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Math Help - limits in trigonometry

  1. #1
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    limits in trigonometry

    I frankly have no idea how to solve this:

    limx-> pi/2 (tanx)^cotx

    any help is appreciated
    Last edited by Hampus; October 1st 2009 at 03:39 PM.
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  2. #2
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    Quote Originally Posted by Hampus View Post
    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})
    Last edited by mr fantastic; October 1st 2009 at 11:21 PM. Reason: Added tags to the website link
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  3. #3
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    huh.

    teacher approved trick question then (swines )
    Last edited by mr fantastic; October 1st 2009 at 11:21 PM.
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  4. #4
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    no, sorry, got it all wrong, its supposed to be x->pi/2
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  5. #5
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    Quote Originally Posted by Hampus View Post
    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.
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  6. #6
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    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.
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Hampus View Post
    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??
    Last edited by DeMath; October 2nd 2009 at 05:17 PM.
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  8. #8
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    yeah I was just thinking out loud.. Thanks a bunch
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