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Math Help - Limit

  1. #1
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    Limit

    Hello, I need some help for two horrific limits...
    <br />
(e ^{(x-1)^2+(1/(x+1))} - (e)^{1/2})  /  (ln(x+1) - ln(2))
    when x -> 1.

    The whole "(x-1)^2+(1/(x+1))" goes with the "e^"

    And, secondly:

    x * ln (x / (x+2))
    when x-> +oo

    Thank you!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Gogol View Post
    Hello, I need some help for two horrific limits...
    <br />
(e ^{(x-1)^2+(1/(x+1))} - (e)^{1/2})  /  (ln(x+1) - ln(2))
    when x -> 1.

    The whole "(x-1)^2+(1/(x+1))" goes with the "e^"

    Thank you!
    Since we have a case of \frac{0}{0}, L'Hopital's Rule works.

    \frac{d}{dx}\left[\frac{e^{(x-1)^2+(x+1)^{-1}}-\sqrt{e}}{\ln(x+1)-\ln(2)}\right]=\frac{e^{(x-1)^2+(x+1)^{-1}}\cdot[2(x-1)-(x+1)^{-2}]}{(x+1)^{-1}}

    Plug in x=1:

    \frac{e^{(1-1)^2+(1+1)^{-1}}\cdot[2(1-1)-(1+1)^{-2}]}{(1+1)^{-1}}=\frac{e^{1/2}\cdot-\frac{1}{4}}{\frac{1}{2}}=\boxed{-\frac{\sqrt{e}}{2}}
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  3. #3
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by Gogol View Post
    Hello, I need some help for two horrific limits...

    And, secondly:

    x * ln (x / (x+2))
    when x-> +oo

    Thank you!
    \lim_{x\to\infty}x\ln\left(\frac{x}{x+2}\right)=\l  im_{x\to\infty}\frac{\ln\left(\frac{x}{x+2}\right)  }{\frac{1}{x}}

    Now we have another \frac{0}{0} case, so we can use L'Hopital's Rule again. Taking the derivatives of the top and bottom gives us:

    \frac{\frac{x+2}{x}\cdot\frac{2}{(x+2)^2}}{-\frac{1}{x^2}}=-\frac{2x}{x+2}=-\frac{2x}{x(1+2/x)}=-\frac{2}{1+2/x}

    Now take the limit:

    \lim_{x\to\infty}-\frac{2}{1+2/x}=-\frac{2}{1+0}=\boxed{-2}
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  4. #4
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    Thank you!
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