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Puzzled by this application of L'Hospital's rule

  1. Jun 5, 2008 #1

    quasar987

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    [SOLVED] Puzzled by this application of L'Hospital's rule

    1. The problem statement, all variables and given/known data
    In an example in my textbook, they define a function f on the real line by f(0)=0 and f(x)=exp(-1/x^2) otherwise.

    They then say that we can evaluate f'(0) by L'Hospital's rule, and they write

    [tex]f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}=0[/tex]

    How did they get that?? It seems if I apply L'Hospital, I get

    [tex]\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}=\lim_{x\rightarrow 0}\frac{-2x^{-3}e^{-1/x^2}}{1}=\lim_{x\rightarrow 0}\frac{-2e^{-1/x^2}}{x^{3}}[/tex]

    and thus I'm not more advanced!
     
  2. jcsd
  3. Jun 5, 2008 #2

    rock.freak667

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    Well I guess you can say that as x tends to zero

    [itex]\frac{1}{x^2}\rightarrow \infty[/itex] faster than [itex]\frac{1}{x}[/itex]
    so that exp(-1/x^2) tends to zero before 1/x (from the denominator) can tend to infinity.
     
  4. Jun 5, 2008 #3
    [itex] \lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}= \lim_{x\rightarrow 0} \frac{x^{-1}}{e^{1/x^2}} [/itex] and now do L'Hospital's rule.
     
    Last edited: Jun 5, 2008
  5. Jun 5, 2008 #4

    Dick

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    Change the variable to s=1/x. Now you have the limit as s->infinity of s/e^(s^2). Now use l'Hopital. Which is what eok is basically saying.
     
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