I frankly have no idea how to solve this:
limx-> pi/2 (tanx)^cotx
any help is appreciated
Doesn't exist. The function is periodic, here is a plot of it from x=0 to x=20:
http://www.wolframalpha.com/input/?i...,+{x,+0,++20})
$\displaystyle \mathop {\lim }\limits_{x \to \pi /2} {\left( {\tan x} \right)^{\cot x}} = \mathop {\lim }\limits_{x \to \pi /2} \exp \ln {\left( {\tan x} \right)^{\cot x}} =$
$\displaystyle = \mathop {\lim }\limits_{x \to \pi /2} \exp \cot x\ln \tan x = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{\ln \tan x}}{{\tan x}} =$
$\displaystyle = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{{{\left( {\ln \tan x} \right)}^\prime }}}{{{{\left( {\tan x} \right)}^\prime }}} = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{1}{{\tan x}} = {e^0} = 1.$
doesn't ln 'end' in a horizontal asymptote (in which case limx->pi/2 ln(tanx) =/= tanx)? the derivative of ln should move towards zero as x approaches infinity , which gives me the feeling that the function is limited. If this is the case then l'Hopitals rule shouldn't be applicable to the case in question, unless I've gotten something wrong somewhere.
$\displaystyle \ldots = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{\ln \tan x}}
{{\tan x}} = {\color{blue}\exp \frac{\infty }{\infty }} = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{{{\left( {\ln \tan x} \right)}^\prime }}}{{{{\left( {\tan x} \right)}^\prime }}}$
Do you understand this??