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Math Help - limit comparison test

  1. #1
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    limit comparison test

    I'm a bit confused about how to use this test. So the theorem for it basically says, if the limit as x -> inf. f(x) / g(x) =L and 0<L<inf.

    then the integral of f(x) and the integral of g(x) both converge or diverge.

    so if i have a particular function that i need to find whether it converges or diverges, and i want to use this test, when i pick another function to compare it to, do i have to pick one that i already know whether or not it converges?
    an example that is shown in my book is:
    integral dx/(1+x^2) (and it's from 1 to inf.) so the function is just 1/(1+x^2).
    they choose to compare it to 1/x^2 because the two functions are essentially the same at infinity. but that's what is confusing me. so should i choose a function that is similar to the given one at infinity, or just pick a function that i already know if it converges or diverges, or both? i think it's a crappy example.
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  2. #2
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    Quote Originally Posted by isuckatcalc View Post
    I'm a bit confused about how to use this test. So the theorem for it basically says, if the limit as x -> inf. f(x) / g(x) =L and 0<L<inf.

    then the integral of f(x) and the integral of g(x) both converge or diverge.
    you also need both f(x) and g(x) to be continuous and positive in the interval.

    so should i choose a function that is similar to the given one at infinity, or just pick a function that i already know if it converges or diverges, or both? i think it's a crappy example.
    both! this being "similar" should be defined as \lim_{x \to\infty} \frac{f(x)}{g(x)}=L, where 0 < L < \infty.
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  3. #3
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    Thanks, that helps a lot. If I use this test, and I get an indeterminate form, I can then use L' Hopital's rule, right?
    I'm having trouble with one in particular though.

    I'm looking at:
    (ln(x))^2 / x^3
    So I compare it to 1/x^2.
    And when I divide the two, I get:
    (ln(x))^2 / x. and as x-> inf, they both go to inf. So use L' hopital's rule and get
    2ln(x)/x. still both go to inf. so use it again to get
    2/x. but that goes to 0. so that doesn't help me at all, does it? What am I doing wrong?
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  4. #4
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    Quote Originally Posted by isuckatcalc View Post
    Thanks, that helps a lot. If I use this test, and I get an indeterminate form, I can then use L' Hopital's rule, right?
    I'm having trouble with one in particular though.

    I'm looking at:
    (ln(x))^2 / x^3
    So I compare it to 1/x^2.
    And when I divide the two, I get:
    (ln(x))^2 / x. and as x-> inf, they both go to inf. So use L' hopital's rule and get
    2ln(x)/x. still both go to inf. so use it again to get
    2/x. but that goes to 0. so that doesn't help me at all, does it? What am I doing wrong?
    you're not doing anything wrong! the point is that we need the complete version of limit comparison test for improper integrals here. i don't know if you'be been taught this or not though!

    so what if L=\lim_{x\to\infty} \frac{f(x)}{g(x)} = 0 \ \text{or} \ \infty? then this is what we can say: if L=0, then convergence of \int_a^{\infty} g(x) \ dx implies convergence of \int_a^{\infty} f(x) \ dx. if L=\infty, then divergence of

    \int_a^{\infty} g(x) \ dx implies divergence of \int_a^{\infty} f(x) \ dx.
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  5. #5
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    That's just what I needed, thank you so much.
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