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Math Help - Quick question on a limit

  1. #1
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    Quick question on a limit

    Hi, Im having a little trouble understanding a limit. The limit as k goes to infinity of k/(k+1), my book says the answer is 1. I tried to divide everything by k and got 1/infinity = 0? Can someone please try and explain how this limit is equal to 1.

    Thank you
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  2. #2
    Moo
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    Hello,

    This is a very classical one

    \frac{x}{x+1}

    Divide above and below by x powered to the biggest coefficient encountered.
    So here, divide by x.

    =\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}=\frac  {1}{1+\frac{1}{x}}

    But if any constant number is divided by something that tends to infinity, the resulting limit will be 0.

    So \lim_{x \to + \infty} \frac{1}{1+\frac{1}{x}}=\frac{1}{1}=1
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hello

    Dividing by k is a good idea, it yields \frac{k}{k+1}=\frac{1}{1+\frac{1}{k}}
    As \lim_{k\to\infty}\frac{1}{k}=0, \lim_{k\to\infty}\frac{1}{1+\frac{1}{k}}=1
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  4. #4
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    thanks for the help
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  5. #5
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    Or just make \frac{k}<br />
{{k + 1}} = \frac{{k + 1 - 1}}<br />
{{k + 1}} = 1 - \frac{1}<br />
{{k + 1}} and as k\to\infty the answer is pretty obvious.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    This is a very classical one

    \frac{x}{x+1}

    Divide above and below by x powered to the biggest coefficient encountered.
    So here, divide by x.

    =\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}=\frac  {1}{1+\frac{1}{x}}

    But if any constant number is divided by something that tends to infinity, the resulting limit will be 0.

    So \lim_{x \to + \infty} \frac{1}{1+\frac{1}{x}}=\frac{1}{1}=1
    O yeah...this one technique Moo taught me is amazing \lim_{k \to {\infty}}\frac{k}{k+1}=\lim_{k \to {\infty}}\frac{1+0}{1}=1...its called L'hopital's rule
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