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.
 A: 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, $$\lim_{x \to 0} \frac{\sin x (1-\cos x)}{x^2}=\lim_{x \to 0} \sin x \times \lim_{x \to 0} \frac{1-\cos x}{x^2}=0 \times \frac{1}{2}=0.$$
A: Hint:
$$\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^2} = \left(\lim_{x\to 0}\frac{\sin x - \sin 0}{x-0}\right)(-1)\left(\lim_{y\to 0}\frac{\cos y - \cos 0}{y-0}\right),$$
assuming the two limits on the right hand side exist (why do they exist?).
A: Applying the l'Hopital rule, if you put: $f=\frac{\sin(x)(1-\cos(x))}{x^2}=\frac{g(x)}{h(x)}$ you get:
$$L=\lim_{x\to 0}f=\lim_{x\to 0}\frac{g'(x)}{h'(x)}$$ wich gives:
$$L=lim_{x\to 0}\frac{\cos(x)-\cos(x)^2+\sin(x)^2}{2x}$$
If you apply l'Hopital again you get:
$$L=\lim_{x\to 0}\frac{\sin(x)(4\cos(x)-1)}{2}=0$$
A: Hint: Use the fundamental trigonometric limit:
$$ 
\lim_{x\to 0}\frac{\sin x}{x}=1
$$
For more informations about $\lim_{x\to 0}\frac{\sin x}{x}$ see  Manipula Math Applets for $\lim_{x\to 0}\frac{\sin x}{x}$. For $\lim_{x\to 0}\frac{1-\cos x}{x}$, 
\begin{align}
\lim_{x\to 0}\frac{1-\cos x}{x}=&
 \lim_{x\to 0}\frac{1-\cos x}{x}\frac{1+\cos x}{1+\cos x}\\
=& \lim_{x\to 0}\frac{1-\cos ^2x}{x\cdot(1+\cos x)}\\
=& \lim_{x\to 0}\frac{\sin^2x}{x\cdot(1+\cos x)}\\
=& \lim_{x\to 0}\frac{\sin x}{x}\frac{\sin x}{(1+\cos x)}\\
=& \lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x \to 0}\frac{1}{(1+\cos x)}\cdot \lim_{x\to 0}(\sin x)\\
=& 1\cdot\frac{1}{2}\cdot0 
\end{align}
