$\lim_{x \to 0} f(x)=\frac{\tan 3x}{\tan 2x}$ without L'Hospital

Question

I am having trouble finding the limit for the following function:

$\lim_{x \to 0} f(x)=\frac{\tan 3x}{\tan 2x}$

I do not want to use L'Hospital's rule.

I know that $\frac{\tan 3x}{\tan 2x} = \frac{\sin 3x \cos 2x}{\sin 2x \cos 3x}$, but that does not help me.

Can you maybe give me a hint? Thanks.

Answer 1

$$\frac{\sin 3x\cos 2x}{\sin 2x\cos 3x}=\frac{3}{2}.\frac{\sin 3x}{3x}.\frac{2x}{\sin 2x}.\frac{\cos 2x}{\cos 3x}$$

Answer 2

Using $\tan (A+B)=\frac {\tan A+\tan B}{1-\tan A\tan B}$ twice we obtain $$\tan 2x =\frac {2\tan x}{1-\tan^2 x}; \tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$$ so that $$\frac {\tan 2x}{\tan 3x}=\frac {2(1-3\tan^2 x)}{(3-\tan^2 x)(1-\tan^2 x)}$$

And (having cancelled a factor of $\tan x$) nothing vanishes and the limit is easy to find, even if you know no other limits.