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L'Hopitals Rule problem

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following limit

    [tex] \lim_{t→2π\mathbb{N}}\frac{sin(t)}{1-cos(t)} [/tex] for all natural numbers [itex]\mathbb{N}[/itex].

    2. Relevant equations



    3. The attempt at a solution
    Plugging in [itex] 2π\mathbb{N} [/itex] gives us 0 for both the numerator and denominator. Thus, we differentiate top and bottom to get [tex] \lim_{t→2π\mathbb{N}}\frac{cos(t)}{sin(t)} [/tex] which evaluates to [itex]\frac{1}{0}[/itex] which does not exist.

    Am I correct?

    BiP
     
  2. jcsd
  3. Sep 24, 2012 #2
    Use l'hopital's rule the second time to get -sin(t)/cos(t) and evaluate that to see what you get. But "1/0" is the correct way of writing 1/0 and that is infinity if you're dealing with limits.
     
  4. Sep 24, 2012 #3

    Mark44

    Staff: Mentor

    You probably should elaborate a bit. Here are two examples of limits where the denominator approaches 0.

    $$ \text{1}~\lim_{x \to 0} \frac{1}{x}$$
    $$ \text{2}~\lim_{x \to 0} \frac{1}{x^2}$$

    For the first, the limit does not exist in any sense, since the left- and right-side limits are different. For the second, the limit also does not exist, in the sense of being a finite number, but we say that the limit is ∞.

    For your problem, is the limit ∞ or -∞, or does it fail to exist at all?
     
  5. Sep 24, 2012 #4
    How can I show which one of these is the case in my problem?

    BiP
     
  6. Sep 24, 2012 #5

    Mark44

    Staff: Mentor

    In the expression cos(t)/sin(t), when t is close to k(2##\pi##), the numerator is close to 1, and the denominator is close to 0. The key is to determine what the denominator is doing when t is also close to this value. If the denominator changes sign, then the limit is undefined. If the denominator doesn't change sign, then the limit is either ∞ or -∞.
     
  7. Sep 24, 2012 #6

    SammyS

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    Staff Emeritus
    Science Advisor
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    You can't use L'Hôpital's rule a second time. after on application, you don't have an indeterminate form.

    Bipolarity,

    Look at [itex]\displaystyle \frac{\cos(t)}{\sin(t)}[/itex] as [itex]\displaystyle \cos(t)\text{csc}(t)\ .[/itex]
     
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