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Limit of square root function.

  1. Oct 9, 2012 #1
    I have to find the limit as x→∞ of √(x2+x)-x


    I can't rearrange this into a form where I can put infinity into the expression and get a meaningful answer. I've tried taking out square roots to get √x( √(x+1)-√x ) but if I put infinity into this I just get ∞(∞-∞) which is meaningless.

    Now I know that the limit approaches 0.5 (simply by plugging large numbers into my calculator) but I have no idea why this happens.
     
  2. jcsd
  3. Oct 9, 2012 #2
    Remember the formula (x+y)(x-y) = x2 - y2? That might prove useful here.
     
  4. Oct 9, 2012 #3
    Are you suggesting [itex](\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)[/itex]? If so, how do I account for the [itex](\sqrt{x^2+x}+x)[/itex] factor?

    Thanks for helping :)
     
  5. Oct 9, 2012 #4

    HallsofIvy

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    Have you considered "multiplying and dividing by the same thing"?
     
  6. Oct 9, 2012 #5
    Multiply and divide by √(x^2+x)+x
     
  7. Oct 10, 2012 #6
    If I do that, then I end up with [itex]\dfrac{x}{\sqrt{x^2+x}+x}[/itex], and I'm not sure how to progress from here. I can use L'Hopital's rule but then I get a horrible mess which doesn't seem to lead to any sort of answer.
     
  8. Oct 10, 2012 #7

    ehild

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    Divide both the numerator and denominator by x.

    ehild
     
  9. Oct 10, 2012 #8
    Ahhh. That makes the whole thing work, thanks :)
     
  10. Oct 10, 2012 #9
    Also, could anyone explain how the limit of [itex](1+\frac{1}{x})^x=e[/itex] as x goes to infinity? I first assumed it went to 1 (as 1/x goes to 0, and (1+0)^∞=1).
     
  11. Oct 10, 2012 #10

    micromass

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    Please start a new thread for each new question.
     
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