Finding $\lim_{x\to 0}\frac{\tan x(1- \cos x)}{3x^2}$ How do I find the following limit:
$$
\lim_{x\to 0}\frac{\tan x(1- \cos x)}{3x^2}
$$
without using L'Hopital's rule? The reason I'm making a point of not using L'Hopital is that if I run the limit through Wolfram Alpha that's the method it uses, but we haven't gone through that yet so I'm guessing I should use something else. 
I don't really know what to do here. I've done quite a number of exercises on limits by now and I almost always get it right and know immediately what to do, but not this time. The only thing I can think of is to use $\tan x=\frac{\sin x}{\cos x}$. But I can't say that helps me much... That only gives me
$$
\lim_{x\to 0}\frac{\tan x(1- \cos x)}{3x^2}=\lim_{x\to 0}\frac{\sin x(\cos^{-1}x- 1)}{3x^2}
$$
Does that help me? What should I do next? Or should I start with something different?

Hint:

$$1-\cos(x) = 2\sin(x/2)^2. $$

