Example 6. Find $\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right)$.
Solution. When $x \rightarrow \mathbf{0}$, we get an indeterminate form of $\infty-\infty$. We will resolve this indeterminacy by converting it to an indeterminate form of $\frac{0}{0}$ and applying L'Hôpital-Bernoulli's rule,

$$
\begin{gathered}
\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right)=\lim _{x \rightarrow 0} \frac{\sin x-x}{x \sin x}=\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin x+x \cos x}= \\
=\lim _{x \rightarrow 0} \frac{-\sin x}{\cos x+\cos x-x \sin x}=0
\end{gathered}
$$