Example 9. Find $\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{\frac{1}{x}}$.
Solution. Since $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1, \lim _{x \rightarrow 0} \frac{1}{x}=\infty$, we have an indeterminate form of $1^{\infty}$ here.

Taking into account identity (A) (see example 8), using L'Hôpital-Bernoulli's rule, we find

$$
\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{\frac{1}{x}}=\lim _{x \rightarrow 0} e^{\frac{1}{x} \ln \frac{\sin x}{x}}=e^{\lim _{x \rightarrow 0} \frac{\ln \frac{\sin x}{x}}{x}}
$$

We need to find $\lim _{x \rightarrow 0} \frac{\ln \frac{\sin x}{x}}{x}$, which is an indeterminate form of $\frac{0}{0}$:

$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{\ln \frac{\sin x}{x}}{x}=\lim _{x \rightarrow 0} \frac{\frac{x}{\sin x} \cdot \frac{x \cos x-\sin x}{x^{2}}}{1}=\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x \sin x}= \\
=\lim _{x \rightarrow 0} \frac{\cos x-x \sin x-\cos x}{\sin x+x \cos x}=-\lim _{x \rightarrow 0} \frac{x \sin x}{\sin x+x \cos x}= \\
=-\lim _{x \rightarrow 0} \frac{\sin x+x \cos x}{\cos x+\cos x-x \sin x}=0
\end{gathered}
$$

Thus, 104

$$
\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{\frac{1}{x}}=e^{0}=1
$$