1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

  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


    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?

  6. Sep 24, 2012 #5


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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


    Look at [itex]\displaystyle \frac{\cos(t)}{\sin(t)}[/itex] as [itex]\displaystyle \cos(t)\text{csc}(t)\ .[/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook