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Math Help - Limits

  1. #1
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    Limits

    Can someone help me with this problem:

    lim (n^2 + n)^(1/n) = 1;


    In (b) you may assume that n^1/n = 1.

    I know how to solve these when it's a number to the power of n but not when it's n to a power.
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  2. #2
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     (n^2 + n)^{1/n} = [n(n+1)]^{1/n} = n^{1/n} (n+1)^{1/n} .

    Can you show that  (n+1)^{1/n} goes to 1 as n goes to infinity?
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  3. #3
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    Quote Originally Posted by Mathsnewbie View Post
    Can someone help me with this problem:

    lim (n^2 + n)^(1/n) = 1;


    In (b) you may assume that n^1/n = 1.

    What does this mean??

    I know how to solve these when it's a number to the power of n but not when it's n to a power.

    If you already know \sqrt[n]{n}\xrightarrow [n\to\infty]{}1 , then it's easy: \sqrt[n]{n^2+n}=\sqrt[n]{n}\sqrt[n]{n+1}\xrightarrow [n\to\infty]{}1\cdot 1=1 .

    Tonio
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  4. #4
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    Quote Originally Posted by JG89 View Post
     (n^2 + n)^{1/n} = [n(n+1)]^{1/n} = n^{1/n} (n+1)^{1/n} .

    Can you show that  (n+1)^{1/n} goes to 1 as n goes to infinity?
    I can't, I know how to do some by sandwich principle but I couldn't get either of these to work.
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  5. #5
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    Quote Originally Posted by Mathsnewbie View Post
    I can't, I know how to do some by sandwich principle but I couldn't get either of these to work.


    \sqrt[n]{a}\xrightarrow [n\to\infty]{}1\,,\,\,\forall a>0 , so just check that 1\leq\sqrt[n]{n+1}\leq \sqrt[n]{2n}=\sqrt[n]{2}\,\sqrt[n]{n} ...

    Tonio
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  6. #6
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    Thank you, I managed to complete
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