## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{\sin ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}$
## Solution

$\lim _{x \rightarrow \pi} \frac{\sin ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}=\lim _{x \rightarrow \pi} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{2} x}-\tan ^{2} x}{(x-\pi)^{4}}=$

$=\lim _{x \rightarrow \pi} \frac{\tan ^{2} x \cdot \cos ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}=\lim _{x \rightarrow \pi} \frac{\tan ^{2} x\left(\cos ^{2} x-1\right)}{(x-\pi)^{4}}=$

$=\lim _{x \rightarrow \pi} \frac{\tan ^{2} x\left(-\sin ^{2} x\right)}{(x-\pi)^{4}}=$

Substitution:

$x=y+\pi \Rightarrow y=x-\pi$

$x \rightarrow \pi \Rightarrow y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{\tan ^{2}(y+\pi)\left(-\sin ^{2}(y+\pi)\right)}{((y+\pi)-\pi)^{4}}=\lim _{y \rightarrow 0} \frac{\tan ^{2} y\left(-(-\sin y)^{2}\right)}{y^{4}}=$

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

Using the substitution of equivalent infinitesimals:

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

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

We get:

$$
=\lim _{y \rightarrow 0} \frac{y^{2}\left(-y^{2}\right)}{y^{4}}=\lim _{y \rightarrow 0} \frac{-1}{1}=-1
$$

## Problem Kuznetsov Limits 13-7