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


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




  You can do it without L'Hospital


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


  $$\large\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}=\frac{9}{2}
$$
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
$$\lim\limits_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
$$=\frac12\lim\limits_{x \to 0} \frac{\sin^2(3x)}{\sin^2(x)}$$ 
$$=\frac12\lim\limits_{x \to 0} \frac{(3x)^2x^2\sin^2(3x)}{(3x)^2x^2\sin^2(x)}$$
$$=\frac92\left(\lim\limits_{x \to 0} \frac{x^2}{x^2}\right)\left(\lim\limits_{x \to 0} \frac{\sin^2(3x)}{(3x)^2}\right)\left(\lim\limits_{x \to 0} \frac{x^2}{\sin^2(x)}\right)$$
$$=\frac92$$