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Proof of the value of e without using L'Hopital's rule?

  1. Nov 30, 2009 #1
    The proofs of [tex]y = \displaystyle\lim_{x \to\infty} (1 + 1/x)^{x} = e[/tex] that I have seen basically involve taking the natural log of both sides and getting the equation in a form where L'Hopital's rule can be applied. The problem I have with this is that it requires taking the derivative of the natural log function, and the proof of the derivative of that requires prior knowledge that the limit above converges to e! It all seems a bit circular - is there a proof of the limit of the compound interest function that doesn't involve using L'Hopital's rule?
  2. jcsd
  3. Nov 30, 2009 #2
    If you expand it by binomial theorem,

    [tex](1+\frac{1}{n})^n=1^n+\begin{pmatrix}n\\ 1\end{pmatrix}1^n\frac{1}{n}+\begin{pmatrix}n\\ 2\end{pmatrix}1^{n-1}(\frac{1}{n})^2+..+\begin{pmatrix}n\\ n\end{pmatrix}(\frac{1}{n})^n[/tex]
    Now we look k. term

    [tex]\begin{pmatrix}n\\ k\end{pmatrix}\frac{1}{n^k}=\frac{1}{k!}\frac{n(n-1)(n-2)..(n-k+1)}{n^k}\Rightarrow \lim_{n\rightarrow \infty}\begin{pmatrix}n\\ k\end{pmatrix}\frac{1}{n^k}=\frac{1}{k!}\lim_{n\rightarrow \infty}\frac{n(n-1)(n-2)..(n-k+1)}{n^k}=\frac{1}{k!}[/tex]

    So our first serie equals to [tex]1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...=\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n=e[/tex]
  4. Dec 1, 2009 #3
    [tex] log(1 + 1/n)^n = nlog(1 + 1/n) = n \int_1^{1 + 1/n} \frac{du}{u} = n*(1/n)*c [/tex] where 1 < c < 1 + 1/n. Note that I used the MVT of integral calculus.

    [tex] n*(1/n)*c = c [/tex] which tends to 1 as n goes to infinity since 1 < c < 1 + 1/n. Thus [tex] \lim_{n \rightarrow \infty} log(1 + 1/n)^n = 1 [/tex] so by the continuity of the exponential function, [tex] \lim_{n \rightarrow \infty} e^{log(1+1/n)^n} = \lim_{n \rightarrow \infty} (1 + 1/n)^n = e^1 = e [/tex]
  5. Dec 1, 2009 #4
    Thanks, everyone!
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