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  1. #1
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    limit

    evaluate \lim_x\to0 (e - (1+x) ^{1/x})/x
    i used l'hopitals rule for the \frac{o}{o} case but i get the limit as infinity but when i graph this function the lim is 1.25... any sort of insight would be helpful.
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  2. #2
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    Quote Originally Posted by bubble86 View Post
    evaluate \lim_x\to0 (e - (1+x) ^{1/x})/x
    i used l'hopitals rule for the \frac{o}{o} case but i get the limit as infinity but when i graph this function the lim is 1.25... any sort of insight would be helpful.
     \displaystyle \lim_{x\to 0} \frac {e-(1+x)^{\frac {1}{x}} } {x} = \frac {e}{2}


    show what have you done
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  3. #3
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    Quote Originally Posted by bubble86 View Post
    evaluate \lim_x\to0 (e - (1+x) ^{1/x})/x
    i used l'hopitals rule for the \frac{o}{o} case but i get the limit as infinity but when i graph this function the lim is 1.25... any sort of insight would be helpful.
    To apply l'Hospital's Rule you need to be able to differentiate (1 - x)^{1/x}. I suggest writing it as e^{\frac{1}{x} \ln (1 - x). Can you differentiate it now?
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  4. #4
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    my work so far

    \lim_{x\to\o} \ (e - ((1+x)^{\frac{1}{x}} ) / x
    = \lim_{x\to\o} \frac{d}{dx} \((1+x)^{\frac{1}{x}}
    = \lim_{x\to\o} \frac{d}{dx} \ e^{\ln\((1+x)^{\frac{1}{x}}}
    = \lim_{x\to\o} \ e^{\ln\((1+x)^{\frac{1}{x}}} \frac{d}{dx} \ln\((1+x)^{\frac{1}{x}}
    = \lim_{x\to\o} \ e^{\ln\((1+x)^{\frac{1}{x}}} \frac{d}{dx} \frac{1}{x} \\ln\((1+x)
    = \lim_{x\to\o} \ e^{\ln\((1+x)^{\frac{1}{x}}} \cdot(\frac{-1}{x^{2}}\ln(1+x)\ +\frac{1}{x}\cdot\frac{1}{1+x})
    = \lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)})
    the term \lim_{x\to\o}\frac{1}{x}}\cdot\frac{1}{1+x} tends to \infty so does that mean the whole expression
    \lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)}) tend to \infty
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  5. #5
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    Quote Originally Posted by bubble86 View Post
    \lim_{x\to\o} \ (e - ((1+x)^{\frac{1}{x}} ) / x
    = \lim_{x\to\o} \frac{d}{dx} \((1+x)^{\frac{1}{x}}
    = \lim_{x\to\o} \frac{d}{dx} \ e^{\ln\((1+x)^{\frac{1}{x}}}
    = \lim_{x\to\o} \ e^{\ln\((1+x)^{\frac{1}{x}}} \frac{d}{dx} \ln\((1+x)^{\frac{1}{x}}
    = \lim_{x\to\o} \ e^{\ln\((1+x)^{\frac{1}{x}}} \frac{d}{dx} \frac{1}{x} \\ln\((1+x)
    = \lim_{x\to\o} \ e^{\ln\((1+x)^{\frac{1}{x}}} \cdot(\frac{-1}{x^{2}}\ln(1+x)\ +\frac{1}{x}\cdot\frac{1}{1+x})
    = \lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)})
    the term \lim_{x\to\o}\frac{1}{x}}\cdot\frac{1}{1+x} tends to \infty so does that mean the whole expression
    \lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)}) tend to \infty
    The derivative is a product of two functions. The limit of the first function in the product is readily recognised as e. The limit of the second function is 1/2: see limit of 1/(x(1+x)) - Log[1+x]/x^2 as x approaches 0 - Wolfram|Alpha (click on show steps).
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  6. #6
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    The taylor series of ln(x+1) is
    ln(1+x)=x-x^2/2+x^3/3-x^4/4+-...
    for |x|<1
    so
    1/x^2 * ln(1+x)-1/x*1/(x+1)=
    1/x^2*(x-x^2/2+x^3/3-x^4/4+-...)-1/x*1/(x+1)=
    1/x-1/2+x*(...)-1/x*1/(x+1)=
    1/x(1-1/(x+1))-1/2+x*(...)=
    1/(x+1)-1/2+x*(...)->1/2 for x->0
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  7. #7
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    Iondor I follow you until the 2nd last line. The taylor series for ln (1+x) is \sum_{n=0}^{\infty}\ (-1)^n\frac{x^{n+1}}{n+1}
    so \frac{-1}{x^2} ln (1+x) = \frac{-1}{x^2}\sum_{n=0}^{\infty}\ (-1)^n\frac{x^{n+1}}{n+1} this expands out to
    \frac{-1}{x^2}\cdot\ ( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} .....)
    \ = (\frac{-1}{x} + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6})
    now \lim_{x\to\o} (\frac{-1}{x} + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})
    \ = \lim_{x\to\o} (\frac{-1}{x}\cdot\frac{1+x}{1+x} + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x}) is this what you mean?
    in which case wouldn't the lim be \ = \lim_{x\to\o} ( \frac{-1}{x}\cdot\frac{1}{1+x}\cdot (1+x) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})
    \ = \lim_{x\to\o} (( \frac{-1}{x}\cdot\frac{1}{1+x} ) + (x\cdot\frac{-1}{x}\cdot\frac{1}{1+x} ) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})
    \ = \lim_{x\to\o} ( (x\cdot\frac{-1}{x}\cdot\frac{1}{1+x} ) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x}) +  ( \frac{-1}{x}\cdot\frac{1}{1+x} )
    \ = \lim_{x\to\o} ( ( \frac{-1}{1+x} ) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...)
    \ = -1 + \frac{1}{2}
    \ = \frac{-1}{2}
    Last edited by bubble86; August 30th 2010 at 10:07 AM.
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  8. #8
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    Quote Originally Posted by bubble86 View Post
    \lim_{x\to\o} \ (e - ((1+x)^{\frac{1}{x}} ) / x
    = \lim_{x\to\o} \frac{d}{dx} \((1+x)^{\frac{1}{x}}
    You missed here a minus sign, because you have to do differentiate the whole nominater, i.e. <br />
\frac{d}{dx} (e-(1+x)^{\frac{1}{x}})<br />
=-\frac{d}{dx}(1+x)^{\frac{1}{x}}

    and from there you go as in your last post. You just have to multiply everything with -1 and then you get 1/2.
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  9. #9
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    thanks a lot guys would have not taught the taylor series expansion on my own
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  10. #10
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    \displaystyle{\lim_{x\to 0} \frac{e-(1+x)^{\frac{1}{x}}}{x} = -\lim_{x\to 0} \frac{d}{dx} (1+x)^{\frac{1}{x}}= \lim_{x\to 0} (1+x)^{\frac{1}{x}} \frac{\ln(1+x)-x/(1+x)}{x^2}}

    \displaystyle{=e \lim_{x\to 0} \frac{(1+x)^{-1}-(1+x)^{-2}}{2x}} = \frac{e}{2}
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