# Evaluate the limit for the following if exists. $\;\lim\limits_{x \to 0} \large\frac{\cot x}{\cot 2x}$

Toolbox:
• L'Hopital's rule: Let $f$ and $g$ be continous real valued functions defined on the closed interval $[a,b], f,g$ be differentiable on $(a,b)$ and $g'(c) \neq 0$
• Then if $\lim\limits_{x \to c}\; f(x)=0, \lim \limits_{x \to c}\; g(x)=0$ and
• $\lim\limits_{x \to c} \large\frac{f'(x)}{g'(x)}$$=L it follows that • \lim \limits_{x \to c} \large\frac{f(x)}{g(x)}$$=L$
$\;\lim\limits_{x \to 0} \large\frac{\cot x}{\cot 2x}$ is of the form $\large\frac{\infty}{\infty}$
This can be rewritten as $\lim \limits_{x \to 0} \large\frac{\Large\frac{1}{\cot 2x}}{\Large\frac{1}{\cot x}}$
$\;\lim\limits_{x \to 0} \large\frac{\tan 2x}{\tan x}$ which is of the form $\large\frac{0}{0}$
Step 2:
We have, by applying L'Hopital's rule
$\;\lim\limits_{x \to 0} \large\frac{\cot x}{\cot 2x}$$=\lim\limits_{x \to 0}\large\frac{2 \sec ^2 2x}{\sec ^2 x}$$=2$

edited Jul 29, 2013 by meena.p