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Math Help - Evaluate Limit

  1. #1
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    Evaluate Limit

    40. \lim_{x\to\infty}(2^n + 3^n)^\frac{1}{n}

    The books says to try the sandwich theorem, but I have no idea what I am doing anymore. Been doing homework for like 48 hours straight .

    Thanks for help
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  2. #2
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    Do you mean as n approaches infinity not x?
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  3. #3
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    Yah I am sorry about that, as n--> inf
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    \displaystyle \lim_{n \to \infty}(2^n + 3^n)^{\frac{1}{n}} = \lim_{n \to \infty}e^{\ln{(2^n + 3^n)^{\frac{1}{n}}}}

    \displaystyle = e^{\lim_{n \to \infty}\ln{(2^n + 3^n)^{\frac{1}{n}}}}

    \displaystyle = e^{\lim_{n \to \infty}\frac{1}{n}\ln{(2^n + 3^n)}}

    \displaystyle = e^{\lim_{n \to \infty}\frac{\ln{(2^n + 3^n)}}{n}}

    \displaystyle = e^{\lim_{n \to \infty}\frac{\frac{2^n\ln{2} + 3^n\ln{3}}{2^n + 3^n}}{1}} by L'Hospital's Rule

    \displaystyle = e^{\lim_{n \to \infty}\frac{2^n\ln{2} + 3^n\ln{3}}{2^n + 3^n}}

    \displaystyle = e^{\lim_{n \to \infty}\frac{\frac{2^n\ln{2}}{3^n} + \ln{3}}{\frac{2^n}{3^n} + 1}}

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

    \displaystyle = e^{\frac{0 + \ln{3}}{0 + 1}}

    \displaystyle = e^{\ln{3}}

    \displaystyle = 3.
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    Quote Originally Posted by Warrenx View Post
    40. \lim_{x\to\infty}(2^n + 3^n)^\frac{1}{n}

    The books says to try the sandwich theorem, but I have no idea what I am doing anymore. Been doing homework for like 48 hours straight .

    Thanks for help

    With sandwich rule it is much simple!

    Notice that:

    ({3^n})^\frac{1}{n}<(2^n + 3^n)^\frac{1}{n}<(2\cdot 3^n)^\frac{1}{n}

    Now:

    lim_{n\to\infty} ({3^n})^\frac{1}{n}=lim_{n\to\infty} 3=3

    lim_{n\to\infty} ({2\cdot 3^n})^\frac{1}{n}=lim_{n\to\infty} {2}^\frac{1}{n}\cdot lim_{n\to\infty} ({3^n})^\frac{1}{n}=1\cdot 3=3

    Hence, by the sandwich rule lim_{n\to\infty} (2^n + 3^n)^\frac{1}{n}=3
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