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Math Help - Prove the definition of e

  1. #1
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    Prove the definition of e

    Prove that
    lim_{h-->0} (1+h)^{1/h} = e

    given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
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  2. #2
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    Quote Originally Posted by wopashui View Post
    Prove that
    lim_{h-->0} (1+h)^{1/h} = e

    given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
    By definition, we want e^x to be the function that is its own derivative. We want to try to find what value e is.

    So let f(x) = e^x

    f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}.


    So f'(x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}

     = \lim_{h \to 0}\frac{e^xe^h - e^x}{h}

     = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}

     = e^x \lim_{h \to 0}\frac{e^h - 1}{h}.


    So for f(x) to be equal to f'(x), we would require

    \lim_{h \to 0}\frac{e^h - 1}{h} = 1

    \lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h

    \lim_{h \to 0}e^h - \lim_{h \to 0}1 = \lim_{h \to 0}h

    \lim_{h \to 0}e^h = \lim_{h \to 0}1 + \lim_{h \to 0}h

    \lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)

    \lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}

    \lim_{h \to 0}e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}

    e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}.
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  3. #3
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    Quote Originally Posted by wopashui View Post
    Prove that
    lim_{h-->0} (1+h)^{1/h} = e
    given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
    I find that definition rather bazaar. By that I mean beginning students do not have the tools to deal with that abstraction.
    My all time favorite calculus textbook is the first edition of Gillman & McDowell, 1972.
    It is normal is size( not the dictionary size), does integral as betweeness, and has this definition of e.
    The number e is defined as the number such that  \int_1^e {\frac{{dx}}{x}}  = 1
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