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Any other way?

  1. Dec 19, 2005 #1

    dextercioby

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    of doing this

    [tex] \lim_{x\rightarrow +\infty }\cos^{x}\frac{\pi}{x} [/tex]

    than using

    [tex] \cos x \simeq 1- \frac{x^{2}}{2} \mbox {when "x" goes to zero}[/tex]

    ?

    Daniel.
     
  2. jcsd
  3. Dec 19, 2005 #2

    arildno

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    Hmm.. shouldn't it work to evaluate the limit of the expression's logarithm first?
    Just an idea..
     
  4. Dec 19, 2005 #3

    dextercioby

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    Ouch, that hurt :((

    Daniel.
     
  5. Dec 19, 2005 #4

    TD

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    That would work, together with applying L'Hopital once.

    [tex]\mathop {\lim }\limits_{x \to \infty } \left( {\cos \frac{\pi }{x}} \right)^x = \mathop {\lim }\limits_{x \to \infty } \exp \left( {x\ln \left( {\cos \frac{\pi }{x}} \right)} \right) = \exp \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{{1/x}}\ln \left( {\cos \frac{\pi }{x}} \right)} \right) = \exp \mathop {\lim }\limits_{x \to \infty } \left( { - \pi \tan \frac{\pi }{x}} \right) = e^0 = 1[/tex]

    Edit: didn't see you already replied :smile:
     
    Last edited: Dec 19, 2005
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