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Math Help - limit and integration

  1. #1
    Senior Member Sambit's Avatar
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    Question limit and integration

    Evaluate  {\lim }\limits_{x \to 0} \frac{\int_0^x\,e^{t^2}(e^t-1)^2\, dt}{x(1-cos x)}
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by Sambit View Post
    Evaluate  {\lim }\limits_{x \to 0} \frac{\int_0^x\,e^{t^2}(e^t-1)^2\, dt}{x(1-cos x)}
    Apply l'Hospital's Rule (and recall the Fundamental Theorem of Calculus when doing so).
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hi, I will continue mr fantastic idea...


    Why you can apply l'Hospital's rule?

    Let we see what happens to the whole expession when x\to 0:



    \frac{\int_0^0\,e^{t^2}(e^t-1)^2\, dt}{0(1-cos 0)}=\frac{0}{0}

    Next, {x (1-cos(x))}' = {x sin(x)-cos(x)+1}\neq 0 for every x\neq0

    Applying l'Hospital's rule:
    Now by the Fundamental Theorem of Calculus:

    {\int_0^x\,e^{t^2}(e^t-1)^2\ dt}}'=e^{x^2}(e^x-1)^2

    {\lim }\limits_{x \to 0} {\frac{(\int_0^x\,e^{t^2}(e^t-1)^2\, dt)'}{(x(1-cos x))'}={\lim }\limits_{x \to 0} \frac{e^{x^2}(e^x-1)^2}{x sin(x)-cos(x)+1}

    The limit is still from the form of \frac{0}{0}, applying l'Hospital's rule again...
    Last edited by mr fantastic; October 20th 2010 at 05:04 AM. Reason: I would like to see the OP do some work before the complete solution is posted.
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  4. #4
    Senior Member Sambit's Avatar
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    ...............in this manner, if L'Hospitals rule is applied, at the  3rd oreder differentiation, we get:-
    \frac{((8x^3+24x^2+36x+20)e^{2x}+(-16x^3-24x^2-36x-14)e^x+8x^3+12x)e^{x^2}}{3cos x-xsin x}

    putting x=0, we get:-  6/3 = 2

    am i right?
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  5. #5
    Flow Master
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    Quote Originally Posted by Sambit View Post
    ...............in this manner, if L'Hospitals rule is applied, at the  3rd oreder differentiation, we get:-
    \frac{((8x^3+24x^2+36x+20)e^{2x}+(-16x^3-24x^2-36x-14)e^x+8x^3+12x)e^{x^2}}{3cos x-xsin x}

    putting x=0, we get:-  6/3 = 2

    am i right?
    No.

    Here is the rest of post #3:

    Quote Originally Posted by Also sprach Zarathustra
    {\lim }\limits_{x \to 0} \frac{e^{x^2}(e^x-1)^2)'}{(x sin(x)-cos(x)+1)'}={\lim }\limits_{x \to 0} \frac{2xe^{x^2}(e^x-1)^2+2e^{x^2}e^x(e^x-1)}{2sin(x)+xcos(x)}

    ... l'Hospital's rule again...

    {\lim }\limits_{x \to 0} \frac{(2xe^{x^2}(e^x-1)^2+2e^{x^2}e^x(e^x-1))'}{(2sin(x)+xcos(x))'}=2{\lim }\limits_{x \to 0} \frac{-(e^{x^2} (2 x^2+e^{2 x} (2 x^2+4 x+3)-e^x (4 x^2+4 x+3)+1)}{xsin(x)-3cos(x)}...=\frac{2}{3}
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  6. #6
    Senior Member Sambit's Avatar
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    Lightbulb

    oh yes... actually my denominator was right, but i differentiated the numerator 4 times instead of 3 times....now i have got it correct...
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