## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 2}\left(2 e^{x-2}-1\right)^{\frac{3 x+2}{x-2}}$
## Solution

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

Substitution:

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

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

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{3(y+2)+2}{(y+2)-2} \cdot \ln \left(2 e^{(y+2)-2}-1\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{3 y+8}{y} \cdot \ln \left(2 e^{y}-1\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{3 y+8}{y} \cdot \ln \left(1+2\left(e^{y}-1\right)\right)\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \frac{3 y+8}{y} \cdot 2\left(e^{y}-1\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$e^{y}-1 \sim y$, as $y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{3 y+8}{y} \cdot 2 y\right\}=\exp \left\{\lim _{y \rightarrow 0} 2(3 y+8)\right\}= \\
& =\exp \{2(3 \cdot 0+8)\}=\exp \{16\}=e^{16}
\end{aligned}
$$

## Problem Kuznetsov Limits 19-21