Derivative of trigonometric function help I am having trouble computing derivatives at $x=2k \pi$. 
Let $f(x)=x \sqrt{1-\cos(x)}$. 
Then $\displaystyle f'(x)=\frac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}$. 
Then $\displaystyle f'(2k\pi)=\lim_{x \rightarrow 2k\pi}\frac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}$. 
I know limits such as $\displaystyle \lim_{x \rightarrow o}\frac{1-\cos x}{x^{2}}=\frac{1}{2}$, but I don't know how to use it here correctly.

As $x$ approaches $2k\pi$ from the negative side, $\sin x$ is negative.  From the positive side, $\sin x$ is positive.  We have
$$\lim_{x\rightarrow2k\pi^-}\dfrac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}=\lim_{x\rightarrow2k\pi^-}\dfrac{2(1-\cos x)-x\sqrt{1-\cos^2x}}{2\sqrt{1-\cos x}}=$$
$$\lim_{x\rightarrow2k\pi^-}\sqrt{1-\cos x}-\dfrac{x\sqrt{1+\cos x}}2=-k\pi\sqrt2$$
The positive side is similar except $\sin x=\sqrt{1-\cos^2x}$ instead of $-\sqrt{1-\cos^2x}$ yielding $k\pi\sqrt2$.  The limit as $x$ approaches $2k\pi$ exists when these 2 are equal, when $k=0$.
