Problem condition

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{\cos ^{2} x}-1}{\ln (\sin x)}$
Solution

Substitution:

$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$

We obtain:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_62fa08db54895d8ac1b3g-08.jpg?height=217&width=1279&top_left_y=843&top_left_x=152)

We get:

$=\lim _{y \rightarrow 0} \frac{\ln 2 \cdot \sin ^{2} y}{-2 \sin ^{2} \frac{y}{2}}=$

Using the substitution of equivalent infinitesimals:

$\sin y \sim y_{\text {, as }} y \rightarrow 0$

$\sin \frac{y}{2} \sim \frac{y}{2}_{\text {, as }} y \rightarrow 0\left(\frac{y}{2} \rightarrow 0\right)$

We get:

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

## Problem Kuznetsov Limits 14-1