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Nasty Limit

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate: [tex]\lim_{\ x \to 0^+} \frac{\ e^{-1/x}}{x}[/tex]

    2. Relevant equations
    L'Hopitals' Rule maybe? Taylor series?

    3. The attempt at a solution

    Before I did anything, my guess was this tends to 0, as exponentials decay faster than rational funtions.

    The limits as -1/x tends to 0 from the right is [tex]-\infty[/tex]
    [tex]\lim_{\ x \to 0^+} \ e^{-1/x} = 0[/tex]

    and ofcourse the limit of x as x tends to 0 is 0. So it is of indeterminate form 0/0
    Using L'Hopital's Rule:

    [tex]\lim_{\ x \to 0^+} \frac{\ e^{-1/x}}{x^2}[/tex]

    And that is still of indeterminate form 0/0. In fact, you can see that using L'Hopitals Rule will always generate an indeterminate form 0/0. Taylor series are no better: after the exapansion you get
    [tex]\frac{1}{x^2} +\frac{1}{2x^5} +...[/tex] or something of that nature, point is, it diverges.
    Help please?
  2. jcsd
  3. Nov 1, 2007 #2

    D H

    Staff: Mentor

    Look at it from the other end of the reals: Let [itex]u=1/x[/itex]. Then

    [tex]\frac{e^{-\;1/x}}{x} = ue^{-u} = \frac{u}{e^u}[/tex]
  4. Nov 1, 2007 #3


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    You might take the logarithm of the fraction, and try to evaluate:
  5. Nov 1, 2007 #4
    try subs x = 1/ ln y and then try L'hopital..
  6. Nov 1, 2007 #5
    thats correct.
  7. Nov 1, 2007 #6
    Thank-you everyone for your answers, but I do not like substitutions, like put x=(...),
    is there any way to solve it without the substitution?

    I do not get this substitution, or any other substitution for that matter! Why, and how, look at it from the other end? Or is it merely a substitution one must use to solve the problem, but has no real meaning. Am I making sense?
  8. Nov 1, 2007 #7

    D H

    Staff: Mentor

    Making substitutions is a very important tool. You had better learn to love them if you want to proceed far in mathematics.

    In the case of my example, u=1/x, the substitution changes an indeterminate expression of the form [itex]0/0[/itex] as [itex]x\to0[/itex] to an indeterminate expression f the form [itex]\infty/\infty[/itex] as [itex]u\to\infty[/itex]. The advantage of the substitution is that using L'Hopital's Rule yields a value.
  9. Nov 1, 2007 #8
    Ok, thank you again everyone, otherwise I would have really struggled with this one!
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