## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(5-\frac{4}{\cos x}\right)^{\frac{1}{\sin ^{2} 3 x}}$
## Solution

$\lim _{x \rightarrow 0}\left(5-\frac{4}{\cos x}\right)^{\frac{1}{\sin ^{2} 3 x}}=\lim _{x \rightarrow 0}\left(e^{\ln \left(5-\frac{4}{\cos x}\right)}\right)^{\frac{1}{\sin ^{2} 3 x}}=$
$=\lim _{x \rightarrow 0} e^{\frac{\ln \left(5-\frac{4}{\cos x}\right)}{\sin ^{2} 3 x}}=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \left(1+\left(4-\frac{4}{\cos x}\right)\right)}{\sin ^{2} 3 x}\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\left(4-\frac{4}{\cos x}\right)\right) \sim 4-\frac{4}{\cos x}$, as $\quad x \rightarrow 0\left(4-\frac{4}{\cos x} \rightarrow 0\right)$

$\sin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{4-\frac{4}{\cos x}}{(3 x)^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{4\left(\frac{\cos x}{\cos x}-\frac{1}{\cos x}\right)}{9 x^{2}}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{-4(1-\cos x)}{9 x^{2} \cdot \cos x}\right\}=$

Using the substitution of equivalent infinitesimals:

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

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{-4 \cdot \frac{x^{2}}{2}}{9 x^{2} \cdot \cos x}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{-2}{9 \cdot \cos x}\right\}= \\
& =\exp \left\{\frac{-2}{9 \cdot \cos 0}\right\}=\exp \left\{\frac{-2}{9}\right\}=e^{-\frac{2}{9}}
\end{aligned}
$$

## Problem Kuznetsov Limits 17-6