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L'Hopitals's Rule?

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the lim as x approaches infinity of [tex] \frac{sin x}{x-\pi} [/tex]

    3. The attempt at a solution
    This was in the section for L'Hopital's Rule, but if you substitute infinity in the functions you don't get an indeterminate form. I don't know what to do next.
  2. jcsd
  3. Feb 16, 2010 #2
    Why not? With infinity, any number added to or subtracted from it is negligible. The pi becomes negligible and vanishes as x tends to infinity.
  4. Feb 16, 2010 #3
    Yes but what about sinx?
  5. Feb 16, 2010 #4
    I'm not sure how to use L'Hopital's rule but I can use common sense. The maximum magnitude for sinx is 1. so say that we have the maximum value. Then there is infinity - pi at the bottom. which is going to be infinity. then you have 1 / infinity. giving 0.
  6. Feb 16, 2010 #5
    Oops, an oversight on my part there. I apologise.

    Hmm....the standard method is by the "Sandwich theorem" or "Squeeze theorem":

    [tex]-\frac{1}{x - \pi} \leq \frac{sin\,x}{x - \pi} \leq \frac{1}{x -\pi}[/tex]

    Taking the limits as x tends to infinity of the upper and lower bounds gives 0, so this necessarily implies that your limit is 'squeezed' to zero as well.

    It would appear that this question is not L'Hopitals; sin x is a periodic oscillating function.
    Last edited: Feb 16, 2010
  7. Feb 16, 2010 #6
    Oh! I completely forgot about that theorem. Good call. It's like they show it to you in the way beginning of calc I but you don't really use it that much.
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