## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{1+\cos (x-\pi)}{\left(e^{3 x}-1\right)^{2}}$
Solution

$\lim _{x \rightarrow 0} \frac{1+\cos (x-\pi)}{\left(e^{3 x}-1\right)^{2}}=\lim _{x \rightarrow 0} \frac{1+\cos (\pi-x)}{\left(e^{3 x}-1\right)^{2}}=$

Using the reduction formula:

$=\lim _{x \rightarrow 0} \frac{1-\cos x}{\left(e^{3 x}-1\right)^{2}}=$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& 1-\cos x \sim \frac{x^{2}}{2} \\
& e^{3 x}-1 \sim 3 x, \text { as } x \rightarrow 0 \\
& \text {, as } x \rightarrow 0(3 x \rightarrow 0)
\end{aligned}
$$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\cos x}{\left(e^{3 x}-1\right)^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\frac{x^{2}}{2}}{(3 x)^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{1}{2 \cdot 3^{2}}=\frac{1}{18}
\end{aligned}
$$

## Problem Kuznetsov Limits 12-24