Help with $\lim\limits_{x \to 0} \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}$ I'm preparing for my first exam in university (just recently enrolled in computer science) and I'm having difficulties working out this limit. I either currently lack the proper reasoning process to get it done or they haven't yet explained us all the theorems needed. I'd be really grateful if someone could point me in the right direction. Thank you!
$\lim\limits_{x \to 0} \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}$
 A: We have that
$$ \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}=\frac{(\sin(3x))^3}{\log (1+(\sin3x)^3)}\cdot \frac{(3x)^3 }{(\sin3x)^3}\cdot \frac{ \sin(2x)}{2x}\cdot  \frac2{27}$$
then refer to standard limits as $u \to 0$


*

*$\frac{\log (1+u)}{u}\to 1$

*$\frac{\sin u}{u}\to 1$
A: Hint:
By L'Hôpital ("$\frac{0}{0}$"),
\begin{align}
\lim_{x \to 0} \frac{x^2 \sin(2x)}{\log(1 + \sin^3(3x))}
& = \lim_{x \to 0} \frac{2(x^2 \cos(2x)+ x \sin(2x))(\sin^3(3x) + 1)}{9 \cos(3x) \sin^2(3x)} \\
& = \frac{2}{9} \left( \lim_{x \to 0} \left(\frac{x}{\sin(3x)}\right)^2 + \lim_{x \to 0} \left(\frac{x \sin(2x)}{\sin^2(3x)}\right)\right) \\
&= \frac{2}{9} \left( \frac{2}{9} + \frac{1}{9}\right)
= \frac{2}{27}.
\end{align}
A: Let use two results if $t \to 0$ then ${\log (1+t) \over t}=1$ and  ${\sin t \over t}=1$
Now give everyone what they want 
$$x^2.2x.{\sin 2x \over 2x}.\frac{\sin^3x}{\log (1+\sin^33x)}. \frac{(3x)^3}{\sin^33x}.{1 \over (3x)^3}\\={2 \over 27}$$
Assume $sin^3x$ or $2x$ as t . You will see . For exam you can remember this like sin zero upon same zero = 1 .
Or $\lim_{f(x) \to 0}{\sin(f(x)) \over f(x)}=1$ and $\lim_{f(x) \to 0}{\log(1+f(x)) \over f(x)}=1$ 
