## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1} \frac{\sqrt[3]{1+\ln ^{2} x}-1}{1+\cos \pi x}$
## Solution

Substitution:
$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt[3]{1+\ln ^{2} x}-1}{1+\cos \pi x}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{1+\ln ^{2}(y+1)}-1}{1+\cos \pi(y+1)}= \\
& =\lim _{y \rightarrow 0} \frac{\sqrt[3]{1+\ln ^{2}(y+1)}-1}{1+\cos (\pi y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{\left(\sqrt[3]{1+\ln ^{2}(y+1)}-1\right)\left(\sqrt[3]{\left(1+\ln ^{2}(y+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(y+1)}+1\right)}{(1-\cos \pi y)\left(\sqrt[3]{\left(1+\ln ^{2}(y+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(y+1)}+1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1+\ln ^{2}(y+1)-1}{(1-\cos \pi y)\left(\sqrt[3]{\left(1+\ln ^{2}(y+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(y+1)}+1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln ^{2}(y+1)}{(1-\cos \pi y)\left(\sqrt[3]{\left(1+\ln ^{2}(y+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(y+1)}+1\right)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln (1+y) \sim y_{\text {, as }} y \rightarrow 0$

$\cos \pi y \sim \frac{(\pi y)^{2}}{2}$, as $y \rightarrow 0(\pi y \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{y^{2}}{\frac{(\pi y)^{2}}{2}\left(\sqrt[3]{\left(1+\ln ^{2}(y+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(y+1)}+1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{2}{\pi^{2}\left(\sqrt[3]{\left(1+\ln ^{2}(y+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(y+1)}+1\right)}= \\
& =\frac{2}{\pi^{2}\left(\sqrt[3]{\left(1+\ln ^{2}(0+1)\right)^{2}}+\sqrt[3]{1+\ln ^{2}(0+1)}+1\right)}= \\
& =\frac{2}{\pi^{2}\left(\sqrt[3]{(1+0)^{2}}+\sqrt[3]{1+0}+1\right)}=\frac{2}{3 \pi^{2}}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-15