Q:

How can evaluate $\lim_{x\to0}\frac{3x^2}{\tan(x)\sin(x)}$?

I know this
$$\lim_{x\to0}\frac{3x^2}{\tan(x)\sin(x)}$$
But I have no idea how make a result different of:
$$\lim_{x\to0}\frac{3x}{\tan(x)}$$
I would like understand this calculation without using derivation or L'hôpital's rule. Thank you.

A:

By Taylor series we have $\tan x\sim_0 x$ hence
$$\lim_{x\to0}\frac{3x}{\tan(x)}=\lim_{x\to0}\frac{3x}{x}=3$$

A:

Since $\tan x = \dfrac{\sin x}{\cos x}$, we have the identity:
$$\frac{3x^2}{\tan x\sin x} = \frac{3x^2\cos x}{\sin^2 x}$$
valid in at least an open, punctured neighbourhood around $0$.
Now using the product rule for limits, and that $\lim\limits_{x\to 0}\dfrac{\sin x}x = 1$ (proofs here, also some without differentiation):
$$\lim_{x\to0}\frac{3x^2}{\tan x\sin x} = \lim_{x\to0}\cos x\cdot\lim_{x\to 0}\frac{3x^2}{\sin^2 x} = 3$$

