Answer
The solution is
$$\lim_{x\to0}\csc6x\sin7x=\frac{7}{6}.$$
Work Step by Step
To calculate this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to0}\csc6x\sin7x=\lim_{x\to0}\frac{1}{\sin 6x}\cdot \sin7x=\lim_{x\to0}\frac{\sin7x}{\sin 6x}=\left[\frac{\sin (7\cdot0)}{\sin(6\cdot0)}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(\sin7x)'}{(\sin6x)'}=\lim_{x\to0}\frac{\cos7x(7x)'}{\cos6x(6x)'}=\lim_{x\to0}\frac{7\cos7x}{6\cos6x}=\frac{7\cos(7\cdot0)}{6\cos(6\cdot0)}=\frac{7}{6}.$$
