## Problem Statement

Calculate the limit of the function:

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

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

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

We obtain:

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

Using the substitution of equivalent infinitesimals:

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

$\ln \left(1+\frac{2 y}{\pi}\right) \sim \frac{2 y}{\pi}$, as $y \rightarrow 0\left(\frac{2 y}{\pi} \rightarrow 0\right)$

We obtain:

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

Using the substitution of equivalent infinitesimals:

$$
\tan 2 y \sim 2 y, \text{ as } y \rightarrow 0(2 y \rightarrow 0)
$$

We obtain:

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