Answer
\begin{aligned} \lim _{y \rightarrow 0} \frac{\sin 3 y \cot 5 y}{y \cot 4 y}=\frac{12}{5}
\end{aligned}
Work Step by Step
Given $$\lim _{y \rightarrow 0} \frac{\sin 3 y \cot 5 y}{y \cot 4 y}$$
So, we get
\begin{aligned} L&=\lim _{y \rightarrow 0} \frac{\sin 3 y \cot 5 y}{y \cot 4 y}\\
&=\lim _{y \rightarrow 0} \frac{\sin 3 y }{y }. \frac{ \cot 5 y}{ \cot 4 y}\\
&=3\lim _{y \rightarrow 0} \frac{\sin 3 y }{3y }. \frac{ \cot 5 y}{ \cot 4 y}\\
&=3\lim _{y \rightarrow 0} \frac{\sin 3 y }{3y }.\lim _{y \rightarrow 0} \frac{ \cot 5 y}{ \cot 4 y}\\
&=3\cdot 1\cdot \lim _{y \rightarrow 0} \frac{ \cot 5 y}{ \cot 4 y}\\
&=3\lim _{y \rightarrow 0} \frac{ \cot 5 y}{ \cot 4 y}\\
&=3 \lim _{y \rightarrow 0} \frac{\sin 4 y}{\sin 5 y} \cdot \frac{\cos 5 y}{\cos 4 y}\\
&=\frac{1}{3} \lim _{y \rightarrow 0} \frac{\sin 4 y}{\sin 5 y} \cdot \lim _{y \rightarrow 0} \frac{\cos 5 y}{\cos 4 y}\\
&=3 \lim _{y \rightarrow 0} \frac{\sin 4 y}{\sin 5 y} \cdot \frac{\cos 0}{\cos 0}\\
&=3 \lim _{y \rightarrow 0} \frac{\sin 4 y}{\sin 5 y} \cdot 1\\
&=3 \lim _{y \rightarrow 0} \frac{\sin 4 y}{\sin 5 y} \\
&=3 \lim _{y \rightarrow 0} \frac{\frac{\sin 4 y}{y}}{\frac{\sin 5 y}{y}} \\
&=3 \lim _{y \rightarrow 0} \frac{4\frac{\sin 4 y}{4y}}{5\frac{\sin 5 y}{5y}} \\
&=\frac{12}{5}\frac{ \lim\limits _{y \rightarrow 0} \frac{\sin 4 y}{4y}}{ \lim \limits_{y \rightarrow 0} \frac{\sin 5 y}{5y}} \\
&=\frac{12}{5}\frac{1}{1} \\
&=\frac{12}{5}
\end{aligned}
