Trigonometry limit problem? 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. 
 A: Hint: Multiply top and bottom by $1+\cos(3x)$. Then the limit should follow from what you know about $\lim_{t\to 0}\frac{\sin t}{t}$. 
Or else more mechanically use L'Hospital's Rule twice.
Remark: The mathematically most natural approach is none of the above. By the usual power series expansion for $\cos t$, we have 
$$1-\cos(3x)=\frac{(3x)^2}{2!}-\frac{(3x)^4}{4!}+\cdots.$$
Divide the right side by $2x^2$, and let $x\to 0$.  
A: Multiply top and bottom by $1+\cos(3x)\;$ (which is called the conjugate of $1 - \cos(3x))$. 
$$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2} \cdot \frac{1 + \cos(3x)}{1 + \cos(3x)} $$ $$= \lim_{x \to 0} \frac{1 - \cos^2(3x)}{2x^2(1 + \cos(3x)}$$
$$= \lim_{x\to 0}\; \frac{\sin^2(3x)}{2x^2(1 + \cos(3x))}$$
$$=\lim_{x\to 0} \frac{\sin^2(3x)}{3^2\cdot x^2}\cdot \frac{3^2}{2[1+\cos(3x)]}$$
$$=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\frac{9}{2(1+\cos(3x))}$$
$$=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\cdot \frac{9}{2}\cdot \frac{1}{(1+\cos(3x))}=1\cdot\frac{9}{2\cdot 2} = \frac94$$
I'm assuming you know the value of $\lim_{x\to 0}\dfrac{\sin ax}{ax} = 1\,$ where $a$ is a nonzero constant.

If you know L'hospital, you can apply that twice (differentiate each of the numerator and denominator twice), and evaluate the resulting limit. If you haven't learned it yet, you will likely learn it soon, and it can greatly simplify problems like this!
A: $$1- \cos(3x) = 2 \sin^2(3x/2)$$
Hence, $$\dfrac{1-\cos(3x)}{2x^2} = \dfrac{\sin^2(3x/2)}{x^2} = \left(\dfrac{\sin(3x/2)}x\right)^2$$
Now recall that $\displaystyle \lim_{t \to 0}\dfrac{\sin(at)}t = a$ and if $\lim_{t \to b} f(t) = \tilde{f} \in \mathbb{R}$, then $\lim_{t \to b} f^2(t) = \tilde{f}^2$
A: $$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}=\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}\frac{1+\cos(3x)}{1+\cos(3x)}=\lim_{x\to 0} \frac{1-\cos^2(3x)}{2x^2}\frac{1}{1+\cos(3x)}=$$
$$=\lim_{x\to 0} \frac{\sin^2(3x)}{2{(3x)}^2}\frac{9}{1+\cos(3x)}=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\frac{9}{2(1+\cos(3x))}=\frac{9}{4}$$
Using L'Hopital rule 
$$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}=\lim_{x\to 0} \frac{3\sin(3x)}{4x}=\lim_{x\to 0} \frac{9\cos(3x)}{4}$$
