$\lim_{x\to 0}\frac{\tan3x}{\sin2x}$= $\lim_{x\to 0}\frac{\frac{\sin(3x)}{\cos(3x)}}{\sin2x}=\lim_{x\to 0}\frac{\sin3x}{1}\cdot\frac{1}{\cos(3x)}\cdot\frac{1}{\sin(2x)}$
From this point I am lost. I believe I can pull a 3 and 2 out but I am not sure how. Can someone give me detailed instructions for a person in Calculus 1?
$\lim_{x\to 0}\frac{\tan3x}{\sin2x}=\lim_{x\to 0}\frac{\tan3x}{3x}\frac{2x}{\sin2x}\frac{3x}{2x}=\frac{3}{2}$
or by using L'Hôpital's rule $\lim_{x\to 0}\frac{\tan3x}{\sin2x}=\lim_{x\to 0}\frac{3(1+\tan^2 3x)}{2\cos2x}=\frac{3(1+\tan^2 3(0))}{2\cos2(0)}=\frac{3}{2}$
Hint: $\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\tan x}{x} = 1$
Rewrite your expression as
$$\frac{\tan 3x}{x} \cdot \frac{x}{\sin 2x} = \frac 32 \cdot \frac{\tan 3x}{3x} \cdot \frac{2x}{\sin 2x}$$
See what to do now?
$$\lim_{x\to 0}\frac{\sin3x}{1}\cdot\frac{1}{\cos(3x)}\cdot\frac{1}{\sin(2x)}$$ $$\lim_{x\to 0}\frac{\sin3x}{3x}\cdot\frac{1}{\cos(3x)}\cdot\frac{3x}{\sin(2x)}$$ $$\lim_{x\to 0}\frac{\sin3x}{3x}\cdot\frac{1}{\cos(3x)}\cdot\frac{\frac{3}{2}2x}{\sin(2x)}$$ $$\lim_{x\to 0}\frac{\sin3x}{3x}\cdot\frac{1}{cos(3x)}\cdot\frac{\frac{3}{2}}{\frac{\sin(2x)}{2x}}$$
