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Limit Comparison Test

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Use direct comparison test or limit comparison test to determine if the integral converges.


    2. Relevant equations
    [itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]



    3. The attempt at a solution

    If i were to use the limit comparison test, would these integrals fit the criteria.
    ** if the positive functions f and g are continues on [a,∞)

    Note: what does it mean by positive functions?

    limit[itex]_{x->infinity} \frac{f(x)}{g(x)} = L [/itex] 0 < L < ∞

    then [itex]\displaystyle\int_a^∞ {f(x) dx}[/itex] and

    [itex]\displaystyle\int_a^∞ {g(x) dx}[/itex]

    both converge or diverge.

    f(x) = [itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]

    g(x) = [itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex]

    since these functions are not continuous at a=0, then is the limit comparison test not an option here?



    If not, how would i go about choosing a function for the direct comparison test?
     
    Last edited: Mar 13, 2013
  2. jcsd
  3. Mar 13, 2013 #2

    Dick

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    No need to hyperbold on the font. Your integral already has a problem way before infinity. Where is it?
     
  4. Mar 13, 2013 #3
    i was stating the theorem from the book. it states must be continuous from [a,∞), where the a=0 in the case, and (sorry forgot to include the comparing function).

    [itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex] is not continuous at a=0.

    f(x) = [itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]

    g(x) = [itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex]


    further, i don't think it is stating the integral must be to infinity, just that it must be continuous on, or am i incorrect?

    how should i solve with direct comparison test, if easier. please help. test in 2 days.
     
  5. Mar 13, 2013 #4

    Dick

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    Do direct comparison. The problem with your integral happens at x=3, doesn't it? It's not continuous there. Is 1/(9-x^2) integrable on the interval [0,3]? Try to think of a comparison function on just that interval. Factor 9-x^2.
     
  6. Mar 14, 2013 #5
    is (3+x) or (3-x) < (3+x)(3-x) meaning [itex]\frac{1}{3+x} or \frac{1}{3-x} > \frac{1}{9-x^2}[/itex] ??

    thus, since

    [itex]\displaystyle\int_0^6 {\frac{1}{3+x} dx} [/itex] converges to a finite number.

    and [itex]\frac{1}{3+x}> \frac{1}{9-x^2} [/itex]

    [itex]\displaystyle\int_0^6 {\frac{1}{9-x^2} dx} [/itex] it would conclude that this also converges?

    By Direct Comparison Test.
     
  7. Mar 14, 2013 #6

    Dick

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    The comparison you want is something like ##\frac{1}{9-x^2} > C \frac{1}{3-x}## on the interval 0<x<3 for some constant C. Because ##\frac{1}{3-x}## diverges on that interval allowing you to conclude the original integral diverges. Can you find such a constant C?
     
  8. Mar 14, 2013 #7
    i'm sorry. i am still confused. is what i stated above incorrect? the integral is from 0 to 6 not 0 to 3.

    what if i choose 9 for c? i just chose that at random.
     
  9. Mar 14, 2013 #8

    Dick

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    Yes, what you stated before is incorrect. I know your integral is 0 to 6. If you can show it diverges on 0 to 3 then it will diverge on 0 to 6. Why don't you try making a nonrandom choice of C? Use the factorization and figure out what range of values 1/(3+x) takes on [0,3].
     
  10. Mar 14, 2013 #9
    are we working with 1/(3-x) or 1/(3+x) or does it matter because we have switched up a few times?


    if we are using 1/(3+x), do i need to integrate on [0,3] to find the range of values?
     
  11. Mar 14, 2013 #10
    Why don't you just split this integral into two parts and solve for both using limits from the right and left.
    [tex]\int_0^6 \frac{dx}{9-x^{2}}=\lim_{t \to 3^{-}}\int_0^t \frac{dx}{9-x^{2}}+\lim_{t \to 3^{+}}\int_t^6 \frac{dx}{9-x^{2}}[/tex]
    See what I'm saying?
     
  12. Mar 14, 2013 #11
    if it were, 1/(9+x^2) i would choose 1/x^2 and if it converged i would know the other converged.


    the -x^2 is throwing me off in 1/(9-x^2).

    (-1/16) < (1/16) when we get into negatives, correct, i know this sounds elementary, but in this application when we are determining smaller and larger functions are negatives always smaller than positives?
     
  13. Mar 14, 2013 #12
    i want to know how to solve using a limit comparison test or direct comparison test, not by evaluating.
     
  14. Mar 14, 2013 #13

    Dick

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    Negatives are always smaller than positives, but when you doing a comparison test you usually want to compare things of the same sign. Just think about it this way. 1/(9-x^2)=(1/(3+x))(1/(3-x)). If you replace the 1/(3+x) with its minimum on [0,3], then you can also replace the '=' sign with '>='.
     
  15. Mar 14, 2013 #14
    Oh my bad sorry.
     
  16. Mar 14, 2013 #15
    what do you mean by the minimum on [0,3] ? 1/6 would be the minimum, or do you mean integrate on [0,3] ?
     
  17. Mar 14, 2013 #16

    Dick

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    Yes, I mean 1/6. So 1/(9-x^2)>=(1/6)(1/(3-x)). Does the integral of 1/(3-x) converge on [0,3)?
     
  18. Mar 14, 2013 #17
    [itex]\displaystyle\int_0^3 {\frac{1}{3-x} dx}[/itex]

    [itex]\lim_{b \to 3^-} ln|3-x|]^{0}_{b}[/itex]

    [itex]\lim_{b \to 3^-} ln|3-b| - ln|3-0| --> ln|3-3| = ln|0| = ∞ [/itex]

    Thus, the smaller diverges, so the original function diverges.
     
  19. Mar 14, 2013 #18

    Dick

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    Right!
     
  20. Mar 14, 2013 #19
    so explain one last time how you got: 1/(9-x^2)>=(1/6)(1/(3-x))
     
  21. Mar 14, 2013 #20

    Dick

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    It's all in post 13. I factored 1/(9-x^2) and then replaced one of the factors by something smaller.
     
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