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Derivative question

  1. May 31, 2007 #1
    Let [itex]g(x)=e^{-\frac{1}{x}} for x > 0[/itex] and [itex]g(x)=9 for x \le 0[/itex]. I want to prove that derivatives of all orders exist.

    Now I know that the only possible problem is at 0. The limit of the difference quotient from the left is obviously going to be 0. The limit from the right is going to be [itex]\frac{1}{x^2}e^{-\frac{1}{x}}[/itex].

    But I don't see how this limit exists.

    edit: I just checked on maple and it says the limit as x->0 of the above derivative is undefined. What am I doing wrong?

    Also, I figure I'm going to use induction to prove f^n, but I can't even get the case where n=1.
    Last edited: May 31, 2007
  2. jcsd
  3. May 31, 2007 #2


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    The exponent ->-oo, so the exponential goes to 0. This will be true for any inverse power of x as a coefficient.
  4. May 31, 2007 #3
    Taking the limit from both sides, when the above expression is only the correct value for f one side.
  5. May 31, 2007 #4
    Can you elaborate? As x goes to 0, -2ln(x)-1/x gives the indeterminate oo-oo. I don't see how to manipulate it to get -oo.

    I already know the left side derivative is 0. I just need to show the right side is also 0.
  6. Jun 1, 2007 #5


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    Since xln(x)->0 as x->0, my assertion follows.
  7. Jun 4, 2007 #6
    [tex]\lim_{x\rightarrow 0}\frac{1}{x^2}e^{-\frac{1}{x}}[/tex]

    To evaluate this, let [tex]y=\frac{1}{x}[/tex] and then you have

    [tex]\lim_{x\rightarrow 0}\frac{1}{x^2}e^{-\frac{1}{x}}=\lim_{y\rightarrow \infty}y^2 e^{-y}=\lim_{y\rightarrow \infty}\frac{y^2}{ e^{y}}[/tex]

    Which I think you can see does go to zero. If you still need something even more solid, you can use l'Hopital's rule twice for the y limit
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