## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Using the substitution of equivalent infinitesimals:

$$
\operatorname{arctg} \sqrt{x} \sim \sqrt{x} \text{ as } x \rightarrow 0(\sqrt{x} \rightarrow 0)
$$

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{-2 \ln 3 \cdot(\sqrt{x})^{2}}{x}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{-2 \ln 3 \cdot x}{x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-2 \ln 3\right\}=\exp \{-2 \ln 3\}= \\
& =\exp \left\{\ln 3^{-2}\right\}=e^{\ln 3^{-2}}=3^{-2}=\frac{1}{9}
\end{aligned}
$$

## Problem Kuznetsov Limits 17-4