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Homework Help: 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

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    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.
     
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