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?"

You are comparing two different functions because 
$$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\sqrt{\tan^2\left(\frac{x}{2}\right)} = \Bigg|\tan\left(\frac{x}{2}\right)\Bigg| \not = \tan\left(\frac{x}{2}\right) $$
