Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two doubts

  1. Aug 12, 2005 #1
    Hi, Today i had a test, and i was wondering if what i did is correct:

    I had to tell if the [tex]:lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}[/tex] exists. What i did was to say [tex]lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}=lim_{(x,y) \rightarrow (0,0)} \frac {x} {|x|+|y|}*y=0[/tex] because it is bounded multiplied by y that tends to 0.
    Is what i did correct???Something tells me it is not because it was too easy.

    The second doubt i have is about the convergence of the Integral:
    [tex]\int_{0}^{\infty}\frac {sin(x)} {cos(x)+x^2}[/tex]
    My doubt came since the integrand of the series is not always positive. Can i just calculate the convergence of:
    [tex]\int_{0}^{\infty}|\frac {sin(x)} {cos(x)+x^2}|[/tex] and if it converges, then the original integral also converges???
    I would check the convergence of the second integral dividing it between 0-1 and 1-infinity. The integral on the first interval(between 0 and 1) would converge because the integrand is bounded and continuous for x between 0 and 1, and the integral of the second interval(between 1 and infinity) would also converge by comparison with [tex]\int_{1}^{\infty}\frac {1} {x^2-1}[/tex].

    I would really appreciate any help.
    Regards, Paul.
     
  2. jcsd
  3. Aug 13, 2005 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    I'm rusty on stuff like convergence, but what you did with the second problem seems right. However, are you sure that [itex]\int _1 ^{\infty }\frac{1}{x^2 - 1}\, dx[/itex] converges?
     
    Last edited: Aug 13, 2005
  4. Aug 13, 2005 #3

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Since the numerator is zero along that path, the limit will clearly also be zero along that path.

    pbialos, I think you have the right idea. It is indeed not difficult, but the phrasing was a bit vague. It's not clear to me what you mean by "because it is bounded multiplied by y that tends to 0."

    Try something like:

    [tex]\left|\frac {xy} {|x|+|y|}\right|=|y| \frac{|x|}{|x|+|y|} \leq |y|[/tex]
    because [itex]|x|/(|x|+|y|) \leq 1[/itex]. Since |y| approaches zero as (x,y) approaches (0,0), the limit is zero.
     
  5. Aug 13, 2005 #4

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, I don't know what I was thinking... edited.
     
  6. Aug 13, 2005 #5
    thank you

    Many thanks for your responses. Yes, what Galileo said was my idea, although i expressed my self terribly bad.

    Regards, Paul.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook