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Thread: Limits question

  1. #1
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    Limits question

    hi

    i am lost in this question. please help:


    thanks
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  2. #2
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    Quote Originally Posted by champrock View Post
    hi

    i am lost in this question. please help:


    thanks
    Set the entire limit equal to y:

    $\displaystyle y = \lim_{x\rightarrow \infty} (3^x + 3^{2x})^{\frac{1}{x}}$

    Take the natural log of both sides:

    $\displaystyle lny = \lim_{x\rightarrow \infty}[\frac{ln(3^x + 3^{2x}}{x}]$

    You get indeterminate $\displaystyle \frac{\infty}{\infty}$, so use l'Hopital

    $\displaystyle lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}$

    $\displaystyle lny = \frac{1}{\infty} $

    $\displaystyle lny = 0$

    $\displaystyle e^0 = y$

    $\displaystyle y = 1$
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  3. #3
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    isnt the differentiation of
    $\displaystyle

    lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}
    $ wrong? wont we have to apply the chain rule in this so it will become:

    [( 3^x * ln 3 ) + (3^2x * ln 3 * 2) ] / [3^x +3^2x) ?
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  4. #4
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    Quote Originally Posted by champrock View Post
    isnt the differentiation of
    $\displaystyle

    lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}
    $ wrong? wont we have to apply the chain rule in this so it will become:

    [( 3^x * ln 3 ) + (3^2x * ln 3 * 2) ] / [3^x +3^2x) ?
    Yes, the differentiation was wrong.

    You can make life a bit easier by noting that $\displaystyle \left(3^x + 3^{2x}\right)^{1/x} = \left(3^x + \left[3^{x}\right]^2\right)^{1/x} = 3 \left(1 + 3^x\right)^{1/x}$.

    Then note that $\displaystyle \left(1 + 3^x\right)^{1/x} = e^{\ln \left[1 + 3^x\right]^{1/x}} = e^{\frac{\ln \left[1 + 3^x\right]}{x}} $.

    Now apply l'Hopital's Rule on $\displaystyle \frac{\ln \left[1 + 3^x\right]}{x} $.
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  5. #5
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    $\displaystyle
    \frac{\ln \left[1 + 3^x\right]}{x}
    $

    Applying that rule to the above gives (3^x * ln 3) / (1 + 3^x) . Now, I take ln 3 out of the limit and add and subtract 1 from numerator.

    So it becomes 3* ln3 * lim [1 - 1/(1+3^x) ] .

    Now the 1/(1+3^x) is zero when we apply limit to the above so, it becomes ln y = ln 3^3 the answer thus comes as 27 whcih is not even given as an option.
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  6. #6
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    Quote Originally Posted by champrock View Post
    $\displaystyle
    \frac{\ln \left[1 + 3^x\right]}{x}
    $

    Applying that rule to the above gives (3^x * ln 3) / (1 + 3^x) . Now, I take ln 3 out of the limit and add and subtract 1 from numerator.

    So it becomes 3* ln3 * lim [1 - 1/(1+3^x) ] .

    Now the 1/(1+3^x) is zero when we apply limit to the above so, it becomes ln y = ln 3^3 the answer thus comes as 27 whcih is not even given as an option.
    If you do the calculation correctly you will get $\displaystyle 3 e^{\ln 3} = 9$ as the answer.
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  7. #7
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    Does this suprise you?
    $\displaystyle 9 = \left( {3^{2x} } \right)^{\frac{1}
    {x}} \leqslant \left( {3^x + 3^{2x} } \right)^{\frac{1}
    {x}} \leqslant \left( {2\left( {3^{2x} } \right)} \right)^{\frac{1}
    {x}} = 9\left( 2 \right)^{\frac{1}
    {x}} $
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