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L'hopitals theorem (Limits)

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem is:
    A function in the form of f(x)/g(x) is given: [x^2 - pi^2/4 ]/tan^2(x) as x [itex]\rightarrow[/itex] pi/2.


    2. Relevant equations



    3. The attempt at a solution

    Surely I can't solve this question using L'hopital's theorem since it's applicable to indeterminate in the form of 0/0, infinity/infinity, 0 (+- infinity),etc. The above function gives 0/infinity... or am I missing something.. Could I perhaps use trig identity for the denominator ?


    Another problem: Also would differentiating [itex]x^{1/x}[/itex]/x-1 as x -> 1 give me an answer of e^1 only ?

    Here's how I attempted to solve this problem:
    I made y = x^1/x which gave me lny = lnx/x ( which's 0/1) that eventually gives me lny = 1 ? :s
     
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2
    You are missing that [itex]\frac{0}{\infty}[/itex] is not a form of indetermination:
    [itex]\frac{0}{\infty}=0[/itex]
     
  4. Oct 12, 2011 #3

    SammyS

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    0/∞ shouldn't pose a problem.

    DiracR beat me !
     
  5. Oct 12, 2011 #4
    So I can't solve this problem using the aforementioned theorem ?
     
  6. Oct 12, 2011 #5
    There is no reason to evaluate this limit using a theorem. You can solve it by "substitution", and the form you get is not indeterminate.
     
  7. Oct 12, 2011 #6
    Thanks for being patient with me but when you say 'substitution' are you referring to trig. identity = sec^2(x) =1+ tan^2 (x) ?
    I was handed over this question sheet, where we have to basically evaluate functions using 'hopitals theorem.
     
  8. Oct 12, 2011 #7
    no, with substitution I mean simply to put the value x=pi/2 in the limits.

    [itex]\lim_{x\rightarrow\frac{\pi}{2}}\frac{x^2-\frac{\pi^2}{4}}{\tan(x)^2}=\frac{x^2-\frac{\pi^2}{4}}{\tan(x)^2}|_{x=\frac{\pi}{2}}[/itex][itex]=\frac{0^{\pm}}{+\infty}=0^{\pm}[/itex]

    This is the way to solve this limit...
     
    Last edited: Oct 12, 2011
  9. Oct 12, 2011 #8
    Sorry but I am further confused. :(
     
  10. Oct 12, 2011 #9
    I edited my previous post, get a glance at it.

    The thing is, you don't need any theorem.

    The way to solve this limit is the first thing you should have studied, that is: trying substituting the value and see what happens.
     
  11. Oct 12, 2011 #10
    Yes, when I sub. in the values I get the following fraction : 0/infinity..
    I know if it had been 0/0 or infinity/ infinity.. then I could have used L'hopitals theorem

    EDIT: I suppose that's the answer since it's not in the form of 0/0 or infinity/ infinity.

    Thanks for your help DiracRules! btw. dirac indeed does ;)
     
    Last edited: Oct 12, 2011
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