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Math Help - limit proof

  1. #16
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    Quote Originally Posted by alexandros View Post
    Can research show us that:

    \sqrt 2 = \frac{3}{2}??
    Is this related to your earlier question about
    \left(1+\frac{1}{n}\right)^{nh}= 1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+.....
    for n= 1 and h= 1/2?

    The left side is equal to \sqrt{2}. The right hand side is certainly NOT "3/2".
    The first four partial sums are 1, 1.5, 1.375, and 1.4375. Since that is an alternating sum with decreasing terms, its sum lies between any two consecutive partial sums. In particular, the sum lies between 1.375 and 1.4375. \sqrt{2}, approximately 1.414, lies between those two, 3/2= 1.5 does not.
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  2. #17
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    Quote Originally Posted by alexandros View Post
    Can research show us that:

    \sqrt 2 = \frac{3}{2}??
    What I wrote was "truncated"...

    I thought that would be easily deduced.

    Maybe an easier solution to follow would be...

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

    Substitute \displaystyle\ e^h-1=\frac{1}{t}\Rightarrow\ e^h=1+\frac{1}{t}

    \displaystyle\Rightarrow\ h=ln\left[1+\frac{1}{t}\right]

    h\rightarrow\ 0+\ \Rightarrow\ t\rightarrow \infty

    \displaystyle\Rightarrow\lim_{t \to \infty}\frac{1}{tln\left[1+\frac{1}{t}\right]}

    =\displaystyle\lim_{t \to \infty}\frac{1}{ln\left(1+\frac{1}{t}\right)^t}=\f  rac{1}{ln(e)}=1
    Last edited by Archie Meade; December 14th 2010 at 12:07 PM.
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  3. #18
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    Quote Originally Posted by HallsofIvy View Post
    Is this related to your earlier question about
    \left(1+\frac{1}{n}\right)^{nh}= 1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+.....
    for n= 1 and h= 1/2?

    The left side is equal to \sqrt{2}. The right hand side is certainly NOT "3/2".
    The first four partial sums are 1, 1.5, 1.375, and 1.4375. Since that is an alternating sum with decreasing terms, its sum lies between any two consecutive partial sums. In particular, the sum lies between 1.375 and 1.4375. \sqrt{2}, approximately 1.414, lies between those two, 3/2= 1.5 does not.
    Terribly sorry, my mistake
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  4. #19
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    Thread closed due to unproductive antagonism.
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