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Math Help - Limit of A Sequence

  1. #31
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    , I am guessing there is certain circumstances this can be used
    Also, redsox can you explain why you multiplied by what you did?
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  2. #32
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by RockHard View Post
    , I am guessing there is certain circumstances this can be used
    Also, redsox can you explain why you multiplied by what you did?
    In nutshell, all Drexel did was to note that 0<\sqrt{n^2+1}. Add n to both sides to get n<\sqrt{n^2+1}+n. Now divide both sides by n(\sqrt{n^2+1}+n) to get

    \frac{n}{n(\sqrt{n^2+1}+n)}<\frac{\sqrt{n^2+1}+n}{  n(\sqrt{n^2+1}+n)}\implies\frac{1}{\sqrt{n^2+1}+n}  <\frac{1}{n}

    As for why I multiplied by \frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}, I knew that multiplying by it's "conjugate" would clear out the square root sign and put something large in the denominator, which would make it easier to prove that the limit went to zero.
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  3. #33
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    Quote Originally Posted by RockHard View Post
    Can anyone provide me with another example before I deem this thread solved
    Go to your homework questions and attempt a question. Post in in a new thread if you need help.

    Thread closed.
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