Using $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ evaluate the limit Use $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$
to evaluate the limit
$$\lim_{x\to 0} \frac{x\tan (x^2)}{\cos(3x)\sin^3(2x)}$$
I'm really not sure how to go about this apart from trig identities and using the limit $=1$ to simplify it. Can anyone even get me started on it please?
 A: Hint: $\dfrac{x\tan (x^2)}{\cos(3x)\sin ^3(2x)} = \dfrac{1}{8\cos (3x)\cos (x^2)}\cdot \dfrac{1}{\left(\dfrac{\sin(2x)}{2x}\right)^3}\cdot \dfrac{\sin (x^2)}{x^2}$. From this the limit is $\dfrac{1}{8}$ since $\displaystyle \lim_{x \to 0} \cos 3x = \lim_{x \to 0} \cos x^2 = 1$.
A: $\cos(3x)$ is $1$ so it does not matter.
 $\displaystyle \lim_{x \to 0} \frac{(2x)^3}{\sin^3(2x)} = 1$. You can now split the limit into the product of another two limits, first is $1$. And the other can be evaluated using the fact that $\displaystyle \lim_{x \to 0} \frac{\tan(x)}{x} = 1$
A: $$\lim_{x\to0}\frac{x\tan(x^2)}{\cos(3x)\sin^3(2x)}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{x^2}\cdot\frac{\tan(x^2)}{\sin^3(2x)}=$$
$$\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{\sin^3(2x)}\cdot\frac{\tan(x^2)}{x^2}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot[\frac{x}{\sin(2x)}]^3\cdot\frac{\tan(x^2)}{x^2}$$
Now you have that:
$$\lim_{x\to0}\frac{x}{\sin(2x)}=\lim_{x\to0}\frac{1}{2}\frac{2x}{\sin(2x)}=\frac{1}{2}$$
and, setting $\ x^2=t:$
$$\lim_{x\to0}\frac{\tan(x^2)}{x^2}=\lim_{t\to0}\frac{\tan(t)}{t}=\lim_{t\to0}\frac{1}{\cos(t)}\frac{\sin(t)}{t}=1\cdot1=1$$
so the final result should be: $\ [\frac{1}{2}]^3=\frac{1}{8}$
A: Since $$\lim_{x \to 0} \frac{\sin x}{x}=1,$$
then $\sin x$ ~ $x$. It is  not diifficult to conclude that $\tan x$ ~ $x$ as $x\to 0$.
So the limit becomes that: 
$$\lim_{x \to 0} \dfrac{x\tan (x^2)}{\cos(3x)\sin ^3(2x)} = \lim_{x \to 0} \frac {x\cdot x^2}{8x^3}=\frac18.$$
It may be simple:)
