## Problem Statement

Calculate the limit of the function:

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

Substitution:

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

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

We get:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\operatorname{tg} 2 x}-e^{-\sin 2 x}}{\sin x-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2\left(y+\frac{\pi}{2}\right)}-e^{-\sin 2\left(y+\frac{\pi}{2}\right)}}{\sin \left(y+\frac{\pi}{2}\right)-1}$

$=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg}(2 y+\pi)}-e^{-\sin (2 y+\pi)}}{\cos y-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}-e^{\sin 2 y}}{\cos y-1}=$

$=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}-e^{\frac{\sin 2 y \cos 2 y}{\cos 2 y}}}{\cos y-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}-e^{\operatorname{tg} 2 y \cdot \cos 2 y}}{\cos y-1}=$

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

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

Using the substitution of equivalent infinitesimals:

$e^{-2 \operatorname{tg} 2 y \cdot \sin ^{2} y}-1 \sim-2 \operatorname{tg} 2 y \cdot \sin ^{2} y$, as $y \rightarrow 0\left(-2 \operatorname{tg} 2 y \cdot \sin ^{2} y \rightarrow 0\right)$

$1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}\left(-2 \operatorname{tg} 2 y \cdot \sin ^{2} y\right)}{\frac{y^{2}}{2}}=\lim _{y \rightarrow 0} \frac{-4 \operatorname{tg} 2 y \cdot \sin ^{2} y \cdot e^{\operatorname{tg} 2 y}}{y^{2}}=$

Using the substitution of equivalent infinitesimals:

$\sin y \sim y$, as $y \rightarrow 0$

$\operatorname{tg} 2 y \sim 2 y$, as $y \rightarrow 0(2 y \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-4 \cdot 2 y \cdot y^{2} \cdot e^{\operatorname{tg} 2 y}}{y^{2}}=\lim _{y \rightarrow 0}-8 y e^{\operatorname{tg} 2 y}= \\
& =-8 \cdot 0 \cdot e^{\operatorname{tg} 2 \cdot 0}=0
\end{aligned}
$$

## Problem Kuznetsov Limits 14-5