## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}$
## Solution

Substitution:

$x=y^{12} \Rightarrow y=\sqrt[12]{x}$

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

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y^{12}}-1}{\sqrt[4]{y^{12}}-1}= \\
& =\lim _{y \rightarrow 0} \frac{y^{4}-1}{y^{3}-1}=\lim _{y \rightarrow 0} \frac{\left(y^{2}-1\right)\left(y^{2}+1\right)}{(y-1)\left(y^{2}+y+1\right)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{(y-1)(y+1)\left(y^{2}+1\right)}{(y-1)\left(y^{2}+y+1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{(y+1)\left(y^{2}+1\right)}{\left(y^{2}+y+1\right)}=\frac{(1+1)\left(1^{2}+1\right)}{\left(1^{2}+1+1\right)}= \\
& =\frac{2 \cdot 2}{3}=\frac{4}{3}
\end{aligned}
$$

## Problem Kuznetsov Limits 13-22