Use the l'Hopital's rule, two times.
[tex]\lim_{x\rightarrow0} \frac{\cot(2x)}{\cot(x)}=\lim_{x\rightarrow0} \frac{2\sin^2(x)}{\sin(x)}=\lim_{x\rightarrow0} \frac{4(\cos^2(x)-\sin^2(x))}{8(\cos^2(2x)-sin^2(2x))} = \frac{1}{2}[/tex]
let me know if you have questions
