Condition of the problem

Calculate the limit of the function:

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

Substitution:

$x=y+\frac{1}{2} \Rightarrow y=x-\frac{1}{2}$

$x \rightarrow \frac{1}{2} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)^{2}}{e^{\sin \pi x}-e^{-\sin 3 \pi x}}=\lim _{y \rightarrow 0} \frac{\left(2\left(y+\frac{1}{2}\right)-1\right)^{2}}{e^{\sin \pi\left(y+\frac{1}{2}\right)}-e^{-\sin 3 \pi\left(y+\frac{1}{2}\right)}}= \\
& =\lim _{y \rightarrow 0} \frac{(2 y+1-1)^{2}}{e^{\sin \left(\pi y+\frac{\pi}{2}\right)}-e^{-\sin \left(3 \pi y+\frac{3 \pi}{2}\right)}}= \\
& =\lim _{y \rightarrow 0} \frac{4 y^{2}}{e^{\cos \pi y}-e^{\cos 3 \pi y}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{4 y^{2}}{-e^{\cos \pi y}\left(e^{\cos 3 \pi y-\cos \pi y}-1\right)}=
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_b7cebc4f34037103b026g-09.jpg?height=165&width=976&top_left_y=313&top_left_x=152)

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

Using the substitution of equivalent infinitesimals:

$e^{-2 \cdot \sin 2 \pi y \cdot \sin \pi y}-1 \sim-2 \cdot \sin 2 \pi y \cdot \sin \pi y$, as $y \rightarrow 0(-2 \cdot \sin 2 \pi y \cdot \sin \pi y \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{4 y^{2}}{-e^{\cos \pi y}(-2 \cdot \sin 2 \pi y \cdot \sin \pi y)}=$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_b7cebc4f34037103b026g-09.jpg?height=70&width=800&top_left_y=1255&top_left_x=153)

$\sin \pi y \sim \pi y$, as $y \rightarrow 0(\pi y \rightarrow 0)$

We get:

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

## Problem Kuznetsov Limits 14-2