## Problem Statement

Calculate the limit of the function:

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

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

Substitution:

$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

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

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

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:

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

$=\exp \{2(0+1)\}=\exp \{2\}=e^{2}$

## Problem Kuznetsov Limits 19-17