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Derive the limit of an expression (1+1/n)^n

  1. Mar 6, 2006 #1
    i know that that the limit of this expression is e, but i dont know how to derive it...
    i will be really grateful if someone could help me solve this problem
     
  2. jcsd
  3. Mar 6, 2006 #2

    quasar987

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    You would need a definition of e other than that. Otherwise, you can just show that the limit converge and call it e.

    So what definition of e do you use?
     
  4. Mar 6, 2006 #3

    HallsofIvy

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    If a is any positive number then
    [tex]lim_{h\rightarrow0}\frac{a^{x+h}-a^x}{h}= lim_{h\rightarrow0}\frac{a^xa^h- a^x}{h}[/tex]
    [tex]= a^x lim{h\rightarrow0} \frac{a^h- 1}{h}[/tex]

    In other words, ax has the nice property that its derivative is just a number (that limit) time ax itself.

    We define e to be the number such that the derivative of ex is just ex- in other words so that that
    [tex]lim_{x\rightarrow0}\frac{e^h-1}{h}= 1[/tex].

    That means that for small h,
    [tex]\frac{e^h-1}{h}[/tex]
    is approximately 1.

    Let n= 1/h (more accurately the next integer larger than 1/h). Then
    [tex]n\left(e^{\frac{1}{n}}-1\right)= 1[/tex]
    approximately for large n with 1 being the limit as n goes to infinity.

    Solve that for e: e is approximately
    [tex]\left(1- \frac{1}{n}\right)^n[/tex]
    with the limit being e as n goes to infinity.
     
  5. Mar 6, 2006 #4
    i want to show that 2,718... [which is e] is the limit converge of that expression, i've posted my question in this 'strange' way to avoid getting answers such as "under the definition of e the limit converge of that expression is e".
    hope u understand what im talking about

    thats what i trying to ask about
     
    Last edited: Mar 6, 2006
  6. Mar 6, 2006 #5
    Well, you can always proceed [tex]limn\rightarrow\infty(1+\frac{1}{n})^n=1+n(1/n)+(n)(n-1)(1/n^2)1/2!+n(n-1)(n-2)(1/n^3)1/3!+++[/tex]

    Notice here that the numerator of each term we have the power of n that corresponds to the power in the demoninator, other than that, the numenator has lower powers of n and these can be eliminated since

    [tex]limn\rightarrow\infty\frac{n-1}{n}=(n/n-1/n)\rightarrow1[/tex] (Here we have to consider that, for example, [tex]\frac{n(n-1)(n-2)}{n^33!}\leq1/3![/tex] We proceed to do this with every term in the series.

    This leaves us with the series 1+1+1/2!+1/3!+1/4!+++

    Now as my professor once said, e gets to its limit very fast. He added on the blackboard 1+1+1/2+1/6+1/24 +1/120+1/720+1/5040+1/43202=2.718.. (He did this all using decimals way before calculators.)
     
    Last edited: Mar 6, 2006
  7. Mar 6, 2006 #6
    got it, thanks
     
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