## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{ctg}^{2} x}$
## Solution

$\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{ctg}^{2} x}=\lim _{x \rightarrow 0}\left(e^{\ln \left(6-\frac{5}{\cos x}\right)}\right)^{\frac{1}{\operatorname{tg}^{2} x}}=$

$$
=\lim _{x \rightarrow 0} e^{\ln \left(6-\frac{5}{\cos x}\right) / \operatorname{tg}^{2} x}=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \left(1+\left(5-\frac{5}{\cos x}\right)\right)}{\operatorname{tg}^{2} x}\right\}=
$$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\left(5-\frac{5}{\cos x}\right)\right) \sim 5-\frac{5}{\cos x}$, as $x \rightarrow 0\left(5-\frac{5}{\cos x} \rightarrow 0\right)$ $\operatorname{tg} x \sim x$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{5-\frac{5}{\cos x}}{x^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{5\left(\frac{\cos x}{\cos x}-\frac{1}{\cos x}\right)}{x^{2}}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{-5(1-\cos x)}{x^{2} \cdot \cos x}\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
1-\cos x \sim \frac{x^{2}}{2}, \text { as } x \rightarrow 0
$$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{-5 \cdot \frac{x^{2}}{2}}{x^{2} \cdot \cos x}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{-5}{2 \cdot \cos x}\right\}= \\
& =\exp \left\{\frac{-5}{2 \cdot \cos 0}\right\}=\exp \left\{\frac{-5}{2}\right\}=e^{-\frac{5}{2}}
\end{aligned}
$$

## Problem Kuznetsov Limits 17-21