1. ## limits in trigonometry

I frankly have no idea how to solve this:

limx-> pi/2 (tanx)^cotx

any help is appreciated

2. Originally Posted by Hampus
I frankly have no idea how to solve this:

limx-> +oo (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})

3. huh.

teacher approved trick question then (swines )

4. no, sorry, got it all wrong, its supposed to be x->pi/2

5. Originally Posted by Hampus
I frankly have no idea how to solve this:

limx-> pi/2 (tanx)^cotx

any help is appreciated
$\mathop {\lim }\limits_{x \to \pi /2} {\left( {\tan x} \right)^{\cot x}} = \mathop {\lim }\limits_{x \to \pi /2} \exp \ln {\left( {\tan x} \right)^{\cot x}} =$

$= \mathop {\lim }\limits_{x \to \pi /2} \exp \cot x\ln \tan x = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{\ln \tan x}}{{\tan x}} =$

$= \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{{{\left( {\ln \tan x} \right)}^\prime }}}{{{{\left( {\tan x} \right)}^\prime }}} = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{1}{{\tan x}} = {e^0} = 1.$

6. 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.

7. Originally Posted by Hampus
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.
$\ldots = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{\ln \tan x}}
{{\tan x}} = {\color{blue}\exp \frac{\infty }{\infty }} = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{{{\left( {\ln \tan x} \right)}^\prime }}}{{{{\left( {\tan x} \right)}^\prime }}}$

Do you understand this??

8. yeah I was just thinking out loud.. Thanks a bunch