<s>[INST] 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?"
 [/INST] 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).
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