Answer
$$\lim_{y\to0}\frac{\sin3y\cot5y}{y\cot4y}=\frac{12}{5}$$
Work Step by Step
$$A=\lim_{y\to0}\frac{\sin3y\cot5y}{y\cot4y}$$
$$A=\lim_{y\to0}\frac{\sin3y\times\frac{\cos5y}{\sin5y}}{y\times\frac{\cos4y}{\sin4y}}$$
$$A=\lim_{y\to0}\frac{\frac{\sin 3y\cos5y}{\sin5y}}{\frac{y\cos4y}{\sin4y}}$$
$$A=\lim_{y\to0}\frac{\sin3y\sin4y\cos5y}{y\cos4y\sin5y}$$
Multiply both numerator and denominator by $y (y\ne0)$:
$$A=\lim_{y\to0}\frac{y\sin3y\sin4y\cos5y}{y^2\cos4y\sin5y}$$
$$A=\lim_{y\to0}\Big(\frac{y}{\sin5y}\times\frac{\sin3y}{y}\times\frac{\sin4y}{y}\times\frac{\cos5y}{\cos4y}\Big)$$
$$A=\lim_{y\to0}\frac{y}{\sin5y}\times\lim_{y\to0}\frac{\sin3y}{y}\times\lim_{y\to0}\frac{\sin4y}{y}\times\lim_{y\to0}\frac{\cos5y}{\cos4y}$$
$$A=X\times Y\times W\times Z$$
1) With $X$: $$X=\lim_{y\to0}\frac{y}{\sin5y}=\Big(\lim_{y\to0}\frac{\sin5y}{y}\Big)^{-1}=\Big(5\lim_{y\to0}\frac{\sin5y}{5y}\Big)^{-1}$$ $$X=(5\times1)^{-1}=\frac{1}{5}$$ (Theorem 7 with $\theta=5y$)
2) With $Y$: $$Y=\lim_{y\to0}\frac{\sin3y}{y}=3\lim_{y\to0}\frac{\sin3y}{3y}=3\times1=3$$ (Theorem 7 with $\theta=3y$)
3) With $W$: $$W=\lim_{y\to0}\frac{\sin4y}{y}=4\lim_{y\to0}\frac{\sin4y}{4y}=4\times1=4$$ (Theorem 7 with $\theta=4y$)
4) With $Z$: $$Z=\lim_{y\to0}\frac{\cos5y}{\cos4y}=\frac{\cos0}{\cos0}=\frac{1}{1}=1$$
Therefore, $$A=\frac{1}{5}\times3\times4\times1=\frac{12}{5}$$
