is $\cos x + \cos x - x\sin x = 2\cos x - x\sin x$. Now substituting $x=0$ gives $\frac{-\sin(0)}{2\cos(0)-0} = \frac{0}{2} = 0$, which is no longer indeterminate.

Therefore $\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right) = 0.$