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

Question

How to calculate this limit:

$$\lim_{x\rightarrow 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right)$$

All I know is:

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$

$$\lim_{x\rightarrow 0} \, x = 0$$

$$\lim_{x\rightarrow 0} \,\sin x = 0$$

Answer 1

Hint: Write it as $$\lim_{x\to0}\dfrac{\sin x-x}{x\sin x},$$ and apply L'Hôpital's rule twice.

Answer 2

Hint: $\dfrac{1}{x} - \dfrac{1}{\sin x} = \dfrac{\sin x - x}{x\sin x}$, and use L'hopitale rule !