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Limit of Integrals

  1. Jun 8, 2006 #1
    Ok, i have this problem:

    Calculate: Limit x->0 of ((integral of cos t² dt between 0 and x²) / (integral of e^(-t²) dt))

    So, the limit of both integrals is 0 since the interval between both integrals tends to 0. I used L'Hopital then, so:

    Limit x->0 F'(x)/G'(x) = (cos (x^4) * 2x)/(e^(-x²)), F(x) and G(x) being both integrals.

    I know that this is right, i just don't remember why F'(x) = cos (x^4) * 2x and not F'(x) = cos (x^4)... I know it have something to do with the fact that 2x is the derivative of x² but there's something missing in this explanation.

    Tks in advance.
  2. jcsd
  3. Jun 8, 2006 #2


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    [tex]A(x) = \int _0 ^x \cos t^2\, dt[/tex]

    [tex]B(x) = x^2[/tex]


    F = A o B, so F'(x) = B'(x)A'(B(x)) by chain rule, giving:

    F'(x) = 2xA'(x2) = 2xcos(x4) by FTC
  4. Jun 8, 2006 #3
    Tks for that... but what i really wanted to know is why F'(x) = B'(x)A'(B(x))... IOW a proof of the chain rule. I guess was naturally inclined to think that F'(x) would equal A'(B(x)).
  5. Jun 8, 2006 #4


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    Here's a sketch of a proof of the chain rule:

    [tex] \frac{d}{dx}(f(g(x)))= \lim_{h\rightarrow 0} \frac{f(g(x+h))-f(g(x))}{h} [/tex]

    Now define:

    [tex]k=g(x+h)-g(x) [/tex]

    Then as h goes to zero, k goes to zero, and:

    [tex] \lim_{h \rightarrow 0} \frac{k}{h} = g'(x)[/tex]

    Then we have:

    [tex] \lim_{h\rightarrow 0} \frac{f(g(x+h))-f(g(x))}{h} =\lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{h}[/tex]

    And as k gets very small, h can be replaced by k/g'(x), and:

    [tex] \lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{h}= \lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{k/g'(x)} = \lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{k}g'(x) =f'(g(x)) g'(x) [/tex]

    There are some details to fill in if you want to be rigorous.
  6. Jun 8, 2006 #5
    Great, Tks...
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