## Task 3/70

It is $x^{0}=1$ and $0^{x}=0$ for $x \neq 0$. What is the value of $\lim _{x \rightarrow 0} x^{x}$ ?
It is expedient to calculate the natural logarithm of the limit. Once this is found, the limit itself can be determined. It is

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

The limit is subject to L'Hôpital's rule, as $\lim _{x \rightarrow 0} \ln x=-\lim _{x \rightarrow 0} x^{-1}=-\infty$:

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

Thus, the limit $\lim _{x \rightarrow 0} x^{x}=e^{0}=1$.