847. Find the limits:
1) $\lim _{x \rightarrow 0} x^{x}$; 2) $\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}$; 3) $\lim _{x \rightarrow 1} x^{\frac{1}{x-1}}$
Solution. In the given examples, we have indeterminate forms of the type $0^{0}, \infty^{0}, 1^{\infty}$.

In each of these cases, the logarithm of the function takes an indeterminate form of the type $0 \cdot \infty$. By resolving this indeterminate form, we find the limit of the logarithm of the function, from which we obtain the limit of the function itself.

1) Let $y=x^{x}$ and take the logarithm of both sides of this equation: $\ln y=x \ln x$. We find

$$
\lim _{x \rightarrow 0} \ln y=\lim _{x \rightarrow 0} x \ln x=\lim _{x \rightarrow 0} \frac{\ln x}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}}=-\lim _{x \rightarrow 0} x=0
$$

hence $\lim _{x \rightarrow 0} y=e^{0}=1$.

2) Let $y=\left(\frac{1}{x}\right)^{\sin x}$ and take the logarithm of both sides of the obtained equation:

$$
\ln y=\sin x \ln \frac{1}{x}=-\sin x \ln x
$$

We find the limit of $\ln y$:

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

Transform the obtained indeterminate form of the type $0 \cdot \infty$ to an indeterminate form of the type $\frac{\infty}{\infty}$, and then apply L'Hôpital's rule:

$$
\begin{aligned}
\lim _{x \rightarrow 0} \ln y & =-\lim _{x \rightarrow 0} \frac{\ln x}{\frac{1}{\sin x}}=-\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{-\frac{\cos x}{\sin ^{2} x}}=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{x \cos x}= \\
& =\lim _{x \rightarrow 0}\left(\frac{1}{\cos x} \cdot \frac{\sin x}{x} \cdot \sin x\right)=1 \cdot 1 \cdot 0=0
\end{aligned}
$$

hence $\lim y=e^{0}=1$.

$$
x \rightarrow 0
$$

3) $y=x^{\frac{1}{x-1}}, \ln y=\frac{1}{x-1} \ln x$

$$
\lim _{x \rightarrow 1} \ln y=\lim _{x \rightarrow 1} \frac{\ln x}{x-1}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{1}=1
$$

hence $\lim _{x \rightarrow 1} y=e$, i.e., $\lim _{x \rightarrow 1} x^{\frac{1}{x-1}}=e$.

Note that L'Hôpital's rule is applicable when the conditions of the theorem are met, and in particular, when the limit of the ratio of the derivatives exists. However, it may happen that the limit of the ratio of the functions exists, but the limit of the ratio of the derivatives does not exist, in which case the indeterminate form must be resolved by another method.

Let's provide examples. 1) $\lim _{x \rightarrow \infty} \frac{x-\sin x}{x+\sin x}=\lim _{x \rightarrow \infty} \frac{1-\frac{\sin x}{x}}{1+\frac{\sin x}{x}}=1$, while the ratio of the derivatives $\frac{1-\cos x}{1+\cos x}$ does not have a limit as $x \rightarrow \infty$. Additionally, the derivative of the denominator $\varphi^{\prime}(x)=1+\cos x$ equals zero at $x=\pi(2 k-1)$ ( $k= \pm 1 ; \pm 2 ; \ldots$), meaning that the derivative $\varphi^{\prime}(x)$ is zero at some points in the interval [c, へ), which also violates the condition of the theorem.
2) $\lim _{x \rightarrow \infty} \frac{x+\sin x}{x}=\lim _{x \rightarrow \infty} \frac{1+\frac{\sin x}{x}}{1}=1$, while the ratio of the derivatives $\frac{1+\cos x}{1}$ does not have a limit as $x \rightarrow \infty$.

Find the limits: