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Math Help - limits

  1. #1
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    limits

    Hi i need help finding limts!
    (2(^n)+ 3 (^n))^2/n
    from this i am suppose to get
    3^2(1 +2/3^n)^(2/n)
    Do not know how to get there though. The eventual limit is 9!
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by studentsteve1202 View Post
    Hi i need help finding limts!
    (2(^n)+ 3 (^n))^2/n
    from this i am suppose to get
    3^2(1 +2/3^n)^(2/n)
    Do not know how to get there though. The eventual limit is 9!
    (2^n+3^n)^{2/n}

    [2^n(1+(3/2)^n)]^{2/n}

    because 3^n = 2^n * (3/2)^n
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  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by studentsteve1202 View Post
    Hi i need help finding limts!
    (2(^n)+ 3 (^n))^2/n
    from this i am suppose to get
    3^2(1 +2/3^n)^(2/n)
    Do not know how to get there though. The eventual limit is 9!
    \lim_{n \to \infty} \left(2^n + 3^n\right)^{\frac2{n}}

    =\lim_{n \to \infty} \left(3^n\left(\frac{2^n}{3^n} + 1\right)\right)^{\frac2{n}}

    =\lim_{n \to \infty} 3^2 \left(\frac{2^n}{3^n} + 1\right)^{\frac2{n}}

    =\lim_{n \to \infty} 3^2 \left(\left(\frac{2}{3}\right)^n + 1\right)^{\frac2{n}}

    = 9 \,\,\,\,\,\left[\because n \to \infty \Rightarrow \left(\frac{2}{3}\right)^n \to 0\right]
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  4. #4
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    \left( {2^n  + 3^n } \right)^{2/n}  = \left( {\sqrt[n]{{2^n  + 3^n }}} \right)^2  \to 9<br />

    0 < a < b\quad  \Rightarrow \quad \left( {\sqrt[n]{{a^n  + b^n }}} \right) \to b
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  5. #5
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    Hello, studentsteve1202!

    \left(2^n+ 3^n\right)^{\frac{2}{n}}

    from this i am suppose to get: . 3^2\cdot\left[1 + \left(\frac{2}{3}\right)^n\right]^{\frac{2}{n}}

    We have: . \left(2^n + 3^n\right)^{\frac{2}{n}}


    Multiply the first term by \frac{3^n}{3^n}\!:\quad\left({\color{blue}\frac{3^  n}{3^n}}\cdot2^n + 3^n\right)^{\frac{2}{n}} \;=\;\left(3^n\cdot\frac{2^n}{3^n} + 3^n\right)^{\frac{2}{n}}


    Factor: . \left[3^n\left(\frac{2^n}{3^n} + 1\right)\right]^{\frac{2}{n}} \;=\;\left(3^n\right)^{\frac{2}{n}}\left[\left(\frac{2}{3}\right)^n + 1\right]^{\frac{2}{n}} \;=\;3^2\cdot\left[\left(\frac{2}{3}\right)^n + 1\right]^{\frac{2}{n}}

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  6. #6
    Super Member PaulRS's Avatar
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    Given two positive real numbers a and b

    We have \lim_{n \to +\infty}(a^n+b^n)^{\frac{1}{n}}=\max(a;b) (1)

    The proof of this is almost the same Isomorphism provided.

    And: \lim_{x \to -\infty}(a^x+b^x)^{\frac{1}{x}}=\min(a;b) (you can show this limit by applying (1) )

    Interesting
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by studentsteve1202 View Post
    Hi i need help finding limts!
    (2(^n)+ 3 (^n))^2/n
    from this i am suppose to get
    3^2(1 +2/3^n)^(2/n)
    Do not know how to get there though. The eventual limit is 9!
    A more inelegant way of doing this since everyone else did it the fun way would be to take the ln of both sides and then use L'hopitals
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  8. #8
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    Thankyou very much everyone. Very helpful
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