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Math Help - 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:

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

    Take the natural log of both sides:

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

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

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

    lny = \frac{1}{\infty}

    lny = 0

    e^0 = y

    y = 1
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  3. #3
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    isnt the differentiation of
    <br /> <br />
lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}<br />
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
    <br /> <br />
lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}<br />
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 \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 \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 \frac{\ln \left[1 + 3^x\right]}{x} .
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  5. #5
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    <br />
\frac{\ln \left[1 + 3^x\right]}{x}<br />

    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
    <br />
\frac{\ln \left[1 + 3^x\right]}{x}<br />

    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 3 e^{\ln 3} = 9 as the answer.
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  7. #7
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    Does this suprise you?
    9 = \left( {3^{2x} } \right)^{\frac{1}<br />
{x}}  \leqslant \left( {3^x  + 3^{2x} } \right)^{\frac{1}<br />
{x}}  \leqslant \left( {2\left( {3^{2x} } \right)} \right)^{\frac{1}<br />
{x}}  = 9\left( 2 \right)^{\frac{1}<br />
{x}}
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