limits in trigonometry

**Hampus** · October 1st 2009, 02:18 PM

I frankly have no idea how to solve this:

limx-> pi/2 (tanx)^cotx

any help is appreciated

**pomp** · October 1st 2009, 03:21 PM

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})

**Hampus** · October 1st 2009, 03:37 PM

huh.

teacher approved trick question then (swines

)

**Hampus** · October 1st 2009, 03:39 PM

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

**DeMath** · October 1st 2009, 03:50 PM

Originally Posted by Hampus

I frankly have no idea how to solve this:

limx-> pi/2 (tanx)^cotx

any help is appreciated

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

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

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

**Hampus** · October 1st 2009, 04:11 PM

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.

**DeMath** · October 1st 2009, 04:22 PM

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.

$\ldots = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{\ln \tan x}}<br /> {{\tan x}} = {\color{blue}\exp \frac{\infty }{\infty }} = \mathop {\lim }\limits_{x \to \pi /2} \exp \frac{{{{\left( {\ln \tan x} \right)}^\prime }}}{{{{\left( {\tan x} \right)}^\prime }}}$

Do you understand this??

**Hampus** · October 1st 2009, 04:31 PM

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

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