# Limit using L'Hopital rule

$\lim\limits_{x \to 0}\log (\cot x)^{\tan x}$

I was tryng to evaluate this limit, but i really cannot do it using L'Hopital rule.

Its very often of the form Which doesn't correspond to this

• Is that limit supposed to be $x \to 0^+$ (from the right) instead of $x \to 0$ (from both sides)? If not, then $\cot x < 0$ for small negative values of $x$, which makes $\ln(\cot x)$ undefined. – JimmyK4542 Sep 9 '14 at 4:43

## 1 Answer

First, take the natural log and do a bit of manipulation:

$\displaystyle\lim_{x \to 0}\ln\left[(\ln \cot x)^{\tan x}\right] = \lim_{x \to 0}\tan x \ln(\ln(\cot x)) = \lim_{x \to 0}\dfrac{\ln(\ln(\cot x))}{\cot x}$

Now, you can use L'Hopital's rule if you wish, to evaluate this limit.

Finally, note that if $\displaystyle\lim_{x \to 0}\ln\left[(\ln \cot x)^{\tan x}\right] = L$, then $\displaystyle\lim_{x \to 0}(\ln \cot x)^{\tan x} = e^L$

• Thanks a lot,This single idea helped me solve a lot of similar problems. – Adk Sep 9 '14 at 4:55