6.59. $\lim _{x \rightarrow 0}(1-\cos x)^{x}$.
Solution.

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

## Find separately

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

Here, the first remarkable limit was used

$$
\lim _{\alpha \rightarrow 0} \frac{\alpha}{\sin \alpha}=1
$$

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

## INDETERMINACY OF THE FORM « 1 *