Finding the Limit $\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$ Here's the problem.
$$\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$$
I really don't know where to start with this. Please help.

I assume you know that $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$
Now, $$\lim_{x \to 0} \frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x^2 (1+\cos x)}=\lim_{x \to 0} \frac{\sin^2 x}{x^2(1+\cos x)},$$
and therefore $$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \times\lim_{x \to 0}\frac{1}{1+\cos x}=\frac{1}{2}.$$
Finally, $$\lim_{x \to 0} \frac{\sin x (1-\cos x)}{x^2}=\lim_{x \to 0} \sin x \times \lim_{x \to 0} \frac{1-\cos x}{x^2}=0 \times \frac{1}{2}=0.$$
