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Proof: limit=0 for any positive integer n

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex] \lim_{x\to0}\frac{e^\frac{-1}{x^2}}{x^n}=0 [/tex] for any positive integer n.

    2. Relevant equations



    3. The attempt at a solution
    I've tried using a combination of induction and l'hopital's rule to no avail. Perhaps im over complicating it?

    All help is appreciated

    Thanks in advance
     
  2. jcsd
  3. May 12, 2012 #2

    micromass

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    Try to write it as

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

    and apply l'Hopital now.
     
  4. May 12, 2012 #3
    I applied l'hopitals and simplified a bit and now ive got

    [tex]\frac{n}{2}\lim_{x\to0}\frac{x^2}{e^\frac{1}{x^2}x^n}=0 [/tex]

    now what?
     
  5. May 12, 2012 #4

    micromass

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    Simplify more and apply induction.
     
  6. May 12, 2012 #5

    hunt_mat

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    Write [itex]e^{-\frac{1}{x^{2}}}[/itex] as a power series and then perform the limiting process.
     
  7. May 12, 2012 #6
    so im trying to show its true for n=k+1 assuming n=k

    i.e i need to show that [tex] \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=0 [/tex]

    assuming that [tex] \lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=0 [/tex]

    im not sure how to manipulate this now?
     
  8. May 12, 2012 #7
    Split the first term into the one you know(assumed), and its factor. Then apply basic limits.
     
  9. May 12, 2012 #8
    so,

    [tex] \lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}} [/tex]

    but i cant simply say that
    [tex]\lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}}= \lim_{x\to0}{x^{-1}} × \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}} [/tex]

    because [tex]\lim_{x\to0}{x^{-1}}[/tex] doesnt exist, right?
     
  10. May 12, 2012 #9

    micromass

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    You did

    [tex]x^{2-k}=x^{-1}x^{1-k}[/tex]

    But this is false.
     
  11. May 12, 2012 #10
    Uhh, this is wrong. Does multiplying x-1 and x1-k give x2-k??
     
  12. May 12, 2012 #11
    oh sorry, that should be the other way around
     
  13. May 12, 2012 #12
    it should be
    [tex] \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{2-k}}{e^\frac{1}{x^2}} [/tex]
     
  14. May 12, 2012 #13

    micromass

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    Uuuh, or slightly more useful

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

    Can you split up the limits now?
     
  15. May 12, 2012 #14

    Dick

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    I would take the absolute value and then look at the log of your expression.
     
  16. May 12, 2012 #15
    oh! how didn't I see this before

    thanks so much <3
     
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