How can evaluate $\lim_{x\to0}\frac{\sin(3x^2)}{\tan(x)\sin(x)}$ I know this:
$$\lim_{x\to0}\frac{\sin(3x^2)}{\tan(x)\sin(x)}$$
But I have no idea how make a result different of: $$\lim_{x\to0}\frac{3(x)}{\tan(x)}$$
Any suggestions? 
 A: $$\frac{\sin 3x^2}{\tan x\sin x}=3\cos x\frac x{\sin x}\frac{\sin 3x^2}{3x^2}\frac x{\sin x}\xrightarrow [x\to 0]{}3\cdot 1\cdot1\cdot 1\;\ldots$$
A: Use the fact that 
$${\sin(3x^2)\over 3x^2}\to 1 $$
as $x\to 0$.
A: Hint : The given limit is of the 0/0 form. Simply apply L' Hospitals rule (Differentiate Nr and Dr ) wrt x. 
No, you can write  Above limit in your reduced form. It's wrong because you can't sin function applies applied over 3x^2 and x too. 
A: I'm guessing from your reduction to $\frac {3x}{\tan x}$ that you are trying to use Taylor's theorem to apply a first-order approximation (as all the functions' Taylor series converge in a neighborhood of zero). 
In that case, notice that $\tan x \sim x$ as $x\to 0$. 
A: $$\lim_{x\to0}\frac{\sin(3x^2)}{\tan(x)\sin(x)} =3\lim_{x\to0}\frac{\sin(3x^2)}{3x^2}\cdot\frac{x}{\sin(x)}\cdot\frac{x}{\tan(x)} =3$$
Given that $$\lim_{h\to0}\frac{\sin  h}{h} =\lim_{h\to 0}\frac{\tan  h}{h} =1$$
A: \begin{align}
\ \lim_{x\to 0}\frac{\sin(3x^2)}{\tan(x) \sin^2(x)}=\lim_{x\to 0}\cos(x) \lim_{x\to 0} \frac{\sin(3x^2)}{\sin^2(x)}\\
&=\lim_{x\to 0}\frac{\sin(3x^2)}{\sin^2(x)}\\
&=\Biggl(\lim_{x\to 0}\frac{\sin(3x^2)}{3x^2}3x^2\Biggl)\Biggl(\lim_{x\to 0}\frac{x^2}{\sin^2(x)}\frac{1}{x^2}\Biggl)\\
&=\Biggl(\lim_{x\to 0}\frac{\sin(3x^2)}{3x^2}\Biggl) \Biggl(\lim_{x\to 0}3x^2\Biggl)\Biggl(\lim_{x\to 0}\frac{x^2}{\sin^2(x)}\Biggl)\Biggl(\lim_{x\to 0}\frac{1}{x^2}\Biggl)\\
&=\lim_{x\to 0}3x^2 \Biggl(\lim_{x\to 0}\frac{1}{x^2}\Biggl)=\lim_{x\to 0}\frac{3x^2}{x^2}\\
&=3
\end{align}
