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Math Help - limit of a sequence as n approaching infinity

  1. #1
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    limit of a sequence as n approaching infinity

    H'Lopital Application
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  2. #2
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    Yes, its a L'Hospital's Rule's Application since its an indeterminate power ..
    Personally, I prefer substitute x=\sqrt{n} to make the differentiation step easier ..

    If x=\sqrt{n}, clearly x\to\infty as n\to\infty

    so \lim_{n\to\infty} (\sqrt{n}-1)^{\frac{1}{\sqrt{n}}}=\lim_{x\to\infty} (x-1)^{\frac{1}{x}} ..

    You can evaluate the last one by the L'Hospital's Rule method for the indeterminate powers..
    Or you can substitute t=\frac{1}{x} to get a well-known limit ..
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  3. #3
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    Quote Originally Posted by General View Post
    Yes, its a L'Hospital's Rule's Application since its an indeterminate power ..
    Personally, I prefer substitute x=\sqrt{n} to make the differentiation step easier ..

    If x=\sqrt{n}, clearly x\to\infty as n\to\infty

    so \lim_{n\to\infty} (\sqrt{n}-1)^{\frac{1}{\sqrt{n}}}=\lim_{x\to\infty} (x-1)^{\frac{1}{x}} ..

    You can evaluate the last one by the L'Hospital's Rule method for the indeterminate powers..
    Or you can substitute t=\frac{1}{x} to get a well-known limit ..

    Is it e^0 = 1?
    Last edited by WartonMorton; April 15th 2010 at 09:32 AM. Reason: mistake
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  4. #4
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    Correct.
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