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
 A: 
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

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

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

A: A simple trig identity applied at the outset will simplify your work greatly.
$\displaystyle \sin^2 3x = \frac{1}{2}(1-\cos 6x)$
Apply that and then apply L' Hopital's Rule.
You'll immediately get the limit to be $\displaystyle \frac{1}{2} \cdot \frac{6\sin 6x}{2 \sin 2x}$, following which you can either apply LHR again or even more simply, use $\displaystyle \lim_{y \to 0} \frac{\sin y}{y} = 1$ immediately.
Actually, from where you ended up, you can apply the double-angle identity for sine to transform it into the form above, i.e. $\displaystyle \sin 3x \cos 3x = \frac{1}{2}\sin 6x$.
A: $3\displaystyle\lim_{x\to 0}\dfrac{\sin 3x\cos 3x}{\sin 2x}=\dfrac{9}{2}\displaystyle\lim_{x\to 0}\dfrac{\left(\dfrac{\sin 3x}{3x}\right)\cos 3x}{\left(\dfrac{\sin 2x}{2x}\right)}=\dfrac{9}{2}\left(\dfrac{\displaystyle\lim_{x\to 0}\left(\dfrac{\sin 3x}{3x}\right)\cdot\displaystyle\lim_{x\to 0}\cos 3x}{\displaystyle\lim_{x\to 0}\left(\dfrac{\sin 2x}{2x}\right)}\right)=\dfrac{9}{2}$
A: As $x\to0$,
$$\frac{6\sin(3x)\cos(3x)}{2\sin(2x)}=\frac62\cdot\frac{\sin(3x)}{3x}\cdot\frac{2x}{\sin(2x)}\cdot\frac32\cdot\cos(3x)\to\frac62\cdot1\cdot1\cdot\frac32\cdot1=\frac92.$$
A: Using L'Hôpital's rule, we have
$$\lim\limits_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
$$=\lim\limits_{x \to 0} \frac{\frac{\mathrm d}{\mathrm dx}\sin^2(3x)}{\frac{\mathrm d}{\mathrm dx}\left[1-\cos(2x)\right]}$$
$$=3\lim\limits_{x \to 0} \frac{\sin(3x)\cos(3x)}{\sin(2x)}$$
$$=3\lim\limits_{x \to 0} \frac{\frac{\mathrm d}{\mathrm dx}\left[\sin(3x)\cos(3x)\right]}{\frac{\mathrm d}{\mathrm dx}\sin(2x)}$$
$$=\frac32\lim\limits_{x \to 0} \frac{\cos(3x)\frac{\mathrm d}{\mathrm dx}\sin(3x)+\sin(3x)\frac{\mathrm d}{\mathrm dx}\cos(3x)}{\cos(2x)}$$
$$=\frac92\lim\limits_{x \to 0} \frac{\cos^2(3x)-\sin^2(3x)}{\cos(2x)}$$
$$=\frac92$$
If you'd like to understand how L'Hôpital's rule is applied, see my answer here. We can also solve this limit without L'Hôpital's rule by using the fact that 
$$ \lim\limits_{x\to 0}\frac{\sin(x)}{x}=1$$
Therefore
$$\lim\limits_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
$$=\frac12\lim\limits_{x \to 0} \frac{\sin^2(3x)}{\sin^2(x)}$$
$$=\frac12\lim\limits_{x \to 0} \frac{(3x)^2x^2\sin^2(3x)}{(3x)^2x^2\sin^2(x)}$$
$$=\frac92\left(\lim\limits_{x \to 0} \frac{x^2}{x^2}\right)\left(\lim\limits_{x \to 0} \frac{\sin^2(3x)}{(3x)^2}\right)\left(\lim\limits_{x \to 0} \frac{x^2}{\sin^2(x)}\right)$$
$$=\frac92$$
