## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{1 /(\sqrt[5]{x}-1)}
$$
## Solution

$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{1}{\sqrt[5]{x}-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2 x-1}{x}\right)}\right)^{\frac{1}{\sqrt[5]{x}-1}}=$

$=\lim _{x \rightarrow 1} e^{\frac{1}{\sqrt[5]{x}-1} \cdot \ln \left(\frac{2 x-1}{x}\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{1}{\sqrt[5]{x}-1} \cdot \ln \left(\frac{2 x-1}{x}\right)\right\}=$

Substitution:

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

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

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sqrt[5]{y+1}-1} \cdot \ln \left(\frac{2(y+1)-1}{y+1}\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sqrt[5]{y+1}-1} \cdot \ln \left(\frac{2 y+1}{y+1}\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sqrt[5]{y+1}-1} \cdot \ln \left(1+\frac{y}{y+1}\right)\right\}=
\end{aligned}
$$

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:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y}{(\sqrt[5]{y+1}-1)(y+1)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y\left(\sqrt[5]{(y+1)^{4}}+\sqrt[5]{(y+1)^{3}}+\sqrt[5]{(y+1)^{2}}+\sqrt[5]{y+1}+1\right)}{(\sqrt[5]{y+1}-1)\left(\sqrt[5]{(y+1)^{4}}+\sqrt[5]{(y+1)^{3}}+\sqrt[5]{(y+1)^{2}}+\sqrt[5]{y+1}+1\right)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y\left(\sqrt[5]{(y+1)^{4}}+\sqrt[5]{(y+1)^{3}}+\sqrt[5]{(y+1)^{2}}+\sqrt[5]{y+1}+1\right)}{(y+1-1)(y+1)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y\left(\sqrt[5]{(y+1)^{4}}+\sqrt[5]{(y+1)^{3}}+\sqrt[5]{(y+1)^{2}}+\sqrt[5]{y+1}+1\right)}{y(y+1)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{\sqrt[5]{(y+1)^{4}}+\sqrt[5]{(y+1)^{3}}+\sqrt[5]{(y+1)^{2}}+\sqrt[5]{y+1}+1}{y+1}\right\}= \\
& =\exp \left\{\frac{\sqrt[5]{(0+1)^{4}}+\sqrt[5]{(0+1)^{3}}+\sqrt[5]{(0+1)^{2}}+\sqrt[5]{0+1}+1}{0+1}\right\}=\exp \left\{\frac{1+1+1+1+1}{1}\right\}= \\
& \left.=\exp \{5\}=e^{5}\right\} \\
& \left.=\frac{5}{y}\right\} \\
& =
\end{aligned}
$$

## Problem Kuznetsov Limits 19-7