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):
$$\lim_{x\to0}\frac{3x^2}{\tan x\sin x} = \lim_{x\to0}\cos x\cdot\lim_{x\to 0}\frac{3x^2}{\sin^2 x} = 3$$
A: Replace $\tan x$ by $\frac{\sin x}{\cos x}$. Our expression becomes
$$3(\cos x)\left(\frac{x}{\sin x}\right)^2. $$
You are undoubtedly familiar with the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$. 
A: Hints
$$\frac{3x^{2}}{\tan x\sin x}=3\times \frac{x}{\tan x}\times \frac{x}{\sin x},$$
$$$$
$$0<x<\frac{\pi }{2}\Rightarrow \sin x<x<\tan x\Leftrightarrow 1<\frac{x}{
\sin x}<\frac{1}{\cos x}.$$
A: We have, 
$$\lim_{x\to0}\frac{ x^2 }{\tan(x)\sin(x)} =3\lim_{x\to0}\frac{x}{\sin(x)}\cdot\frac{x}{\tan(x)} =3$$
Given that $$\lim_{h\to0}\frac{\sin  h}{h} =\lim_{h\to 0}\frac{\tan  h}{h} =1$$
