## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)^{\frac{1}{x^{3}}}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)^{\frac{1}{x^{3}}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\left.\ln \left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)\right)^{\frac{1}{x^{3}}}}\right. \\
& =\lim _{x \rightarrow 0} e^{\frac{1}{x^{3}} \cdot \ln \left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)}{x^{3}}\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\ln \left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right) \sim \ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x} \quad, \text { as } \quad x \rightarrow 0\left(\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x} \rightarrow 0\right)
$$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}}{x^{3}}\right\}=$

Using the substitution of equivalent infinitesimals:

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

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \frac{1}{3} \cdot(\sqrt{x})^{6}}{x^{3}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \frac{1}{3} \cdot x^{3}}{x^{3}}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \ln \frac{1}{3}\right\}=\exp \left\{\ln \frac{1}{3}\right\}=e^{\ln \frac{1}{3}}=\frac{1}{3}
\end{aligned}
$$

## Problem Kuznetsov Limits 17-25