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L'Hospital's Rule

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data


    lim x^(1/x)
    x->inf


    3. The attempt at a solution

    So, as I understand, there are several ways of solving it.
    One of them, is not by using l'Hospital Rule because it is not an undeterminate function. OR IS IT?
    If we rewrite the function as:
    e^[lnx * (1/x)]

    We get exponents: 1/infinity=0 and lnx=infinity, multiply both, it's 0.

    Hence, e^0=1
     
  2. jcsd
  3. Jul 10, 2010 #2
    Careful, you have the right answer, but you can't use infinity like a number and say 0 × ∞ = 0. If instead you had limx→∞ x/ln(x), that would be x→∞ and 1/lnx → 0, so 0 × ∞, but that limit isn't 0, it's infinity.

    [tex]\lim_{x\rightarrow \infty} x^{1/x} = \lim_{x\rightarrow \infty} e^{\ln x \cdot (1/x)} = \lim_{x\rightarrow \infty} e^\frac{\ln x}{x} = e^{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}[/tex]

    So now you need to find lim×→∞ (lnx)/x, which you just did. :smile:
     
  4. Jul 10, 2010 #3

    Mark44

    Staff: Mentor

    This is an indeterminate form, but L'Hopital's Rule is not applicable for this particular indeterminate form.
    [[itex]\infty * 0][/itex] is another indeterminate form, so you can't say with certainty that the limit of the exponent is 0.
    Let y = ln(x)/x. Now take the natural log of both sides, and then take the limit of both sides. Presumably your text has an example that uses this approach.
     
  5. Jul 11, 2010 #4

    Well, your last one is essentially the same as the original when lnx/x. it looks like an indeterminate function.
     
    Last edited: Jul 11, 2010
  6. Jul 11, 2010 #5
  7. Jul 11, 2010 #6
    Thank you!
     
  8. Jul 11, 2010 #7

    Mark44

    Staff: Mentor

  9. Jul 11, 2010 #8
    Well, this shouldn't be so complicated cause this is just a beginner calculus.
     
  10. Jul 11, 2010 #9
    OK, so what I do in this problem is I take a derivative twice or three times... in order to solve this indeterminate function?
     
  11. Jul 11, 2010 #10
    As long as it's 0/0 or ∞/∞, you can keep using l'Hopital's rule until you get a different form that may lead to a limit or infinity.
     
  12. Jul 11, 2010 #11
    Or for those who like it more artisanal without l'Hôpital, define:

    [tex]a_n = n^{1/n} - 1[/tex]

    Note that it suffices to prove that [tex]a_n \to 0[/tex].

    From definition & Newton's binomium: [tex]n = (a_n + 1)^n = \sum \left( \frac{n}{i} \right) a_n^i \geq \frac{n(n-1)}{2} a_n^2^[/tex]

    From which directly follows that:

    [tex]\frac{2}{n-1} \geq a_n^2 \geq 0[/tex] QED

    EDIT: oh just noticed how the original poster stressed the use of l'Hôpital, my apologies, this renders my post useless, wasn't aware
     
    Last edited: Jul 11, 2010
  13. Jul 11, 2010 #12
    That is a very important note to make. That is what I was not sure of. Thank you very much.
     
  14. Jul 11, 2010 #13
    Thanks a lot for your help!

     
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