## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{1}{\sqrt[3]{x}-1}}
$$
## Solution

$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{1}{\sqrt[3]{x}-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2 x-1}{x}\right)}\right)^{\frac{1}{\sqrt[3]{x}-1}}=$

$=\lim _{x \rightarrow 1} e^{\frac{1}{\sqrt[3]{x}-1} \cdot \ln \left(\frac{2 x-1}{x}\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{1}{\sqrt[3]{x}-1} \cdot \ln \left(\frac{2 x-1}{x}\right)\right\}=$

## Substitution:

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

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

We get:

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

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

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\frac{y}{y+1}\right) \sim \frac{y}{y+1}_{, \text {as }} y \rightarrow 0\left(\frac{y}{y+1} \rightarrow 0\right)$

We get:

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

## Problem Kuznetsov Limits 19-3