Limit: $\lim_{x\to 0}\frac{\tan3x}{\sin2x}$

$\lim_{x\to 0}\frac{\tan3x}{\sin2x}=\lim_{x\to 0}\frac{\tan3x}{3x}\frac{2x}{\sin2x}\frac{3x}{2x}=\frac{3}{2}$

or by using L'Hôpital's rule $\lim_{x\to 0}\frac{\tan3x}{\sin2x}=\lim_{x\to 0}\frac{3(1+\tan^2 3x)}{2\cos2x}=\frac{3(1+\tan^2 3(0))}{2\cos2(0)}=\frac{3}{2}$

$$\lim_{x\to 0}\frac{\sin3x}{1}\cdot\frac{1}{\cos(3x)}\cdot\frac{1}{\sin(2x)}$$ $$\lim_{x\to 0}\frac{\sin3x}{3x}\cdot\frac{1}{\cos(3x)}\cdot\frac{3x}{\sin(2x)}$$ $$\lim_{x\to 0}\frac{\sin3x}{3x}\cdot\frac{1}{\cos(3x)}\cdot\frac{\frac{3}{2}2x}{\sin(2x)}$$ $$\lim_{x\to 0}\frac{\sin3x}{3x}\cdot\frac{1}{cos(3x)}\cdot\frac{\frac{3}{2}}{\frac{\sin(2x)}{2x}}$$