## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0}x\csc x=1.$$
We will use L'Hopital's rule to solve this limit. "LR" will stand for "Use L'Hopital's rule": $$\lim_{x\to0}x\csc x=\lim_{x\to0}x\cdot\frac{1}{\sin x}=\lim_{x\to0}\frac{x}{\sin x}=\left[\frac{0}{\sin 0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(x)'}{(\sin x)'}=\lim_{x\to0}\frac{1}{\cos x}=\frac{1}{\cos 0}=\frac{1}{1}=1.$$