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Math Help - killing limits, please

  1. #1
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    Unhappy killing limits, please

    Please, I need step by step solution of this two limits. I'd be so greatful for help. please don't mind my stupidity of not knowing L'Hospital rule...
    So, this's killing me:
    1. lim (x/x-1 - 1/ln x)
    x->1

    2. lim (x+3/x)^x
    x->infinity
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  2. #2
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    2) is there a typo?
    is the limit ((x+3)/x)^x?
    <br />
\lim_{x\to\infty} = (\frac {x+3}{x})^x =  \lim_{x\to\infty} = (\frac {x}{x} + \frac {3}{x})^x = \lim_{x\to\infty} = (1 + \frac {3}{x})^x<br />

    <br />
(1 + \frac {1}{t})^t = e<br />

    <br />
\frac {1}{t} = \frac {3}{x} \Rightarrow x = 3t<br />

    <br />
\lim_{x\to\infty} = (1 + \frac {3}{x})^x = \lim_{t\to\infty} = (1 + \frac {1}{t})^{3t} = \lim_{t\to\infty} ((1 + \frac {1}{t})^t)^3 = e^3<br />
    Last edited by metlx; January 30th 2009 at 12:53 PM.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by metlx View Post
    2) is there a typo?
    if its limit of ((x+3)/x)^x then i know how to solve it. if its (x+3/x)^x then i'm clueless.
    If the limit is indeed  \left(x+\frac{3}{x}\right)^x , it approaches  \infty as  x approaches  \infty .
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Note that  \frac{x}{x-1}-\frac{1}{\ln(x)} = \frac{x\ln(x)-(x-1)}{(x-1)\ln(x)}. Since the numerator and denominator approach  0 as  x approaches  1 , l'H˘pital's rule can be applied.

     \lim_{x \to 1} \frac{x\ln(x)-(x-1)}{(x-1)\ln(x)} = \lim_{x \to 1} \frac{\ln(x)}{\ln(x)+\frac{x-1}{x}}

    Notice for the same reasons as above we can apply l'H˘pital's rule again.

     \lim_{x \to 1} \frac{\ln(x)}{\ln(x)+\frac{x-1}{x}} = \lim_{x \to 1} \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}

    Simplifying yields  \lim_{x \to 1} \frac{x}{x+1} = \frac{1}{2}
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  5. #5
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    Hello, RinoST!

    Let me baby-step through #2 . . .


    \lim_{x\to\infty}\left(\frac{x+3}{x}\right)^x

    We're expected to know this: . \lim_{z\to\infty}\left(1 + \frac{1}{z}\right)^z \;=\;e

    And we will hammer the problem into that form . . .



    We have: . \left(\frac{x+3}{x}\right)^x\;=\;\left(\frac{x}{x} + \frac{3}{x}\right)^x \;=\;\left(1 + \frac{3}{x}\right)^x<br />

    Raise it to the "power 1": . \bigg[\left(1 + \frac{3}{x}\right)^x\bigg]^{\frac{3}{3}} \;=\;\bigg[\left(1 + \frac{3}{x}\right)^{\frac{x}{3}}\bigg]^3 \;=\;\bigg[\left(1 + \frac{1}{\frac{x}{3}}\right)^{\frac{x}{3}}\bigg]^3

    . . So we have: . \lim_{x\to\infty}\bigg[\left(1 + \frac{1}{\frac{x}{3}}\right)^{\frac{x}{3}}\bigg]^3


    Let z \,=\,\frac{x}{3}

    and we have: . \lim_{z\to\infty}\bigg[\left(1 + \frac{1}{z}\right)^z\bigg]^3 \;=\;\left[\lim_{x\to\infty}\left(1 + \frac{1}{z}\right)^z\right]^3 \;=\;\boxed{e^3}

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  6. #6
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    You can treat [(x+3)/x]^x=y
    Of you ln both side, you will get:
    ln (y)= x ln[(x+3)/x]
    = [(x+3)/x] / (1/x)
    Then you can apply L'hopital rule, that is differentiate the numerator and denominator separately
    Eventually you will get:

    lim{x to infinity} ln y = -3/(x+3)
    = 0
    lim{x to infinity} y = 1
    lim{x to infinity} [(x+3)/x]^x=1

    sorry, i do not know how to type the math notation.
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