## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(3-\frac{2}{\cos x}\right)^{\operatorname{cosec}^{2} x}
$$
## Solution

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

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\left(2-\frac{2}{\cos x}\right)\right) \sim 2-\frac{2}{\cos x}$, as $\quad x \rightarrow 0\left(2-\frac{2}{\cos x} \rightarrow 0\right)$ $\sin x \sim x$, as $x \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_f6f54646985984cd50ccg-12.jpg?height=155&width=588&top_left_y=1267&top_left_x=153)

We get:

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

## Problem Kuznetsov Limits 17-22