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Math Help - Limit question

  1. #1
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    Limit question

    Calculate the following limit
    lim x-> Infinity (1+a/x)^x
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  2. #2
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    Quote Originally Posted by Monster32432421 View Post
    Calculate the following limit
    lim x-> Infinity (1+a/x)^x

    \forall\,a\in\mathbb{R}\,,\,\,\lim_{x\to\infty}\le  ft(1+\frac{a}{x}\right)^x=e^a

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    \forall\,a\in\mathbb{R}\,,\,\,\lim_{x\to\infty}\le  ft(1+\frac{a}{x}\right)^x=e^a

    Tonio
    yea i know the answer.... how did you do it?
    Last edited by Monster32432421; April 29th 2010 at 06:07 AM.
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  4. #4
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    anyone?
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  5. #5
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    hint:

    Quote Originally Posted by Monster32432421 View Post
    anyone?
    lim (1+x)^(1/x)=e
    x->inf
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  6. #6
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    We know that if f(x) = e^x then f'(x) = e^x by definiton.


    Notice that if

    f(x) = a^x

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

     = \lim_{h \to 0} \frac{a^xa^h - a^x}{h}

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

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


    If we defined a = e so that f'(x) = f(x), then that would mean

    \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}}.


    Notice that if h = \frac{1}{X} then for h \to 0, X \to \infty.

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

    \lim_{X \to \infty}\left(1 + \frac{1}{X}\right)^{X} = e.
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    Quote Originally Posted by Prove It View Post
    We know that if f(x) = e^x then f'(x) = e^x by definiton.


    Notice that if

    f(x) = a^x

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

     = \lim_{h \to 0} \frac{a^xa^h - a^x}{h}

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

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


    If we defined a = e so that f'(x) = f(x), then that would mean

    \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}}.


    Notice that if h = \frac{1}{X} then for h \to 0, X \to \infty.

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

    \lim_{X \to \infty}\left(1 + \frac{1}{X}\right)^{X} = e.


    I'm not sure as whether this proof is valid in most basic calculus courses that I know of: you use stuff way more advanced than what is usually assumed for this kind of questions (e.g., derivatives which are defined by means of limits, or the basic characterization of the exponential function as the only real one (up to a scalar multiple) which equals its own derivative ... ).

    One approach is to define e:=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n , after we've proved the limit exists (for example by proving the seq. is monotone increasing and bounded above).

    Then we can prove the limit is between 2 and 3, irrational and etc.


    After this it's already easy to prove that \left(1+\frac{1}{a_n}\right)^{a_n}\xrightarrow [n\to\infty]{}e for any sequence \{a_n\}\,\,\,s.t.\,\,\,a_n\xrightarrow [n\to\infty]{}\infty, and from here we can get \left(1+\frac{1}{x}\right)^x \xrightarrow [x\to\infty]{}e (i.e., pass

    from discrete to continuous variable once we know a little about limits of functions)

    So, if we already know this, we can use a little limits arithmetic:

    \left(1+\frac{a}{x}\right)^x=\left[\left(1+\frac{1}{x/a}\right)^{x/a}\right]^a\xrightarrow [x\to\infty] {}[e]^a=e^a , as x\xrightarrow [x\to\infty]{}\infty \Longrightarrow x/a\xrightarrow [x\to\infty]{}\infty\,\,\,\forall\,0<a\in\mathbb{R} .

    For a<0 the process is similar with \left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1} and etc. , though here some other things must be proved...


    Tonio
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