## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow-2} \frac{\tan \left(e^{x+2}-e^{x^{2}-4}\right)}{\tan x+\tan 2}$
## Solution

Substitution:
$x=y-2 \Rightarrow y=x+2$

$x \rightarrow-2 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\tan\left(e^{x+2}-e^{x^{2}-4}\right)}{\tan x+\tan 2}=\lim _{y \rightarrow 0} \frac{\tan\left(e^{(y-2)+2}-e^{(y-2)^{2}-4}\right)}{\tan(y-2)+\tan 2}= \\
& =\lim _{y \rightarrow 0} \frac{\tan\left(e^{y}-e^{y^{2}-4 y+4-4}\right)}{\left(\frac{\sin ((y-2)+2)}{\cos (y-2) \cos 2}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\cos (y-2) \cdot \cos 2 \cdot \tan\left(e^{y}-e^{y^{2}-4 y}\right)}{\sin y}= \\
& =\lim _{y \rightarrow 0} \frac{\cos (y-2) \cdot \cos 2 \cdot \tan\left(e^{y^{2}-4 y}\left(e^{-y^{2}+5 y}-1\right)\right)}{\sin y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_42745aa634bb73565023g-09.jpg?height=76&width=461&top_left_y=1290&top_left_x=152)

$\tan\left(e^{y^{2}-4 y}\left(e^{-y^{2}+5 y}-1\right)\right) \sim e^{y^{2}-4 y}\left(e^{-y^{2}+5 y}-1\right)_{\text {, as }} y \rightarrow 0\left(e^{y^{2}-4 y}\left(e^{-y^{2}+5 y}-1\right) \rightarrow 0\right)$

We get:

$=\lim _{y \rightarrow 0} \frac{\cos (y-2) \cdot \cos 2 \cdot e^{y^{2}-4 y}\left(e^{-y^{2}+5 y}-1\right)}{y}=$

Using the substitution of equivalent infinitesimals:

$e^{-y^{2}+5 y}-1 \sim -y^{2}+5 y_{\text { as }} y \rightarrow 0\left(-y^{2}+5 y \rightarrow 0\right)$

We get:

$=\lim _{y \rightarrow 0} \frac{\cos (y-2) \cdot \cos 2 \cdot e^{y^{2}-4 y}\left(-y^{2}+5 y\right)}{y}=$

$=\lim _{y \rightarrow 0} \cos (y-2) \cdot \cos 2 \cdot e^{y^{2}-4 y}(-y+5)=$

$=\cos (0-2) \cdot \cos 2 \cdot e^{0^{2}-4 \cdot 0}(-0+5)=$

$=\cos 2 \cdot \cos 2 \cdot e^{0} \cdot 5=5 \cos ^{2} 2$

## Problem Kuznetsov Limits 14-20