Finding $\lim_{x\rightarrow 0}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}$ We know that $$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\tan({x}/{2})$$ so we can change the above function to another form as follow $$\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=\frac{(x/2)}{\tan(x/2)}$$ If we use the Taylor series of $\tan(x/2)$ we will get
$$\frac{(x/2)}{\tan(x/2)}=\frac{(x/2)}{\frac{x}{2}+\frac{x^3}{24}+\frac{x^5}{240}+....}=\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}$$ Now we will find the limit for the two same functions
$$\lim_{x\rightarrow 0^{+}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^{-}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^+}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=1$$
$$\lim_{x\rightarrow 0^-}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=-1$$
My question is "Which solution is right?"
 A: To begin with, it's $\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}=\left| \tan{\left( \frac{x}{2}\right)} \right|$.
Anyway, you could have answered without expanding into Taylor series.
$$ \frac{\frac{x}{2}}{\left| \tan{\left( \frac{x}{2}\right)} \right|} = \frac{\frac{x}{2}\left| \cos{\frac{x}{2}} \right|}{\left| \sin{\left( \frac{x}{2}\right)} \right|} = \frac{\frac{x}{2}}{\left| \sin{\frac{x}{2}} \right|}\left| \cos{\frac{x}{2}} \right|$$
You know that $\lim_{x \to 0}\left| \cos{\frac{x}{2}} \right| = 1$ and it is easy to prove that $\lim_{x \to 0}\frac{x/2}{\left| \sin{x/2} \right|}$ doesn't exists. In fact, its lateral limits exist, and are 1 from one side and -1 from the other!
A: Any factors in the limit that don't approach zero or infinity can simply be evaluated:
$$\begin{align}
\lim_{x\to0}\frac{x}{2}\sqrt{\frac{1+\cos x}{1-\cos x}}
&=\frac{\sqrt2}{2}\lim_{x\to0}\frac{x}{\sqrt{1-\cos x}}\\
&=\frac{\sqrt2}{2}\lim_{x\to0}\frac{x}{\sqrt{1-\cos x}}\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}\\
&=\frac{\sqrt2}{2}\lim_{x\to0}\frac{x\sqrt{1+\cos x}}{\sqrt{1-\cos^2 x}}\\
&=\frac{\sqrt2}{2}\sqrt{2}\lim_{x\to0}\frac{x}{\sqrt{1-\cos^2 x}}\\
&=\frac{2}{2}\lim_{x\to0}\frac{x}{\sqrt{\sin^2 x}}\\
&=\lim_{x\to0}\frac{x}{\left|\sin x\right|}\\
\end{align}$$
I think we can see now that the limit does not exist (that it's $1$ from the right and $-1$ from the left).
A: You are comparing two different functions because 
$$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\sqrt{\tan^2\left(\frac{x}{2}\right)} = \Bigg|\tan\left(\frac{x}{2}\right)\Bigg| \not = \tan\left(\frac{x}{2}\right) $$
