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A limit

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Evaluate [tex]\lim_{x\rightarrow0}\frac{e^{-\frac{1}{x}}}{x}[/tex]


    3. The attempt at a solution
    I tried taking logarithms and applying l'Hôpital's rule, but that just led to an expression which diverged, and as far as I understand, the limit in l'Hôpital's rule MUST exist.
     
  2. jcsd
  3. Jun 19, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    You could also use the following hand waving argument. The numerator will go to zero much faster than the denominator will so the limit will be 0.
     
  4. Jun 19, 2009 #3

    Mark44

    Staff: Mentor

    How about rewriting the original expression as
    [tex]\frac{1/x}{e^{1/x}}[/tex]
    and then using l'Hopital's rule on that?
     
  5. Jun 19, 2009 #4
    Beautiful, Mark44, that works.
     
  6. Jun 19, 2009 #5

    Mark44

    Staff: Mentor

    Wouldn't it be nice if you could just cancel the 1/x's?:biggrin:
     
  7. Jun 19, 2009 #6

    zcd

    User Avatar

    You could, in a way. Replace 1/x = u and take u to infinity instead.
     
  8. Jun 20, 2009 #7
    By the way, this limit does not exist when we approach zero from the negative numbers, because then the result is infinity/0. I guess both limits from the positive and the negative numbers have to be equal in order for the limit to exist?
     
  9. Jun 20, 2009 #8
    Nevermind.
     
    Last edited: Jun 20, 2009
  10. Jun 20, 2009 #9
    JG89: I don't get it. Why do you say that x=1 ?
     
  11. Jun 20, 2009 #10
    Ah, I made a mistake. Please disregard that post.
     
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