## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 2} \frac{\ln \left(9-2 x^{2}\right)}{\sin 2 \pi x}$
## Solution

Substitution:

$x=y+2 \Rightarrow y=x-2$

$x \rightarrow 2 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\ln \left(9-2 x^{2}\right)}{\sin 2 \pi x}=\lim _{y \rightarrow 0} \frac{\ln \left(9-2(y+2)^{2}\right)}{\sin 2 \pi(y+2)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(9-2\left(y^{2}+4 y+4\right)\right)}{\sin (2 \pi y+4 \pi)}=\lim _{y \rightarrow 0} \frac{\ln \left(9-2 y^{2}+8 y+8\right)}{\sin (2 \pi y)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (1-2 y(y+4))}{\sin (2 \pi y)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin 2 \pi y \sim 2 \pi y$, as $y \rightarrow 0(2 \pi y \rightarrow 0)$

$\ln (1-2 y(y+4)) \sim-2 y(y+4)$, as $y \rightarrow 0(-2 y(y+4) \rightarrow 0)$

We get:

$$
=\lim _{y \rightarrow 0} \frac{-2 y(y+4)}{2 \pi y}=\lim _{y \rightarrow 0} \frac{-(y+4)}{\pi}=\frac{-(0+4)}{\pi}=-\frac{4}{\pi}
$$

## Problem Kuznetsov Limits 13-20