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Tricky limit. taylor series?

  1. Dec 22, 2012 #1
    1. The problem statement, all variables and given/known data

    compute the following limit:

    ## \displaystyle{\lim_{x\to +\infty} x \left((1+\frac{1}{x})^{x} - e \right)} ##

    3. The attempt at a solution

    i wanted to use the taylor expansion, but didn't know what ##x_0## would be correct, as the x goes to ## \infty##.

    also, i tried to use de l'hopital's theorem but it wouldn't work.
    how can i do that?
     
  2. jcsd
  3. Dec 22, 2012 #2
    Is it supposed to be e or e^x?
     
  4. Dec 22, 2012 #3

    Hurkyl

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    Surely e; this limit would quantify, to first order, how rapidly [itex](1 + 1/x)^x[/itex] converges to its limit e.
     
  5. Dec 22, 2012 #4

    vela

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    Try rewriting the limit in terms of z=1/x.
     
  6. Dec 22, 2012 #5

    pasmith

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    [tex]x\left(\left(1 + \frac1x\right)^x - e\right) = x\left(\exp\left(x\ln \left(1 + \frac1x\right)\right) - e\right)[/tex]

    At this point, one can proceed to substitute the Maclaurin series for [itex]\ln(1 + t)[/itex] with [itex]t = x^{-1}[/itex] (we want the limit as [itex]x \to \infty[/itex], so we can assume [itex]0 < t = x^{-1} < 1[/itex]) to get
    [tex]x\ln\left(1 + \frac1x\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)x^n}[/tex]
    and then substitute that into the series for [itex]\exp(x)[/itex] to get
    [tex]\exp\left(x\ln\left(1 + \frac1x\right)\right) =
    \sum_{k=0}^{\infty} \frac1{k!} \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)x^n}\right)^k = \sum_{m=0}^{\infty} \frac{a_m}{x^m}[/tex]
    (everything here is absolutely convergent, so I can add terms in whatever order I want) so that
    [tex]x\left(\left(1 + \frac1x\right)^x - e\right) = \left(\sum_{m=0}^{\infty} \frac{a_m}{x^{m-1}} - ex\right) = x(a_0 - e) + a_1 + \sum_{m=2}^{\infty} \frac{a_m}{x^{m-1}}[/tex]
    It's clear that the series on the right tends to 0 as [itex]x \to \infty[/itex], so all you need to work out to determine the limit (if any) is [itex]a_0[/itex] and [itex]a_1[/itex]. (You'll need to consider, for each [itex]k[/itex], what the coefficients of [itex]x^0[/itex] and [itex]x^{-1}[/itex] are and add them all together. Actually you can truncate the inner series after [itex]n = 1[/itex], because including [itex]n \geq 2[/itex] doesn't give you any more terms of order [itex]x^0[/itex] and [itex]x^{-1}[/itex] then you already have.

    But that is a brute-force method and I'm sure there's a more elegant solution.
     
  7. Dec 22, 2012 #6

    SammyS

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    L'Hôpital's rule works just fine.

    [itex]\displaystyle \frac{d}{dx}\left(f(x)\right)^x=\frac{d}{dx}e^{x \ln(f(x))}[/itex]
    [itex]\displaystyle =\left(\ln(f(x))+\frac{x}{f(x)}\right)e^{x \ln(f(x))}[/itex]

    [itex]\displaystyle =\left(\ln(f(x))+\frac{xf'(x)}{f(x)} \right)\left(f(x)\right)^x[/itex]

    Fixed in Edit.
     
    Last edited: Dec 22, 2012
  8. Dec 22, 2012 #7

    vela

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    You forgot to apply the chain rule when differentiating ln(f(x)).
     
  9. Dec 22, 2012 #8

    SammyS

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    DUH !

    Thank you vela !
     
  10. Dec 23, 2012 #9
    what if i use the taylor series for ##x_0=1##?
    it should become:
    ##\displaystyle \lim_{x \to \infty}\ x [(2+(x-1)log(4)-1+o((x-1)^2)-e] = \infty * \infty = \infty ##

    is it correct?
     
  11. Dec 23, 2012 #10

    vela

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    Nope
     
  12. Dec 23, 2012 #11
    ..maybe it was just de l'hospital then, does this look ok to you?

    ##\displaystyle \lim{x \to \infty}\ \frac{1}{\frac{1}{x}} ((1+ \frac{1}{x})^x-e) = \frac{0}{0}##
    using de l'hopital
    ##\displaystyle \lim{x \to \infty}\ \frac{1}{\frac{-1}{x^2}} (x(1+\frac{1}{x})^{x-1}*\frac{-1}{x^2}##
    which is
    ## = \infty * \frac{e}{1+0} = \infty ##
     
  13. Dec 23, 2012 #12

    vela

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    You didn't differentiate correctly. See post #6. You should get a finite answer.
     
    Last edited: Dec 23, 2012
  14. Dec 23, 2012 #13
    i can't solve it with derivatives. i eep using de l'hopital but it gets worse and worse.
    i thought i could split it into:
    ##\displaystyle \lim x \to \infty\ x(1+1/x)^x - \displaystyle \lim x \to \infty\ xe ##
    and use taylor series for x=0 of the second part, which becomes:
    ##x*\displaystyle\sum\limits_{k=0}^n \frac{1}{k!} ##
    however, i can't figure out the series of the first part
     
  15. Dec 23, 2012 #14

    vela

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    You don't want to use a Taylor series about x=0. You want to be able to neglect high-order terms which you can't because the limit is for ##x \to \infty##.

    Why don't you try my suggestion of rewriting the limit in terms of z=1/x? It'll make applying L'Hopital's more straightforward, and because the limit will be for ##z \to 0##, you can use a Taylor series about ##z=0## if necessary.
     
  16. Dec 30, 2012 #15
    okay, I think I've solved it:
    ##\lim_{y \to 0}\frac{(1+y)^{\frac{1}{y}}-e}{y}=\lim_{y \to 0} \frac{e^{\frac{1}{y}log(1+y)}-e}{y}##
    I find the Maclaurin extension for log(1+y), stopping at the third power

    ##log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}##

    ##\displaystyle{\lim_{y \to 0} \frac{e^{1-\frac{y}{2}-\frac{y^2}{3}}-e }{y}}(=\frac{0}{0})##

    ## \stackrel{\text{H}}{=} \displaystyle{ \lim_{y \to 0} (-\frac{1}{2}+2\frac{y}{3})e^{1-\frac{y}{2}+\frac{y^2}{2}}}##

    ##=-\frac{e}{2}##

    should be ok now
     
  17. Dec 30, 2012 #16

    Dick

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    Yes, that's ok now. There's a couple of what are obviously just typos. But nothing important.
     
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