## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\sin5x\csc3x=\frac{5}{3}$$
$$A=\lim_{x\to0}\sin5x\csc3x$$ 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: $$A=\lim_{x\to0}\frac{(\sin 5x)'}{(\sin 3x)'}$$ $$A=\lim_{x\to0}\frac{5\cos 5x}{3\cos 3x}$$ $$A=\frac{5\cos(5\times0)}{3\cos(3\times0)}$$ $$A=\frac{5\cos0}{3\cos0}$$ $$A=\frac{5\times1}{3\times1}$$ $$A=\frac{5}{3}$$