## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 1}\left(\frac{3 x-1}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-1}}
$$
## Solution

$\lim _{x \rightarrow 1}\left(\frac{3 x-1}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{3 x-1}{x+1}\right)}\right)^{\frac{1}{\sqrt[3]{x}-1}}=$

$=\lim _{x \rightarrow 1} e^{\frac{1}{\sqrt[3]{x}-1} \cdot \ln \left(\frac{3 x-1}{x+1}\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{1}{\sqrt[3]{x}-1} \cdot \ln \left(\frac{3 x-1}{x+1}\right)\right\}=$

Substitution:

$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_62fa08db54895d8ac1b3g-12.jpg?height=162&width=1076&top_left_y=735&top_left_x=164)

We get:

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

## Problem Kuznetsov Limits 19-1