## Problem Statement

Calculate the limit of the function:

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

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

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

We get:

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

## Problem Kuznetsov Limits 16-29