## Problem Statement

Calculate the limit of the function:

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

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

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

Substitution:

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

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

We get:

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Using the substitution of equivalent infinitesimals:
$e^{y}-1 \sim y_{\text {, as }} y \rightarrow 0$

We get:

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

$=\exp \{2(3 \cdot 0+2)\}=\exp \{4\}=e^{4}$

## Problem Kuznetsov Limits 19-19