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Limit of function with square roots

  1. Jan 20, 2006 #1
    for sqr root of (n + sqr root (n) ) - sqr root (n),is the answer = zero or infinity so converges or diverges??
  2. jcsd
  3. Jan 20, 2006 #2


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    What would be the argument as to why it should be zero?
  4. Jan 20, 2006 #3
    if i divide each constant n by n itself then limit n to infinity and i'll get 1 - 1 =0 right??
  5. Jan 20, 2006 #4


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    You mean, you'll have


    and since 1/n3 ---> 0 much faster than 1/n, you'll end up with n[sqrt(1/n) - sqrt(1/n)]?
    Last edited: Jan 20, 2006
  6. Jan 21, 2006 #5
    no,just sqr root of 1/n without the power of 3
  7. Jan 21, 2006 #6
    if i divide every thing by n,i'll get sqr root [1 + (sqr root 1/n)] - sqr root (1)
    what is the final answer??
  8. Jan 21, 2006 #7


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    If you divide each term of
    [tex]\sqrt{n+ \sqrt{n}}-\sqrt{n}[/tex]
    by [itex]\sqrt{n}[/itex] you get
    [tex]\sqrt{1+ \frac{1}{\sqrt{n}}}- 1[/tex]
    What is the limit of that as n goes to infinity?

    By the way, why are you posting so many problems involving limits, derivatives, and integrals in the precalculus section?
  9. Jan 21, 2006 #8
    is the answer = to zero??
  10. Jan 21, 2006 #9


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    I have a question. It's not clear to tell that what you are asking for. Are you asking for this:
    [tex]\lim_{n \rightarrow \infty} \sqrt{n - \sqrt{n}} - \sqrt{n}[/tex]
    or this:
    [tex]\lim_{n \rightarrow \infty} \frac{\sqrt{n - \sqrt{n}} - \sqrt{n}}{\sqrt{n}}[/tex]?
    If it's the latter, then you are correct!
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