Problem 1. Calculate $\lim _{x \rightarrow 0} \frac{(\cos x)^{\cos x}-\cos 2 x}{x^{2}}$.
Solution. For any $x$ with $0<|x|<\frac{\pi}{2}$, we have:

Florian Dumitrel

$$
\begin{aligned}
\frac{(\cos x)^{\cos x}-\cos 2 x}{x^{2}} & =\frac{e^{\cos x \cdot \ln (\cos x)}-1}{x^{2}}+\frac{1-\cos 2 x}{x^{2}}= \\
& =\cos x \cdot \frac{e^{\cos x \cdot \ln (\cos x)}-1}{\cos x \cdot \ln (\cos x)} \cdot \ln (\cos x)^{\frac{1}{x^{2}}}+\frac{1-\cos 2 x}{x^{2}}
\end{aligned}
$$

Since

$$
\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^{2}}=2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2}=2, \quad \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x^{2}}}=e^{-\frac{1}{2}}
$$

and

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

it follows that $\lim _{x \rightarrow 0} \frac{(\cos x)^{\cos x}-\cos x}{x^{2}}=\frac{3}{2}$.