TITLE: Trigonometry limit problem?
QUESTION [5 upvotes]: How would I find the following limit:
$$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}$$
I am not sure how to do this one help will be appreciated.

REPLY [3 votes]: $$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}=\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}\frac{1+\cos(3x)}{1+\cos(3x)}=\lim_{x\to 0} \frac{1-\cos^2(3x)}{2x^2}\frac{1}{1+\cos(3x)}=$$
$$=\lim_{x\to 0} \frac{\sin^2(3x)}{2{(3x)}^2}\frac{9}{1+\cos(3x)}=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\frac{9}{2(1+\cos(3x))}=\frac{9}{4}$$
Using L'Hopital rule 
$$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}=\lim_{x\to 0} \frac{3\sin(3x)}{4x}=\lim_{x\to 0} \frac{9\cos(3x)}{4}$$