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Limit (Taylor expansion?)

  1. Jul 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ##x_n## be the solution to the equation
    ##\left( 1+\frac{1}{n} \right)^{n+x} = e##

    Calculate ##\lim_{n\to \infty} x_n##

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Since ##\lim_{n \to \infty} \left(1+ \frac{1}{n} \right) = e## that tells me that ##\lim_{n\to \infty} x_n = 0## but the answer in the book says it should be ##\frac{1}{2}## which I don't understand at all.
    This is also in a section about Taylorexpansion which suggest I should use that somehow. If I expand on x i get something like this
    ##\left( 1+\frac{1}{n} \right)^{n+x} = \left( 1+\frac{1}{n} \right)^{n}\cdot \left( 1+\frac{1}{n} \right)^{x} =\left( 1+\frac{1}{n} \right)^{n}\cdot (1+\ln (1+\frac{1}{n})x + O(x^2) )##
    which doesn't seen to help me and I still don't see why ##x_n\to \frac{1}{2}##.
    Some advice? :)
     
    Last edited: Jul 20, 2015
  2. jcsd
  3. Jul 20, 2015 #2

    pasmith

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    Homework Helper

    What you are given is [tex]
    \left(1 + \frac1n\right)^{n+x_n} = e.[/tex] The first step is to solve for [itex]x_n[/itex] in terms of [itex]n[/itex]. Then you can consider how you might determine the limit.
     
  4. Jul 20, 2015 #3
    Alright I think I got it.
    If I take the natural logarithm on both sides i get
    ##(n+x_n)\left( 1 + \frac{1}{n} \right) = 1##
    seperating ##x_n## i get
    ##x_n = \frac{1-n\ln (1+\frac{1}{n}) }{\ln(1+\frac{1}{n})}##
    and If I use taylor expansion ##\ln (1+ \frac{1}{n} ) = \frac{1}{n} - \frac{1}{2n^2} + O(\frac{1}{n^3})## i get (with ##n\to \infty##)
    ##x_n = \frac{1-n(\frac{1}{n}-\frac{1}{2n^2} + O(\frac{1}{n^3})) } {\frac{1}{n} + O(\frac{1}{n^2})} = \frac{\frac{1}{2n}}{\frac{1}{n}} = \frac{1}{2}##.

    Thanks! Any idea where I go wrong In my initial argument where I thought that since ##\left( 1 + \frac{1}{n} \right)^{n} = e## should mean ##x_n=0##?
     
  5. Jul 20, 2015 #4
    Edit: Accidently posted twice
     
  6. Jul 20, 2015 #5

    RUber

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    Homework Helper

    I think it might be hidden in the word "solution".
    Clearly, the limit of ##\left( 1 + \frac 1n \right)^{n} = e##, but really, any constant in the exponent would have the same limit.
    You have to solve for x in order to determine the pattern.
     
  7. Jul 20, 2015 #6
    Thank you, that makes sense!
     
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