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Math Help - Limit problem.

  1. #1
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    Limit problem.

    Find \lim_{x \to \infty} \left(\frac{x+5}{x+1} \right)^{x+3}
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  2. #2
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    Hi.

    Quote Originally Posted by varunnayudu View Post
    Find \lim_{x \to \infty} \left(\frac{x+5}{x+1} \right)^{x+3}

    It is
     (\frac{x + 5}{x + 1})^{x + 3} = e^{x*LN[(x + 5)/(x + 1)]}*(\frac{x + 5}{x + 1})^3

    and
    lim_{x \to \infty}(\frac{x + 5}{x + 1})^3= 1

    consider lim_{x \to \infty} x*LN[(x + 5)/(x + 1)] = ... = 4

    Therefor

     lim_{x \to \infty} e^{x*LN[(x + 5)/(x + 1)]}*(\frac{x + 5}{x + 1})^3 = e^4
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  3. #3
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    Last edited by Mush; December 29th 2008 at 07:22 AM.
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  4. #4
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    Indeed
    Last edited by Mush; December 29th 2008 at 07:22 AM.
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  5. #5
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    This looks like an 'e' limit.

    Remember the useful and famous \lim_{x\to 0}(1+x)^{\frac{1}{x}}=e?.

    Rewrite:

    \lim_{x\to \infty}\left(1+\frac{4}{x+1}\right)^{x+3}

    Let t=\frac{4}{x+1}

    \lim_{t\to 0}\left(1+t\right)^{\frac{4}{t}+2}

    \lim_{t\to 0}\left(1+t\right)^{\frac{4}{t}}\cdot\lim_{t\to 0}\left(1+t\right)^{2}

    Now, it is obvious the right limit goes to 1.

    So, we have \left[\lim_{t\to 0}\left(1+t\right)^{\frac{1}{t}}\right]^{4}

    See what the inside limit is?. e.

    Therefore, we get e^{4}
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  6. #6
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    Hello, varunnayudu!

    A variation of Rapha's solution . . .


    Find: . \lim_{x \to \infty}\left(\frac{x+5}{x+1} \right)^{x+3}

    We have: . y \;=\;\left(\frac{x+5}{x+1}\right)^{x+1}\left(\frac  {x+5}{x+1}\right)^2


    Use long division on the first fraction.
    In the second fraction, divide top and bottom by x.

    . . y \;=\;\left(1 + \frac{4}{x+1}\right)^{x+1}\left(\frac{1 + \frac{5}{x}}{1 + \frac{1}{x}}\right)^2


    \text{Then: }\;\lim_{x\to\infty}y \;=\;\underbrace{\lim_{x+1\to\infty}\left(1 + \frac{4}{x+1}\right)^{x+1}}_{\text{This is }e^4}\cdot\underbrace{ \lim_{x\to\infty}\left(\frac{1 + \frac{5}{x}}{1 + \frac{1}{x}}\right)^2}_{\text{This is 1}} \;=\;e^4

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