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Lim x^{1/x}, x-> infty

  1. Jun 12, 2007 #1
    This seems to be an obvious problem, but for some reason I'm stumped. So what is the limit of x^{1/x} as x approaches infinity? I know that limit of (1+1/x)^{x} = e (as x-> infty), and limit of (1+x)^{1/x} = e (as x-> 0), but this is slightly different.

    Any help is appreciated :)
     
  2. jcsd
  3. Jun 12, 2007 #2
    its an indeterminate form, do l'hospitale
     
  4. Jun 12, 2007 #3

    mathman

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    One way to solve it is by observing that;

    x1/x=elnx/x.

    Since lnx/x -> 0 as x ->oo, the answer you want is 1.
     
  5. Jun 12, 2007 #4
    no lim lnx/x -> oo/oo as x->oo , you still get an indeterminate form. but i realize applying l'hospitale directly to the first expression is pointless
     
    Last edited: Jun 12, 2007
  6. Jun 12, 2007 #5

    Office_Shredder

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    well, no ice, lim lnx/x as x goes to infinity is just 0. The fact that you can't split it up doesn't change that fact
     
  7. Jun 12, 2007 #6
    what? how is that even true? since when is the ln function bounded?
     
  8. Jun 12, 2007 #7

    morphism

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    We're dealing with ln(x)/x as a single entity, not just ln(x)...
     
  9. Jun 12, 2007 #8
    Thanks guys for clearing this up. It seems the main idea is to rewrite x^{1/x} into the form e^{1/x * ln x}, so that the exponent -> 0 as x -> infinity, and thus e^{0} = 1.
     
  10. Jun 12, 2007 #9
    im very aware of what we're dealing with

    [tex] \frac{ln(x)}{x} [/tex] does not go to zero at infinity, it goes to undeterminate.
     
  11. Jun 12, 2007 #10

    morphism

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    No. It goes to zero.

    Here's a proof that doesn't use L'Hopital.

    If x>0, then [itex]\ln x < x[/itex], so [itex](\ln{\sqrt{x}}) / \sqrt{x} < 1[/itex].

    So if x>1, then:
    [tex]0 < \frac{\ln x}{x} = \frac{2\ln{(\sqrt{x})}}{\sqrt{x}\sqrt{x}} < \frac{2}{\sqrt{x}}[/itex]

    A quick application of the squeeze/sandwich theorem now proves that ln(x)/x goes to zero.

    And what exactly do you mean when you say "goes to undeterminate" anyway? The very fact that you use L'Hopital on lim ln(x)/x to get lim ln(x)/x = lim 1/x = 0 proves that the limit is 0.
     
    Last edited: Jun 12, 2007
  12. Jun 12, 2007 #11
    i don't understand your proof, but what i mean is that as the expression stands [tex]\frac{ln(x)}{x}[/tex] goes the the undeterminate form [tex]\frac{\infty}{\infty}[/tex] and that l'hospitale has to be applied to see that the limit is zero. the initial point was that the rewriting of the expression using logs didn't save him any steps because still had to apply l'hospitale to figure out the limit. i then realized that applying l'hospitale to original expression was pointless.
     
  13. Jun 12, 2007 #12

    morphism

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    The point is: you do not need L'Hopital to do that limit.
     
  14. Jun 12, 2007 #13
    you still used squeeze theorem? direct evaluation leads to an indeterminate form. who cares how you rectify the problem, obviously the limit of the expression as it stands is not just 0

    seriously dude rewrite your proof in tex so i can understand it
     
  15. Jun 12, 2007 #14

    morphism

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    How can you still say that?!?!?! :confused:
     
  16. Jun 12, 2007 #15
    omg dude seriously its not that difficult, what is the limit of [tex]\frac{1}{x}[/tex] as x approaches infinity? in the same sense as that limit is zero the other limit is not, you need an auxiliary scheme to prove it is zero, where this one is obvious by direct inspection. this is seriously down to semantics now. the end.
     
  17. Jun 12, 2007 #16

    morphism

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    You obviously don't know what the definition of a limit is.

    [tex]\lim_{x \to \infty} \frac{\ln{x}}{x} = 0[/tex]

    It's absolutely NOT [itex]\frac{\infty}{\infty}[/itex].


    PS. I tex'd my proof.
     
  18. Jun 12, 2007 #17
    yea terrible proof, and i'll just wait for someone else to refute you because i'm exasperated
     
  19. Jun 12, 2007 #18

    morphism

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    Right. OK. :rolleyes:
     
  20. Jun 13, 2007 #19
    It makes perfect sense to me. What exactly is wrong with it?
     
  21. Jun 13, 2007 #20
    at second glance the proof is fine but he still had to use an auxiliary theorem to prove the limit is 0
     
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