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  1. #1
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    really hard limit of an integral!

    limit as x goes to 0 of (1/x) * the integral from 0 to x of (1-tan(2t))^(1/t) dt
    Last edited by stones44; October 22nd 2009 at 06:05 PM.
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  2. #2
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    Quote Originally Posted by stones44 View Post
    limit as x goes to 0 of (1/x) * the integral from 0 to x of (1-tan(2t))^(1/t) dt
    Is this what it looks like?

    \lim_{x->0}\frac{1}{x}\int_{x}^0(1-tan(2t))^{\frac{1}{t}}dt

    If so, I will help you work on it. I haven't worked it out yet, but I think that we need to use the squeeze theorem. Also remember that the function \int_{x}^0(1-tan(2t))^{\frac{1}{t}}dt=-\int_{0}^x(1-tan(2t))^{\frac{1}{t}}dt
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  3. #3
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    flip the limits and yes

    so

    \lim_{x->0}\frac{1}{x}\int_{0}^x(1-tan(2t))^{\frac{1}{t}}dt
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  4. #4
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    i dont think i know that theorem yet so i doubt that is the way i am expected to solve it
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  5. #5
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    Quote Originally Posted by stones44 View Post
    limit as x goes to 0 of (1/x) * the integral from 0 to x of (1-tan(2t))^(1/t) dt
    \lim_{x \rightarrow 0} \frac{\int_0^x (1 - \tan(2t))^{1/t} \, dt}{x} = \lim_{x \rightarrow 0} (1 - \tan(2x))^{1/x}

    using l'Hopital's Rule and the Fundamental Theorem of Calculus.

    Now note that (1 - \tan(2x))^{1/x} = e^{\ln (1 - \tan(2x))^{1/x}} = e^{\frac{1}{x} \ln (1 - \tan(2x))}

    so you should now use l'Hopital's rule to find \lim_{x \rightarrow 0} \frac{\ln (1 - \tan(2x))}{x} .
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