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Math Help - Limits

  1. #1
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    Limits

    I need some help to find this limit:

    $\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$

    I tried to start doing something like
    $[\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$

    But then I don't know what to do with this 2^n and 3^n

    Help?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hint:


    3^n < {2^n+3^n} < 2*3^n
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  3. #3
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    Quote Originally Posted by cris View Post
    I need some help to find this limit:

    $\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$

    I tried to start doing something like
    $[\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$

    But then I don't know what to do with this 2^n and 3^n

    Help?
    Note that \displaystyle \lim_{n\to \infty} \left(2^n+3^n\right)^{\frac{1}{n}} = 3 \lim_{n\to \infty} \left(1 + \left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}}.
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  4. #4
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    Quote Originally Posted by cris View Post
    $\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$
    Suppose that  0 < a < b then b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{a^n  + b^n }} \leqslant \sqrt[n]{{b^n  + b^n }} \leqslant b\sqrt[n]{2}

    From the above, your limit should be obvious.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Note that \displaystyle \lim_{n\to \infty} \left(2^n+3^n\right)^{\frac{1}{n}} = 3 \lim_{n\to \infty} \left(1 + \left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}}.
    Is it correct if I say that (\frac {2}{3})^n tends to zero, then this limit is 3?

    Quote Originally Posted by Plato View Post
    Suppose that  0 < a < b then b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{a^n  + b^n }} \leqslant \sqrt[n]{{b^n  + b^n }} \leqslant b\sqrt[n]{2}

    From the above, your limit should be obvious.
    I understood what you did... and I would think about doing 0 < a < b, but this second part... Did you just deduce that? Or is it a theorem I should know?
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  6. #6
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    0 < a\, \Rightarrow \,\left( {\sqrt[n]{a}} \right) \to 1
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  7. #7
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    Quote Originally Posted by cris View Post
    I need some help to find this limit:

    $\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$

    I tried to start doing something like
    $[\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$
    Actually, this makes no sense. You can't has a power of 1/n outside the limit as n goes to infinity.

    But then I don't know what to do with this 2^n and 3^n

    Help?
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    3^n < {2^n+3^n} < 2*3^n

    ==>

    {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2*3^n}^{1/n}

    ==>

    {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2}^{1/n} * {3^n}^{1/n}

    ==>


    3 < {2^n+3^n}^{1/n} < {2}^{1/n} * 3


    ==>


    lim{3} <lim[ {2^n+3^n}^{1/n}] < lim[{2}^{1/n} * 3 ]

    ==>

    3 <lim[ {2^n+3^n}^{1/n}] < 3 lim[{2}^{1/n} ]


    { lim 2^1/n =1 }

    ==>

    3 <lim[ {2^n+3^n}^{1/n}] < 3

    Applying sandwich rule :

    lim[ {2^n+3^n}^{1/n}] = 3
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  9. #9
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    Quote Originally Posted by Also sprach Zarathustra View Post
    3^n < {2^n+3^n} < 2*3^n

    ==>

    {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2*3^n}^{1/n}

    ==>

    {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2}^{1/n} * {3^n}^{1/n}

    ==>


    3 < {2^n+3^n}^{1/n} < {2}^{1/n} * 3


    ==>


    lim{3} <lim[ {2^n+3^n}^{1/n}] < lim[{2}^{1/n} * 3 ]

    ==>

    3 <lim[ {2^n+3^n}^{1/n}] < 3 lim[{2}^{1/n} ]


    { lim 2^1/n =1 }

    ==>

    3 <lim[ {2^n+3^n}^{1/n}] < 3

    Applying sandwich rule :

    lim[ {2^n+3^n}^{1/n}] = 3
    What a mess that post is!
    I am willing to bet that were Nietzsche alive today he would have embraced the use of LaTeX.
    Why don’t you learn to use it?
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  10. #10
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by cris View Post
    I need some help to find this limit:

    $\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$

    I tried to start doing something like
    $[\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$

    But then I don't know what to do with this 2^n and 3^n

    Help?
    Let  y = (2^n + 3^n )^{ \frac{1}{n} }

     lny = \frac{1}{n} ln (2^n + 3^n)

     \lim\limits_{ n \to \infty } y = \lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln (2^n + 3^n) }

     \lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln (3^n [ 1 + ( \frac{2}{3} )^n ] ) }

     \lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln (3^n ) + \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ]  }

     \lim\limits_{ n \to \infty } e^{ n \frac{1}{n} ln (3 ) + \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ]  }

     3\lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ]  }

     3\lim\limits_{ n \to \infty } e^{ ln [ 1 + ( \frac{2}{3} )^n ]^{ \frac{1}{n}  }

     3\lim\limits_{ n \to \infty } [ 1 + ( \frac{2}{3} )^n ]^{ \frac{1}{n}

    Wow. And after all that I end up at Mr. Fantastics result. I'm actually depressed.

    In any event, to evaluate Mr. Fantastics limit we will revert one step.

     3\lim\limits_{ n \to \infty } e^{ ln [ 1 + ( \frac{2}{3} )^n ]^{ \frac{1}{n}  }

     3\lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ] }

    Note how the 1 over n will tend towards 0 as n goes to infinity as will what is in the ln portion of the equation. We will end up with  e^{~0} = 1 resulting in the answer 3.
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    What a mess that post is!
    I am willing to bet that were Nietzsche alive today he would have embraced the use of LaTeX.
    Why don’t you learn to use it?
    Dear Plato,

    Do you know who said:

    One man cannot practice many arts with success.
    ?
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