Example 7. Find $\lim _{x \rightarrow 0}[(x-\sin x) \ln x]$.
Solution. Here we have an indeterminate form $0 \cdot \infty$. The given function can be represented as

$$
(x-\sin x) \ln x=\frac{\ln x}{\frac{1}{x-\sin x}}
$$

The resulting indeterminate form $\frac{\infty}{\infty}$ is resolved using L'Hôpital's rule:

$$
\lim _{x \rightarrow 0} \frac{\ln x}{\frac{1}{x-\sin x}}=\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-(1-\cos x)}{(x-\sin x)^{2}}}=\lim _{x \rightarrow 0} \frac{(x-\sin x)^{2}}{x(\cos x-1)}
$$

This is an indeterminate form of $\frac{0}{0}$, so

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(x-\sin x)^{2}}{x(\cos x-1)}=\lim _{x \rightarrow 0} \frac{2(x-\sin x)(1-\cos x)}{\cos x-1-x \sin x}= \\
& =2 \lim _{x \rightarrow 0} \frac{(1-\cos x)^{2}+(x-\sin x) \sin x}{-\sin x-\sin x-x \cos x}= \\
& =2 \lim _{x \rightarrow 0} \frac{1-2 \cos x+\cos ^{2} x+x \sin x-\sin ^{2} x}{-2 \sin x-x \cos x}= \\
& \quad=2 \lim _{x \rightarrow 0} \frac{1-2 \cos x+x \sin x+\cos 2 x}{-2 \sin x-x \cos x}= \\
& =2 \lim _{x \rightarrow 0} \frac{2 \sin x+\sin x+x \cos x-2 \sin 2 x}{-3 \cos x+x \sin x}=0
\end{aligned}
$$

Thus,

$$
\lim _{x \rightarrow 0}(x-\sin x) \ln x=0
$$