## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow \frac{\pi}{6}} \frac{\ln (\sin 3 x)}{(6 x-\pi)^{2}}
$$
## Solution

Substitution:

$x=y+\frac{\pi}{6} \Rightarrow y=x-\frac{\pi}{6}$

$x \rightarrow \frac{\pi}{6} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{6}} \frac{\ln (\sin 3 x)}{(6 x-\pi)^{2}}=\lim _{y \rightarrow 0} \frac{\ln \left(\sin 3\left(y+\frac{\pi}{6}\right)\right)}{\left(6\left(y+\frac{\pi}{6}\right)-\pi\right)^{2}}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(\sin \left(3 y+\frac{\pi}{2}\right)\right)}{((6 y+\pi)-\pi)^{2}}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (1-(1-\cos 3 y))}{(6 y)^{2}}=\lim _{y \rightarrow 0} \frac{\ln (\cos 3 y)}{36 y^{2}}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln (1-(1-\cos 3 y)) \sim-(1-\cos 3 y)$, as $y \rightarrow 0(-(1-\cos 3 y) \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{-(1-\cos 3 y)}{36 y^{2}}=$

Using the substitution of equivalent infinitesimals:

$$
1-\cos 3 y \sim{\frac{(3 y)^{2}}{2}}_{\text {, as }} y \rightarrow 0
$$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-\frac{(3 y)^{2}}{2}}{36 y^{2}}=\lim _{y \rightarrow 0} \frac{-9 y^{2}}{2 \cdot 36 y^{2}}= \\
& =\lim _{y \rightarrow 0} \frac{-1}{2 \cdot 4}=-\frac{1}{8}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-6