Trigonometric Functions Limit: $\lim_{x \to 0} \frac{1-\cos x}{x\sin x}$ In my assignment I have to solve the following question. I know the answer, but I keep getting it wrong, and I don't know how to solve it.
$$\lim_{x \to 0} \frac{1-\cos x}{x\sin x}$$
I have tried several things, but first I tried to multiply by $(1+\cos x)$, both numerator and denumerator, to get $$\lim_{x \to 0} \frac{1-\cos^2x}{(x\sin x)(1+\cos x)}$$  I keep getting that the limit is equal to $0$, by calculating (for example) $$\lim_{x \to 0} \frac{\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}}{\frac{(x\sin x)}{\cos^2x}\frac{(1+\cos x)}{\cos^2x}},$$
which equals $$\frac{1-1}{1-1}$$
However, I know the answer is $\frac{1}{2}$.
Any ideas? 
Thanks
 A: $$
\frac{1-\cos^2{x}}{x\sin x(1+\cos x)}=\frac{\sin^2{x}}{x\sin x(1+\cos x)}=\frac{\sin x}{x(1+\cos x)}=\big(\frac{\sin x}{x}\big)\frac{1}{1+\cos x}
$$
Does that look familiar now?
A: Multiplying by $\displaystyle \frac{\sin x}{x}$ immediately does the job, recalling the first two fundamental trigonometric limits.
A: If you know that 
$$\begin{align}\lim\limits_{x\to0} \frac{\sin x}x&=1 \tag{1}\\
\lim\limits_{x\to0} \frac{1-\cos x}{x^2}&=\frac12 \tag{2}
\end{align}$$
then it is not difficult to combine them to get:
$$\lim\limits_{x\to0} \frac{1-\cos x}{x\sin x} =
\lim\limits_{x\to0} \frac{1-\cos x}{x^2} \cdot \frac{x}{\sin x} = \frac 12 \cdot 1 = \frac12.$$
The limits $(1)$ and $(2)$ are quite well-known. The first one can be found, for example, here. And the second one can be found, for example, here or here.
A: hint:Use these facts:
$$1-\cos x = 2\sin^2\left(\frac{x}{2}\right), \sin x = 2\sin(\frac{x}{2})\cos (\frac{x}{2}),\displaystyle \lim_{x\to 0} \dfrac{\sin (\frac{x}{2})}{\frac{x}{2}}=1$$
A: In order to be comprehensive I add an answer based on Taylor expansions
$$1-\cos{x}=\frac{x^2}{2}+o(x^2)$$
$$x\sin{x}=x^2+o(x^2)$$
This yields 
$$\frac{1-\cos{x}}{x\sin{x}}=\frac{1}{2}+o(1)$$
And the limit is $\frac{1}{2}$
A: You can solve this trigonometric limit
also with the l'Hopital's rule:
$$\lim_{x \to 0} \frac{1-\cos x}{x\sin x}=$$
$$\lim_{x \to 0} \frac{\sin x}{\sin x+\cos x}=$$
$$\lim_{x \to 0} \frac{\cos x}{\cos x+\cos x-x\cdot \sin x}=\frac {1}{2}$$
