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L'Hopital's Rule

  1. Nov 5, 2006 #1
    evaluate the limit as x=>0 of sqrt(x^2+x) - x

    so i multiplied by the conjugate to get:

    x^2 + x - x^2 / sqrt(x^2 +x) + x

    which simplifies to:

    x / sqrt (x^2 + x) + x = infinity/infinity

    taking LH, you get:

    1 / 1/2(x^2+x)^(-1/2) * (2x+1) + 1 = infinity/infinity again

    looking ahead, it doesnt look like LH is going to help

    any suggestions?
     
  2. jcsd
  3. Nov 5, 2006 #2

    VietDao29

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    Are you sure it's not infinity?
    Suppose you mean:
    [tex]\lim_{x \rightarrow \infty} \sqrt{x ^ 2 + x} - x[/tex]
    If [tex]x \rightarrow - \infty[/tex], you will have: [tex]\infty + \infty = \infty[/tex], the limit does not exist for that case, so here, we are going to find the limit of that expression as x tends to positive infinity.
    [tex]\lim_{x \rightarrow + \infty} \sqrt{x ^ 2 + x} - x[/tex]
    This is good.
    Do you need to use L'Hospital rule? If not, you can consider dividing both numerator, and denominator by x to get:
    [tex]\lim_{x \rightarrow + \infty} \sqrt{x ^ 2 + x} - x = \lim_{x \rightarrow + \infty} \frac{x}{\sqrt{x ^ 2 + x} + x}= \lim_{x \rightarrow + \infty} \frac{\frac{x}{x}}{\frac{\sqrt{x ^ 2 + x} + x}{x}}[/tex]
    [tex]= \lim_{x \rightarrow + \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}[/tex], (since x is positive ([tex]x \rightarrow + \infty[/tex]), so [tex]\sqrt{x ^ 2} = x[/tex])
    Can you go from here? :)
     
  4. Nov 5, 2006 #3
    Well, in this case you don't need to multiply by the conjugate because if you just evaluate this function with x being 0 you get 0. This function is continuous in x = 0 so you can just calculate the limit of f(x) by calculating f(0).

    marlon
     
  5. Nov 5, 2006 #4
    sorry to cut into this post but im working atm on a question similar to this, i havent even seen LH rule in my book so that may be why im going to ask this question

    how did you turn all those x's into 1's and 1/x in the numerator?
     
  6. Nov 6, 2006 #5

    HallsofIvy

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    Standard step you should have learned as the first thing in limits: if you are taking limits as x goes to infinity, divide numerator and denominator by the highest power of x. That way all the "x"s become "1/x"s and instead of going to infinity, they go to 0.
     
  7. Nov 6, 2006 #6
    yep that makes sense, thanks
     
  8. Nov 7, 2006 #7
    VietDao, based on your calculations, I get 1/2 as an answer. However, I dont know how your supposed to divide sqrt (x^2 + x) by x. Doesn't this violate BEDMAS rules b/c I thought sqrt(x^2+x) does not equal sqrt(x^2) + sqrt(x).
     
    Last edited: Nov 7, 2006
  9. Nov 7, 2006 #8

    arildno

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    Dearly Missed

    What are BEDMAS rules??
     
  10. Nov 7, 2006 #9
    sorry nvm, i forgot that sqrt(a) divide sqrt(b) = sqrt(a/b). thanks for everyones help.
     
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