Example 8. Find $\lim _{x \rightarrow 1}(x-1)^{\ln x}$.
Solution. When $x \rightarrow 1$, we have the indeterminate form $0^{0}$. We will use the identity

$$
[f(x)]^{\varphi(x)}=e^{\varphi(x) \ln f(x)}
$$

which in this case will be

$$
(x-1)^{\ln x}=e^{\ln x \cdot \ln (x-1)}
$$

We have

$$
\lim _{x \rightarrow 1}(x-1)^{\ln x}=\lim _{x \rightarrow 1} e^{\ln x \cdot \ln (x-1)}=e^{\lim _{x \rightarrow 1}^{\ln x \cdot \lim _{x \rightarrow 1} \ln (x-1)}}
$$

Now we will find $\lim _{x \rightarrow 1}[\ln x \cdot \ln (x-1)]$. This is an indeterminate form of type $0 \cdot \infty$. Resolving this indeterminate form, we get

$$
\begin{gathered}
\lim _{x \rightarrow 1}[\ln x \cdot \ln (x-1)]=\lim _{x \rightarrow 1} \frac{\ln (x-1)}{\frac{1}{\ln x}}=\lim _{x \rightarrow 1} \frac{\frac{1}{1-x}}{-\frac{1}{x(\ln x)^{2}}}= \\
=-\lim _{x \rightarrow 1} \frac{x(\ln x)^{2}}{x-1}=-\lim _{x \rightarrow 1} \frac{(\ln x)^{2}+x 2 \ln x \cdot \frac{1}{x}}{1}= \\
=\lim _{x \rightarrow 1}\left[(\ln x)^{2}+2 \ln x\right]=0
\end{gathered}
$$

Therefore,

$$
\lim _{x \rightarrow 1}(x-1)^{\ln x}=e^{0}=1
$$