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An Interesting Limit

  1. Jun 19, 2010 #1
    I found an interesting limit:


    any ideas? :)
  2. jcsd
  3. Jun 19, 2010 #2


    Staff: Mentor

    This might be helpful.
    [tex]\frac{(1 + 1/x)^{x^2}}{e^x} = \frac{(1 + 1/x)^{x^2}}{(e^{1/x})^{x^2}} = \left(\frac{1 + 1/x}{e^{1/x}}\right)^{x^2}[/tex]

    Let y = the last expression above, then take ln of both sides, then take the limit. Try to get something that you can use L'Hopital's Rule on.

    I think I know what the limit is, but I haven't worked this all the way through.
  4. Jun 19, 2010 #3

    Gib Z

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    Homework Helper

    Remember: [tex]\lim_{x\to\infty} \left( 1+ \frac{1}{x}\right)^x = e[/tex]
  5. Jun 20, 2010 #4
    I can't seem to get to the right indeterminate form (inf/inf, 0/0) to use l'Hospital's rule.

    Can someone please try to get the full result? :)
  6. Jun 20, 2010 #5

    Gib Z

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    We can't give you answers like that.

    You could try my suggestion.

    EDIT: Perhaps my last hint was a bit too cryptic. Try taking the natural log of your limit and expand the log term into a series.
    Last edited: Jun 20, 2010
  7. Jun 20, 2010 #6


    Now take limit of both sides and write down the solution here.

  8. Jun 22, 2010 #7
    njama: if I take the limit of both sides I get [tex]ln(y)=inf*0-inf[/tex] It doesn't make sense and you can't use l'Hopital's Rule on it.

    Does anyone knows how to do it ?
  9. Jun 22, 2010 #8
    Here is a try:

    exp( [ln(1 + 1/x) - 1/x] / [1/x^2] )

    this is (I think) the correct type for l'Hopital's Rule of:

    so I differentiate:
    exp( [(x/[x+1]) + 1/x^2] / [-2/x^3] )

    and this is basically:
    exp(1/0) -> infinity

    so limit should be infinity, but it's NOT.

    I did it with changing the limit to (x goes to 0) and (x=1/x) here i could get the right answer.

    I don't understand why can't I get the right limit in the first form !
    Last edited: Jun 22, 2010
  10. Jun 22, 2010 #9


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    Staff Emeritus
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    You didn't differentiate the log term correctly. It's easiest to do if you rewrite it as

    [tex]\log \left(1+\frac{1}{x}\right) = \log \left(\frac{x+1}{x}\right) = \log (x+1) - \log x[/tex]

    and then differentiate with respect to x.
  11. Jun 22, 2010 #10
    My God, thank you very much vela !

    It was driving me crazy!

    probably the coming PDE test messed with my head a bit :)
  12. Jun 22, 2010 #11

    exp( [ln(1 + 1/x) - 1/x] / [1/x^2] )




    exp( [(-1/[x^2+x]) + 1/x^2] / [-2/x^3] ) =
    = exp( -(1/2)*(x - x^2/x+1) ) = exp( -(1/2)*(x/x+1) ) = exp(-(1/2))


    Which is the correct answer !
  13. Jun 22, 2010 #12


    Staff: Mentor

    Use njama's suggestion to write
    [tex]lim~ln y = x^2 ln(1 + 1/x) - x = lim~x^2(ln(1 + 1/x) - 1/x) = lim~\frac{ln(1 + 1/x) - 1/x}{1/x^2}[/tex]

    (All limits taken as x --> infinity.)

    As x --> infinity, the numerator --> 0 as does the denominator, so L' Hopital's Rule applies.
    [tex]= lim~\frac{\frac{1}{(1 + 1/x)} \cdot (-1/x^2) + 1/x^2}{-2/x^3} [/tex]

    Evalulate the last limit and keep in mind that this is lim ln y, not lim y.
  14. Jun 22, 2010 #13
    Take the log

    x^{2} \, \ln(1 + \frac{1}{x}) - x = x \, \left(x \ln(1 + \frac{1}{x}) - 1\right) \\

    = \frac{x \, \ln(1 + \frac{1}{x}) - 1}{\frac{1}{x}}

    Notice that:

    \lim_{x \rightarrow \infty} \left[ x \ln(1 + \frac{1}{x})\right] \stackrel{t = 1\x}{=} \lim_{t \rightarrow 0} \frac{\ln(1 + t)}{t} = \frac{0}{0} = \lim_{t \rightarrow 0} \frac{1}{1 + t} = 1

    so, you have an indeterminate form 0/0. Use L'Hospital's Rule.

    Don't forget to exponentiate back for the final result.
  15. Jun 23, 2010 #14

    Thank you, but as you see I already did it !
    Limit = exp(-(1/2))
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