## Solution:

$$
\lim _{x \rightarrow \pi} \frac{1+\cos x \cos 2 x \cos 4 x \ldots \cos 2^{n} x}{(x-\pi)^{2}} \text { is found in the case of indetermination } \frac{0}{0}
$$

We make the substitution $y=x-\pi, y \rightarrow 0$ and the limit becomes $\lim _{y \rightarrow 0} \frac{1-\cos y \cos 2 y \cos y x \ldots \cos 2^{n} y}{y^{2}} \ldots$ (1p).

At the numerator, we perform a calculation trick:

$$
\begin{aligned}
& \lim _{y \rightarrow 0} \frac{1-\cos y \cos 2 y \cos y x \ldots \cos 2^{n} y}{y^{2}}=\lim _{y \rightarrow 0} \frac{1-\cos y+\cos y-\cos y \cos 2 y \cos y x \ldots \cos 2^{n} y}{y^{2}}= \\
& \lim _{y \rightarrow 0} \frac{1-\cos y}{y^{2}}+\lim _{y \rightarrow 0} \frac{\cos y\left(1-\cos 2 y \cos y x \ldots \cos 2^{n} y\right)}{y^{2}}=\ldots=\sum_{k=0}^{n} \lim _{y \rightarrow 0} \frac{1-\cos 2^{k} y}{y^{2}}= \\
& =\sum_{k=0}^{n} \lim _{y \rightarrow 0} \frac{2 \sin ^{2} 2^{k-1} y}{y^{2}}=\sum_{k=0}^{n} \lim _{y \rightarrow 0} \frac{2 \sin ^{2} 2^{k-1} y}{\left(2^{k-1} y\right)^{2}} \cdot \frac{2^{2 k-2} y^{2}}{y^{2}}=2 \sum_{k=0}^{n} \lim _{y \rightarrow 0} \frac{2^{2 k-2} y^{2}}{y^{2}}=2 \sum_{k=0}^{n} 2^{2(k-1)}= \\
& =\frac{1}{2} \sum_{k=0}^{n} 2^{2 k}=\frac{2^{2 n+2}-1}{2 \cdot 3}=\frac{2^{2 n+2}-1}{6}
\end{aligned}
$$
## Problem 4.

$$
\text { Calculate } \lim _{x \rightarrow \pi} \frac{1+\cos x \cos 2 x \cos 4 x \ldots \cos 2^{n} x}{(x-\pi)^{2}}
$$

Prof. Dicu Florentina, Rm. Vâlcea