Rewriting an infinity-minus-infinity limit and applying L'Hopital's Rule

First combine the fractions over a common denominator: $\frac{1}{x} - \frac{1}{\sin(x)} = \frac{\sin(x) - x}{x\sin(x)}.$ As $x \to 0^+$, both the numerator $\sin(x)-x \to 0$ and the denominator $x\sin(x) \to 0$, so this is now the indeterminate form $0/0$, and L'Hopital's Rule applies.

Differentiating numerator and denominator separately gives $\lim_{x\to 0^+} \frac{\cos(x) - 1}{\sin(x) + x\cos(x)}.$ Substituting $x=0$ again gives $0/0$, so we apply L'Hopital's Rule a second time: differentiating again gives $\lim_{x\to 0^+} \frac{-\sin(x)}{2\cos(x) - x\sin(x)}.$ Now substituting $x=0$ gives $\frac{0}{2} = 0$, which is no longer indeterminate, so the limit equals $0$.