How to answer Calculus by Michael Spivak Chapter 5 Problem 15.VII I need to evaluate :
$$\lim_{x\to 0} \frac{x \sin(x)}{1-\cos(x)}$$
In terms of the number:
$$\alpha = \lim_{x\to0} \frac{\sin(x)}{x}$$
What I am thinking is that $\cos(x) = 1 - 2\sin^2(\frac{x}{2})$
But I don't know how to proceed.
I know an easy way is to use L'Hopital (the answer would be 2) but the objective is not to use derivates.
 A: $$
\frac{x \sin x}{1-\cos x} = \frac{x \sin x}{2\sin^2(\frac{x}{2})} = \frac{4(\frac{x}{2})^2}{2x}\frac{\sin x}{\sin^2(\frac{x}{2})} = 2\left(\frac{\sin x}{x}\right)\left(\frac{\frac{x}{2}}{\sin (\frac{x}{2})}\right)^2
$$
A: Make the trig substitution you mentioned, then split up the fraction to make fractions in the form of $\frac{\sin(x)}{x}$ or $\frac{x}{\sin(x)}$.  Also, note that the $x$ in these form can be $x/2$ as well.  You will need to bring in some extra $x$'s to make these forms.
A: Use
$$\sin{x}=2\sin{x\over 2}\cos{x\over 2}$$
$$1-\cos{x}=2\sin^2{x\over 2}$$
To get
$${x\sin{x}\over 1-\cos{x}}=2{{x\over 2}\cos{x\over 2}\over \sin{x\over 2}}\to {2\over\alpha}$$
because when $x\to 0$, $\cos{x\over 2}\to 1$ and ${x/2\over \sin{x/2}}\to 1/\alpha$
A: $\lim_{x\rightarrow0}  \frac{xsin(x)}{1-cos(x)}$
$\lim_{x\rightarrow0}  \frac{xsin(x)}{1-cos(x)}* \frac{1+cos(x)}{1+cos(x)}$
$\lim_{x\rightarrow0}  \frac{xsin(x)}{sin^2(x)}* (1+cos(x))$
$\lim_{x\rightarrow0}  \frac{x}{sin(x)}* (1+cos(x))$
$1\times 2 = 2$ 
