## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 8}\left(\frac{2 x-7}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-2}}$
## Solution

$\lim _{x \rightarrow 8}\left(\frac{2 x-7}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-2}}=\lim _{x \rightarrow 8}\left(e^{\ln \left(\frac{2 x-7}{x+1}\right)}\right)^{\frac{1}{\sqrt[3]{x}-2}}=$

$=\lim _{x \rightarrow 8} e^{\frac{1}{\sqrt[3]{x}-2} \cdot \ln \left(\frac{2 x-7}{x+1}\right)}=\exp \left\{\lim _{x \rightarrow 8} \frac{1}{\sqrt[3]{x}-2} \cdot \ln \left(\frac{2 x-7}{x+1}\right)\right\}=$

Substitution:

$x=y+8 \Rightarrow y=x-8$

$x \rightarrow 8 \Rightarrow y \rightarrow 0$

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sqrt[3]{y+8}-2} \cdot \ln \left(\frac{2(y+8)-7}{(y+8)+1}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sqrt[3]{y+8}-2} \cdot \ln \left(\frac{2 y+9}{y+9}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sqrt[3]{y+8}-2} \cdot \ln \left(1+\frac{y}{y+9}\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\frac{y}{y+9}\right) \sim \frac{y}{y+9}, \quad y \rightarrow 0\left(\frac{y}{y+9} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y}{(\sqrt[3]{y+8}-2)(y+9)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}{(\sqrt[3]{y+8}-2)\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)(y+9)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}{(y+8-8)(y+9)}\right\}=
\end{aligned}
$$

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}{y(y+9)}\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4}{y+9}\right\}= \\
& =\exp \left\{\frac{\sqrt[3]{(0+8)^{2}}+2 \sqrt[3]{0+8}+4}{0+9}\right\}=\exp \left\{\frac{2^{2}+2 \cdot 2+4}{9}\right\}= \\
& =\exp \left\{\frac{12}{9}\right\}=\exp \left\{\frac{4}{3}\right\}=e^{\frac{4}{3}}
\end{aligned}
$$

## Problem Kuznetsov Limits 19-5