1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of integral question

  1. Feb 28, 2009 #1
    [tex]
    \lim_{x->+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}
    [/tex]
    i was told to differentiate the integral in order to cancel it
    but i dont have 0/0 infinity/infinity form
    in order to differentiate the numerator and denominator.
     
  2. jcsd
  3. Feb 28, 2009 #2
    Of course you have infinity/infinity form. Obviously, [tex] \lim_{x\to +\infty} e^{2x^2} = +\infty [/tex] and [tex] \lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2 [/tex] is infinity as well, since [tex]e^{x^2}[/tex] diverges as x approaches infinity.
     
  4. Mar 1, 2009 #3
    [tex]
    \lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2
    [/tex]
    so you are saying that when we do an ante derivative to the integral we input infinity there
    so it goes to infinity

    [tex]
    \lim_{x->+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}=\lim_{x->+\infty} \frac{(e^{x^2})^2}{4xe^{2x^2}}=\lim_{x->+\infty} \frac{1}{4x}=0
    [/tex]
    is it correct??
     
  5. Mar 1, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "Anti-derivative to the integral"? An integral IS an anti-derivative. It should be clear that [itex]e^{t^2}[/itex] is greater than 1 for all t> 1 so the integral must be unbounded.

    No, you've differentiated wrong. The derivative of (f(x))2 is 2 f(x) f'(x), not (f'(x))2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook