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Proving the limit for the number e

  1. Jan 15, 2006 #1
    Ok, the problem says:

    Show that [tex]\lim_{n\rightarrow\infty} (1+\frac{x}{n})^n = e^x[/tex] for any [tex]x>0[/tex].

    I thought that I could say that y = 1+x/n...and then use the natural logarithm to narrow it down to [tex]\ln y=n\ln(\frac{x}{n})[/tex] ... I should be getting [tex]x[/tex] so that when I take it back into the original limit, I would have [tex]e^x[/tex] but I can't seem to make it that way..
  2. jcsd
  3. Jan 15, 2006 #2


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    Did you mean to make y a function of both x and n? (So that it's value is not a constant as n goes to infinity)
  4. Jan 15, 2006 #3


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    That [tex]\lim_{n\rightarrow\infty} (1+\frac{x}{n})^n = e^x[/tex] may be proven as follows:

    [Is this for a real analysis class, or a calculus class? If this is for a calculus class, ignore this:]

    by the binomial theorem,

    [tex](1+\frac{x}{n})^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}\left( \frac{x}{n}\right) ^{k} = \sum_{k=0}^{n} \frac{n!}{(n-k)!}\frac{1}{n^{k}} \frac{x^k}{k!}\rightarrow \sum_{k=0}^{\infty} \frac{x^k}{k!}=:e^{x}\mbox{ as }n\rightarrow\infty[/tex]

    since [tex] \frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty[/tex]

    although there are some uniform convergence issues to be handled when taking the limit of the above sum...
    Last edited: Jan 15, 2006
  5. Jan 15, 2006 #4
    Well taking the natural log defeats the purpose of the original proof, because the natural log depends on the value of e, in other words your using the number to find it so it has no meaning at all because you had to have already known it in the first place.
  6. Jan 15, 2006 #5


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    Xcron, do please post what definitions you have been given for e and for ex.
  7. Jan 15, 2006 #6
    Sorry, I didn't completely specify the directions of the problem. I am required to use L'Hospital's Rule to solve this limit (this is a Calculus class). I was going to us ln y in order to simplify the limit a bit.

    The definition for the number e is standard I guess...the base of the natural logarithm..
  8. Jan 15, 2006 #7


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    To prove [tex]\lim_{n\rightarrow\infty} (1+\frac{x}{n})^n = e^x[/tex]

    Let [tex]y=\lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right) ^n[/tex].
    Then [tex]\ln y=\lim_{n\rightarrow\infty} n\ln \left( 1+\frac{x}{n}\right) =\lim_{n\rightarrow\infty} \frac{\ln \left( 1+\frac{x}{n}\right)}{\frac{1}{n}} = \lim_{n\rightarrow\infty} \frac{\frac{1}{1+\frac{x}{n}}\left( -\frac{x}{n^2}\right) }{-\frac{1}{n^2}} = \lim_{n\rightarrow\infty} \frac{x}{1+\frac{x}{n}}=x[/tex]

    hence [tex]y=e^x[/tex].
    Last edited: Jan 15, 2006
  9. Jan 16, 2006 #8
    Could you please explain the mechanics of the two variables?

    I'm not sure how to go about thinking/reasoning the presence of both..
  10. Jan 16, 2006 #9


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    x is a constant, n is "the" variable for l'Hospital's Rule (tripped me up the firsty run through of the problem too.)
  11. Jan 16, 2006 #10
    Ahhhh, that was my final guess that I was making as to how they would work. It seemed like we weren't supposed to touch/work with x, and thus the main variable that we should be concerned with is n (since that was the target of the limit).
  12. Jan 17, 2006 #11


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    I like the rigourous version offered by benorin in post #3, seems like the best proof, and it uses e^x defined in terms of its Taylor series. I understood most of it. The only thing is, I can't remember why that expression ---> 1 as n ---> infinity in the last step.

    Obviously though, post #7 is exactly what the instructor wanted.

    I'm not sure if you guys will approve of this "proof", but I know another way of showing it that depends on defining e^x as the exponential function whose derivative is the function itself. So we want:

    [tex] \frac{d}{dx}(e^x) = e^x [/tex]

    [tex] \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x [/tex]

    [tex] \lim_{h \to 0} \frac{e^{x}e^h - e^x}{h} = e^x [/tex]

    [tex] \lim_{h \to 0} \frac{e^h - 1}{h}e^x = e^x [/tex]

    This requires that the number e is a base of the exponential function that satisfies the condition:

    [tex] \lim_{h \to 0} \frac{e^h - 1}{h} = 1 [/tex]

    [tex] \lim_{h \to 0} e^h - 1 = \lim_{h \to 0} h [/tex]

    [tex] \lim_{h \to 0} e^h = \lim_{h \to 0} h + 1 [/tex]

    [tex] e = \lim_{h \to 0} (h + 1)^{\frac{1}{h}} [/tex]

    Let [itex] n \equiv 1/h [/itex]

    [tex] e = \lim_{n \to \infty} (1 + \frac{1}{n})^{n} [/tex]

    change variables (x doesn't depend on n, can be considered a constant here)

    [tex] e = \lim_{n/x \to \infty} (1 + \frac{x}{n})^{n/x} [/tex]

    [tex] = \lim_{n \to \infty} (1 + \frac{x}{n})^{n/x} [/tex]

    It follows that:

    [tex] e^x = \lim_{n \to \infty} (1 + \frac{x}{n})^{(xn/x)} [/tex]

    [tex] = \lim_{n \to \infty} (1 + \frac{x}{n})^{(n)} [/tex]
  13. Jan 17, 2006 #12


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    This is not rigorous, but it works...

    This is not rigorous, but it works...

    [tex] \frac{n!}{(n-k)!} = \frac{n(n-1)(n-2)\cdots (n-(k-1))(n-k)!}{(n-k)!} =n(n-1)(n-2)\cdots (n-(k-1))[/tex]

    counting the number of terms in the above product: it goes n-0 through n-(k-1) so there are (k-1)-0+1 = k terms so we know that the leading term will be nk when the product is expanded, and hence

    [tex] \frac{n!}{(n-k)!} = n^k + \mbox{ some polynomial of degree k-1 in }n[/tex]

    or rather

    [tex] \frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty[/tex]

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