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L' Hopital's rule

  1. Feb 7, 2013 #1
    [itex]^{}[/itex]1. The problem statement, all variables and given/known data

    lim [itex]_{x\rightarrow\infty}[/itex] [itex]\left(cos\frac{1}{x}\right)[/itex][itex]^{x}[/itex]

    find the limit using L' Hopital's Rule

    2. Relevant equations


    3. The attempt at a solution

    1[itex]^{\infty}[/itex] >> change form to 0/0 by taking ln both sides

    ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] x ln [itex]\left(cos\frac{1}{x}\right)[/itex]
    ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] [itex]\frac{ln \left(cos\frac{1}{x}\right)}{\frac{1}{x}}[/itex] ===> 0/0

    Apply L' Hopital's Rule

    ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] [itex]\frac{\frac{1}{cos\frac{1}{2}}(-sin\frac{1}{x})}{\frac{-1}{x^{2}}}[/itex] ===> 0/0

    the next step do i have to apply L' Hopital's Rule again or just rearrange the fraction and take exponential both sides.

    please help thank you
     
  2. jcsd
  3. Feb 7, 2013 #2
    The math would come out much nicer if you define z=1/x and take limit z→0 instead, before using L'Hopital
     
  4. Feb 7, 2013 #3
    You continue applying L'Hopital's until the limit becomes a determinate value. "0/0" is still indeterminate.
     
  5. Feb 7, 2013 #4

    Dick

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    Science Advisor
    Homework Helper

    You didn't finish the chain rule differentiation in l'Hopital. You still have a derivative of 1/x to take in the numerator.
     
  6. Feb 7, 2013 #5
    Hi all thanks for the comments :)
    yes I didn't finish the chain rule yet. :P it will end up with lim -tan (1/x) which is equal 0 and e^0 = 1 thanks you guys again
     
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