## Problem Statement

Calculate the limit of the function:

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

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

$\ln \left(1+\left(1-3^{\sin ^{2} x}\right)\right) \sim\left(1-3^{\sin ^{2} x}\right)$, as $x \rightarrow 0\left(\left(1-3^{\sin ^{2} x}\right) \rightarrow 0\right)$ $\ln \left(1-2 \sin ^{2} \frac{x}{2}\right) \sim-2 \sin ^{2} \frac{x}{2}$, as $x \rightarrow 0\left(-2 \sin ^{2} \frac{x}{2} \rightarrow 0\right)$

We get:

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

Using the substitution of equivalent infinitesimals:

$e^{\ln 3 \cdot \sin ^{2} x}-1 \sim \ln 3 \cdot \sin ^{2} x$, as $x \rightarrow 0\left(\ln 3 \cdot \sin ^{2} x \rightarrow 0\right)$

$\sin \frac{x}{2} \sim \frac{x}{2}$, as $x \rightarrow 0\left(\frac{x}{2} \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln 3 \cdot \sin ^{2} x}{2\left(\frac{x}{2}\right)^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{2 \ln 3 \cdot \sin ^{2} x}{x^{2}}\right\}=$

Using the substitution of equivalent infinitesimals:

$\sin x \sim x$, as $x \rightarrow 0$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{2 \ln 3 \cdot x^{2}}{x^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} 2 \ln 3\right\}=$

$=e^{2 \ln 3}=\left(e^{\ln 3}\right)^{2}=3^{2}=9$

## Problem Kuznetsov Limits 17-19