Answer
$$\lim_{x\to0}\sin5x\csc3x=\frac{5}{3}$$
Work Step by Step
$$A=\lim_{x\to0}\sin5x\csc3x$$
From trigonometric identities, we have $\csc3x=\frac{1}{\sin3x}$, which means
$$A=\lim_{x\to0}\frac{\sin 5x}{\sin 3x}$$
As $x$ approaches $0$, both $\sin 5x$ and $\sin 3x$ approach $\sin0=0$, leading to an indeterminate form of $\frac{0}{0}$. Hence we would apply L'Hospital's Rule here:
$$A=\lim_{x\to0}\frac{(\sin 5x)'}{(\sin 3x)'}$$
$$A=\lim_{x\to0}\frac{5\cos 5x}{3\cos 3x}$$
$$A=\frac{5\cos(5\times0)}{3\cos(3\times0)}$$
$$A=\frac{5\cos0}{3\cos0}$$
$$A=\frac{5\times1}{3\times1}$$
$$A=\frac{5}{3}$$
