Limit, having trouble applying L'Hospital rule I have been asked to solve for this limit: 
$$\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
I try to find the derivative of the numerator and denominator: 
$$\frac{6\sin(3x)\cos(3x)}{2\sin(2x)}$$ but that still gives me a $\frac{0}{0}$. How do I solve this


Using L'Hospital Again

$$\begin{align}\lim_{x \to 0} \frac{6\sin(3x)\cos(3x)}{2\sin(2x)}
&=\lim_{x \to 0} \frac{18 \cos ^2(3 x)-18 \sin ^2(3 x)}{4\cos(2x)}\\
&=\lim_{x \to 0} \frac{18 \cos ^2(3 x)}{4\cos(2x)}-\frac{18 \sin ^2(3 x)}{4\cos(2x)}\\
&=\frac{18}{4}+0\\
&=\frac{9}{2}\\
\end{align}$$


You can do it without L'Hospital

$$\begin{align}
\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}
&=\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos^2(x)+\sin^2 x}\tag{1}\\
&=\lim_{x \to 0} \frac{\sin^2(3x)}{\sin^2(x)+\sin^2x}\tag{2}\\
&=\lim_{x \to 0} \frac{\sin^2(3x)}{2\sin^2(x)}\tag{3}\\
&=\lim_{x \to 0} \frac{3^2x^2\sin^2(3x)}{2\cdot 3^2x^2\sin^2(x)}\tag{4}\\
&=\lim_{x \to 0} \frac{3^2}2\cdot\frac{\sin^2(3x)}{(3x)^2}\cdot\frac{x^2}{\sin^2(x)}\tag{5}\\
&=\lim_{x \to 0} \frac{3^2}2\left(\frac{\sin(3x)}{3x}\right)^2\left(\frac{x}{\sin(x)}\right)^2\tag{6}\\
&=\lim_{x \to 0} \frac{3^2}2\cdot1^2\cdot1^2\tag{7}\\
\end{align}$$

$$\large\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}=\frac{9}{2}
$$

