# 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?

$$\frac{\sin 3x^2}{\tan x\sin x}=3\cos x\frac x{\sin x}\frac{\sin 3x^2}{3x^2}\frac x{\sin x}\xrightarrow [x\to 0]{}3\cdot 1\cdot1\cdot 1\;\ldots$$

Use the fact that $${\sin(3x^2)\over 3x^2}\to 1$$ as $x\to 0$.

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.

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 $x\to 0$.

$$\lim_{x\to0}\frac{\sin(3x^2)}{\tan(x)\sin(x)} =3\lim_{x\to0}\frac{\sin(3x^2)}{3x^2}\cdot\frac{x}{\sin(x)}\cdot\frac{x}{\tan(x)} =3$$

Given that $$\lim_{h\to0}\frac{\sin h}{h} =\lim_{h\to 0}\frac{\tan h}{h} =1$$