## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\tan(x+1)}{e^{\sqrt[3]{x^{3}-4 x^{2}+6}}-e}$
## Solution

Substitution:

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

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

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\tan(x+1)}{e^{\sqrt[3]{x^{3}-4 x^{2}+6}}-e}=\lim _{y \rightarrow 0} \frac{\tan((y-1)+1)}{e^{\sqrt[3]{(y-1)^{3}-4(y-1)^{2}+6}}-e}= \\
& =\lim _{y \rightarrow 0} \frac{\tan y}{e^{\sqrt[3]{y^{3}-3 y^{2}+3 y-1-4 y^{2}+8 y-4+6}}-e}= \\
& =\lim _{y \rightarrow 0} \frac{\tan y}{e^{\sqrt[3]{y^{3}-7 y^{2}+11 y+1}}-e}= \\
& =\lim _{y \rightarrow 0} \frac{\tan y}{e\left(e^{\sqrt[3]{y^{3}-7 y^{2}+11 y+1}-1}-1\right)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \tan y \sim y \\
& e^{\sqrt[3]{y^{3}-7 y^{2}+11 y+1}-1}-1 \sim \sqrt[3]{y^{3}-7 y^{2}+11 y+1}-1 \\
& y \rightarrow 0\left(\sqrt[3]{y^{3}-7 y^{2}+11 y+1}-1 \rightarrow 0\right)
\end{aligned}
$$

## We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{y}{e\left(\sqrt[3]{y^{3}-7 y^{2}+11 y+1}-1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y\left(\sqrt[3]{\left(y^{3}-7 y^{2}+11 y+1\right)^{2}}+\sqrt[3]{y^{3}-7 y^{2}+11 y+1}+1\right)}{e\left(\sqrt[3]{y^{3}-7 y^{2}+11 y+1}-1\right)\left(\sqrt[3]{\left(y^{3}-7 y^{2}+11 y+1\right)^{2}}+\sqrt[3]{y^{3}-7 y^{2}+11 y+1}+1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y\left(\sqrt[3]{\left(y^{3}-7 y^{2}+11 y+1\right)^{2}}+\sqrt[3]{y^{3}-7 y^{2}+11 y+1}+1\right)}{e\left(y^{3}-7 y^{2}+11 y+1-1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y\left(\sqrt[3]{\left(y^{3}-7 y^{2}+11 y+1\right)^{2}}+\sqrt[3]{y^{3}-7 y^{2}+11 y+1}+1\right)}{e\left(y^{3}-7 y^{2}+11 y\right)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\sqrt[3]{\left(y^{3}-7 y^{2}+11 y+1\right)^{2}}+\sqrt[3]{y^{3}-7 y^{2}+11 y+1}+1}{e\left(y^{2}-7 y+11\right)}= \\
& =\frac{\sqrt[3]{\left(0^{3}-7 \cdot 0^{2}+11 \cdot 0+1\right)^{2}}+\sqrt[3]{0^{3}-7 \cdot 0^{2}+11 \cdot 0+1}+1}{e\left(0^{2}-7 \cdot 0+11\right)}= \\
& =\frac{\sqrt[3]{1^{2}}+\sqrt[3]{1}+1}{11 e}=\frac{3}{11 e}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-24