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Limits of derivatives of an exponential

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the lowest derivative order for which the limit towards 0+ of the nth order derivative of f is nonzero (or otherwise does not exist). f = [itex]e^{\frac{-1}{x^{2}}}[/itex]

    2. Relevant equations

    [itex]lim_{x\rightarrow0+}\frac{d^{n}}{dx^{n}}e^{\frac{-1}{x^{2}}}[/itex]

    3. The attempt at a solution

    [itex]lim_{x\rightarrow0+}\frac{d}{dx}e^{\frac{-1}{x^{2}}}[/itex] = 0

    [itex]lim_{x\rightarrow0+}\frac{d^{2}}{dx^{2}}e^{\frac{-1}{x^{2}}}[/itex] = 0

    [itex]lim_{x\rightarrow0+}\frac{d^{3}}{dx^{3}}e^{\frac{-1}{x^{2}}}[/itex] = 0
     
  2. jcsd
  3. Oct 31, 2012 #2

    SammyS

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    Use the chain rule.

    [itex]\displaystyle \frac{d}{dx}e^{-1/x^2}= \frac{2e^{-1/x^2}}{x^3}\ .[/itex]
     
    Last edited: Oct 31, 2012
  4. Oct 31, 2012 #3
    I tried that at the first three orders and I still had the limit of these derivatives towards 0+ as 0.
     
  5. Oct 31, 2012 #4

    SammyS

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    Right!

    I get zero for the fourth derivative also.

    I don't see how it will ever be anything else, no matter how high the order of the derivative, but I haven't proved that to myself.

    .
     
  6. Oct 31, 2012 #5

    HallsofIvy

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    Any derivative is [itex]e^{-1/x^2}[/itex] over a polynomial and its limit as x goes to 0 will always be 0.
     
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