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Limit of integral

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    are there any rules on how to find the limit of an integral equation?

    for example,

    find x such that the limit as y(x) tends to infinitiy of the integral equation equals 1
    [tex]lim \int_0^x \frac{1}{y(t)-y(x)}dt=1[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Im not sure how to do this, can i simply swap the limit sign with the integral sign?

    Thanks in advance.
     
  2. jcsd
  3. May 23, 2010 #2

    gabbagabbahey

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    Huh?:confused:

    [tex]\lim_{y(x)\to\infty}\int_0^x\frac{dt}{y(t)-y(x)}=0[/tex]

    For all values of [itex]x[/itex], so I'm not sure what you mean here.
     
  4. May 23, 2010 #3
    why does the limit = 0
    ?

    what if we had t in the numerator instead of 1?
     
  5. May 23, 2010 #4
    I think you're asking:

    Let:

    [itex]\lim_{x\to a} y(x)=\infty[/itex]

    is there a function y(x) such that:

    [itex]\lim_{x\to a}\int_0^x \frac{1}{y(t)-y(x)}dt=1[/itex]
     
  6. May 23, 2010 #5
    no, i mean, i need to find an 'x'.

    I agree that the integrand is 0 (if y tends to infinity)

    but what if we have

    [tex]\lim_{y(x)\to\infty}\int_0^x\frac{(x-t)^{-1}dt}{y(t)-y(x)}=1[/tex]

    can we find an x such that the limit is 1?
     
  7. May 23, 2010 #6
    If you are fixing x, how can y(x) tend to infinity?
     
  8. May 23, 2010 #7
    because i need to find an x, say x* such that as y(x*) tends to infinity, the limit is 1
     
  9. May 23, 2010 #8
    y(x*) is some number. It doesn't make sense to say that a number tends to infinity. Perhaps I'm misunderstanding the question?
     
  10. May 23, 2010 #9
    sorry, i think im confusing things.

    forget y(x*).

    I want to find x such that
    [tex]\lim_{y(x)\to\infty}\int_0^x\frac{(x-t)^{-1}dt}{y(t)-y(x)}=1[/tex]

    so y(x) is just a function depending on a variable x. then, after i found the limit (in terms of x), i want to solve the equation for x...
    am i making sense?
     
  11. May 23, 2010 #10
    Would the following interpretation be correct:

    Look for x* such that

    [tex]
    \lim_{x \rightarrow x^*} y(x) = \infty
    \quad\text{ and }\quad
    \lim_{x \rightarrow x^*} \int_0^x \frac{(x-t)^{-1}dt}{y(t)-y(x)} = 1,
    [/tex]

    where we always approach x* from the direction of 0.

    But even if this isn't the right interpretation, it seems important to know what y(x) is as well.
     
  12. May 23, 2010 #11
    yes, this is the right interpretation.
    is it possible to find x* without knowing what the function is?
     
  13. May 23, 2010 #12

    gabbagabbahey

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    Are you sure your integrand isn't [tex]\frac{y(t)-y(x)}{t-x}[/tex] instead?
     
  14. May 23, 2010 #13
    yes, im sure.
    would it be easier if it was?
     
  15. May 23, 2010 #14
    In addition, I don't see how to find x* without knowing the function. I mean, x* appears at a vertical asymptote, but different functions have different asymptotes.
     
  16. May 23, 2010 #15
    lets assume it has a vertical asymptote.
    would we be able to find x* then?
     
  17. May 23, 2010 #16
    Evaluate the improper integral at most at two candidates for x*, one to the left and right of zero. Then, if one of those equals 1, you're good. I don't know another way.
     
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