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Limit comparison test

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    use limit comparison test.


    2. Relevant equations
    [itex]\displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}[/itex]


    3. The attempt at a solution
    I have tried usin 1/x as the comparison function, but when applying the test it
    comes out to 0, not an L -> 0 < L < ∞
     
  2. jcsd
  3. Mar 17, 2013 #2

    Dick

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    Show how you got the limit to be L=0. I get L=1.
     
  4. Mar 17, 2013 #3
    [itex]\displaystyle limit (x->inf) \frac{1/x}{1/\sqrt{x^2 - 1}}[/itex]

    [itex]\displaystyle limit (x->inf) \frac{\sqrt{x^2 - 1}}{x}[/itex] = inf/inf

    [itex]\displaystyle limit (x->inf) \frac{2x}{\sqrt{x^2 - 1}}[/itex] = inf/inf

    [itex]\displaystyle limit (x->inf) \frac{2}{\sqrt[3/2]{x^2 - 1}}[/itex] = 2/inf = 0
     
  5. Mar 17, 2013 #4

    Dick

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    I you try to use l'Hopital on that you are just going to go in circles until you make a mistake and miss a chain rule, like you did. Use algebra to simplify the limit. sqrt(x^2-1)=x*sqrt(1-1/x^2).
     
  6. Mar 17, 2013 #5
    i'm sorry i'm lost, and i don't think i left out the chain rule i just didn't include the simplifications in the work above.
     
  7. Mar 17, 2013 #6

    Dick

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    I don't know what simplifications you made, since you didn't show them, but they aren't right. The first l'Hopital should give you ##\frac{x}{\sqrt{x^2 - 1}}##, the next will give ##\frac{\sqrt{x^2 - 1}}{x}##. Etc, etc.
     
  8. Mar 17, 2013 #7
    how does this look:

    1/sqrt(x^2-1) / (1/x)

    [itex]\frac{x}{\sqrt{x^2 - 1}}[/itex]

    lim (x->inf) [itex]\frac{1}{\sqrt{1-(1/x^2)}}[/itex] = 1

    [itex]\displaystyle\int_2^∞ {(1/x) dx}[/itex]

    [itex]ln|x| ^{∞}_{2}[/itex]

    diverges.
     
  9. Mar 17, 2013 #8

    Dick

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    Looks fine. l'Hopital's isn't the best way to handle every limit.
     
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