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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


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    One way to solve it is by observing that;


    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


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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


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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


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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


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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


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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


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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


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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
  22. Jun 13, 2007 #21
    Seconded-- I mean, look at log(x); its derivative is 1/x where as x's derivative is always 1.
    As you can see, log x's derivative is always less than x's derivative-- that is to say that x goes to infinity faster than log(x) would. (Equivalent to what L'Hopital would tell you I guess)
    That's one way to look at it-- but morphism would've won me over with that elegant application of the squeeze theorem.:biggrin:
  23. Jul 9, 2008 #22

    Maybe we have some misconception due to a different meaning of the word "indeterminate form".
    If you mean that [tex] \frac{\infty}{\infty} [/tex] is always an indeterminate form OK!

    It is an indeterminate form OK, but there are plenty of ways we can solve the indeterminateness.

    Also [tex] \frac{x^2}{x} [/tex] and [tex] \frac{x}{x^2} [/tex] and [tex] \frac{x}{x} [/tex] are the same type of indeterminate limits as they are
    [tex] \frac{\infty}{\infty} [/tex] but it all we know that they HAVE limit that are (respectively) [tex] \infty, 0, 1 [/tex] .

    [tex] \frac{ln(x)}{x} [/tex] tends to 0.
    I hate the De L'hopital tool but you can apply this and solve the matter yourself and see that it indeed goes to 0.

    Hope this helps.
  24. Jul 9, 2008 #23
    Another proof.
    It uses the elementry fact that [tex]e^y > y[/tex].
    When y>0 you can apply the logaritm to that disequation and obtain
    [tex]y > \ln(y)[/tex]
    This hold for all [tex]y > 0[/tex].
    Now when [tex]x \mapsto \infty[/tex] notice that
    [tex]0 < \frac{\ln(x)}{x} = \frac{2\ln(x)}{2x}
    =\frac{2\ln(\sqrt x)}{x} < \frac{2\sqrt x}{x} = \frac{2}{\sqrt x } \mapsto 0[/tex]
  25. Jul 9, 2008 #24
    lol why did you bump this thread
  26. Jul 10, 2008 #25

    Yeah! I just misunderstood everything.
    Now I read carefully there was no need to insist.
    The reason is I thought you were not sure that limit went to 0 so I used my argument, but in the end I understand you are perfecly aware and I just misunderstood the matter of the topic.
    I think I am not the only one who misunderstood.

    My apologize and see you next time!
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