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  1. #1
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    test

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

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

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

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

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

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

    So we have that

    $\displaystyle \displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}$
    =$\displaystyle \displaystyle\exp^\frac{-45}{16}$
    Last edited by hmmmm; Dec 3rd 2010 at 04:15 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hmmmm View Post
    If we write $\displaystyle \displaystyle( \frac{16x}{16x+5})^{9x} $ as $\displaystyle \displaystyle (1-\frac{5}{16x+5})^{9x}$


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

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

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

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

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

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

    $\displaystyle \displaystyle \lim_{x\to \infty}(\frac{16x}{16x+5})^{9x})=e^\frac{-45}{16}$
    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 $\displaystyle \displaystyle\lim_{x\to\infty} \left(\frac{16x}{16x+5}\right)^{9x}$


    Ignore the limit for now . . .

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


    $\displaystyle \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}} $


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


    $\displaystyle \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 \displaystyle \left(\frac{1}{e}\right)^{\frac{45}{16}} \;=\;e^{-\frac{45}{16}} $

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