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Another find the limit

  1. Oct 17, 2004 #1

    quasar987

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    I've tried multiplicating by the conjugate of the denominator and of the numerator but this leads to nothing I can see. How can this limit be evaluated? (the limit is to be taken when n goes to infinity)

    [tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}[/tex]

    The answer is 1.

    Thanks for helping.
     
  2. jcsd
  3. Oct 17, 2004 #2

    Hurkyl

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    Show us what you got when you did this.
     
  4. Oct 17, 2004 #3
    Divide the numerator and the denominator by [tex]\sqrt{n}[/tex].
     
  5. Oct 17, 2004 #4

    quasar987

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    Will do tomorrow. I gotta go to bed urgent.

    If you mean "take [itex]\sqrt{n}[/itex] out of the num and denom", when you take the limit you get the undeterminate form 0/0. If that's not what you mean, I don't know what you mean. :smile:
     
  6. Oct 18, 2004 #5

    quasar987

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    Ok, so

    [tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}} \frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}} = \frac{\sqrt{n+7}\sqrt{n+2}+\sqrt{n+7}\sqrt{n}-\sqrt{n+5}\sqrt{n+2}-\sqrt{n+5}\sqrt{n}}{2}[/tex]

    which may be factorised into...

    [tex]\frac{(\sqrt{n+7}-\sqrt{n+5})(\sqrt{n+2}+\sqrt{n})}{2}[/tex]


    If we multiply by the other conjugate, we get

    [tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}} \frac{\sqrt{n+7}+\sqrt{n+5}}{\sqrt{n+7}+\sqrt{n+5}} = \frac{12}{\sqrt{n+2}\sqrt{n+7}+\sqrt{n+2}\sqrt{n+5} -\sqrt{n+7}\sqrt{n}-\sqrt{n+5}\sqrt{n}}[/tex]

    which may be factorised into

    [tex]\frac{12}{(\sqrt{n+2}-\sqrt{n})(\sqrt{n+5}+\sqrt{n+7})}[/tex]
     
  7. Oct 18, 2004 #6

    arildno

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    You've made quite a few errors here
    Multiply with 1*1 in this manner:
    [tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+7}+\sqrt{n+5}}{\sqrt{n+7}+\sqrt{n+5}}*\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}*\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}[/tex]
    Hence, we get:
    [tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+7}+\sqrt{n+5}}[/tex]
     
    Last edited: Oct 18, 2004
  8. Oct 18, 2004 #7

    quasar987

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    And THEN...
    :tongue:

    Ok I get it now. Thanks everyone!
     
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