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  1. #1
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    test

    we have \displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}

    re-writing this as \displaystyle\lim_{x\to\infty}exp^[\ln(( \frac{16x}{16x+5})^{9x})]

    \displaystyle\lim_{x\to\infty}exp^{9x}[\ln\( \frac{16x}{16x+5})]

    Then using the substitution \frac{1}{t}

    we have \displaystyle\lim_{t\to 0}exp^{\frac{9}{t}\ln\frac{\frac{16}{t}}{\frac{16}  {t}+5}

    Then using L'hopitals rule on \displaystyle\lim_{t\to 0}{\frac{9}{t}\ln\frac{\frac{16}{t}}{\frac{16}{t}+  5}= \displaystyle\lim_{t\to 0}\frac{-45}{16+5t}= \frac{-45}{16}

    So we have that

    \displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}
    =  \displaystyle\exp^\frac{-45}{16}
    Last edited by hmmmm; December 3rd 2010 at 05:15 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hmmmm View Post
    If we write \displaystyle( \frac{16x}{16x+5})^{9x} as \displaystyle (1-\frac{5}{16x+5})^{9x}


    we then have \displaystyle\lim_{x\to\infty} (1-\frac{5}{16x+5})^{9x}

    re-writing this as \displaystyle\lim_{x\to\infty}exp^[\ln((1-\frac{5}{16x+5})^{9x})]

    \displaystyle\lim_{x\to\infty}exp^{9x}[\ln((1-\frac{5}{16x+5}))]

    Then using the substitution \frac{1}{t}

    we have \displaystyle\lim_{t\to 0}exp^({\frac{9}{t}[\ln(1-\frac{5}{16(\frac{1}{t})+5})])

    Then using L'hopitals rule on \displaystyle\lim_{t\to 0}{\frac{9}{t}[\ln(1-\frac{5}{16(\frac{1}{t})+5})])

     \displaystyle \lim_{x\to \infty}(\frac{16x}{16x+5})^{9x})=e^\frac{-45}{16}
    Try elastic brackets \left( and \right) they expand to the size needed for what is enclosed, same for \left[ , \right] and \left\{, \right\}

    CB
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  3. #3
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    Hello, hmmmm!


    Use larger parentheses/brackets as CaptainBlack suggested.

    And here is another approach to the limit . . .



    We have \displaystyle\lim_{x\to\infty} \left(\frac{16x}{16x+5}\right)^{9x}


    Ignore the limit for now . . .

    \text{Divide top and bottom by }16x\!:\;\;\dfrac{1}{1 + \frac{5}{16x}} \;=\;\dfrac{1}{1 + \frac{1}{\frac{16}{5}x}}}


    \text{Then: }\;\left(\dfrac{1}{1 + \frac{1}{\frac{16}{5}x}}}\right)^{9x} \;=\; \left[\left(\dfrac{1}{1 + \frac{1}{\frac{16}{5}x}}}\right)^{\frac{16}{5}x}\r  ight]^{\frac{45}{16}} \;=\; \left[\dfrac{1}{\left(1 + \frac{1}{\frac{16}{5}x}\right)^{\frac{16}{5}x}}\ri  ght]^{\frac{45}{16}}


    \displaystyle \text{Hence: }\;\lim_{x\to\infty}  \left[\dfrac{1}{\left(1 + \frac{1}{\frac{16}{5}x}\right)^{\frac{16}{5}x}}\ri  ght]^{\frac{45}{16}} \;=\; \left[\dfrac{1}{\lim \left(1 + \frac{1}{\frac{16}{5}x}\right)^{\frac{16}{5}x}}\ri  ght]^{\frac{45}{16}} .[1]


    \displaystyle \text{We note that: }\:\lim_{x \to \infty} \left(1 + \frac{1}{\frac{16}{5}x}}\right)^{\frac{16}{5}x} \;=\;e


    Therefore, [1] becomes: . \displaystyle \left(\frac{1}{e}\right)^{\frac{45}{16}} \;=\;e^{-\frac{45}{16}}

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