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Math Help - Help finding limit

  1. #1
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    Help finding limit

    Hi

    Can anyone please show me how to establish: \lim_{n \to \infty} \int_0^1\frac{nx^{n-1}}{x+1}dx =1/2

    Thanks a lot.
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  2. #2
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    Attempt:

    Sub  x = 1 - \frac{t}{n}

     n dx = - dt

    the integral becomes :

     \lim_{n\to\infty} \int_n^0 \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ ( -dt)

      = \lim_{n\to\infty}   \int_0^n \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ dt

      =  \lim_{n\to\infty} \int_0^n e^{-t} \left [ \frac{1}{2 - \frac{t}{n} } - \frac{1}{2} + \frac{1}{2} \right ] ~dt

     =  \lim_{n\to\infty} \int_0^n e^{-t} \left [\frac{1}{2 - \frac{t}{n} } - \frac{1}{2} \right ]~dt   ~+~ \frac{1}{2} \lim_{n\to\infty} \int_0^n e^{-t} ~dt

     = \lim_{n\to\infty} \int_0^n e^{-t} \cdot {-\frac{t}{n}}  \cdot \frac{d}{dy} \left( \frac{1}{y} \right)|_{y\to2} ~ dt ~+~ \frac{1}{2}

     =  \lim_{n\to\infty} \int_0^n e^{-t} \cdot {\frac{t}{n}} \cdot \frac{1}{4}~dt ~+~ \frac{1}{2}

     = 0 + \frac{1}{2} = \frac{1}{2}

    Another simplier method that i've just thought of :

    Suppose the limit exists ,

     I_n = \int_0^1 \frac{ n x^{n-1} }{ 1 + x }~dx

     = n  \int_0^1 \frac{ x^{n-1} + x^{n-2 } - x^{n-2}  }{ 1 + x }~dx

     = n \int_0^1 x^{n-2}~dx -  \frac{n}{n-1} I_{n-1}


     I_n = n  \frac{1}{n-1} - \frac{n}{n-1} I_{n-1}

    Let  L = \lim_{n\to\infty} I_n

    we have  L = \lim_{n\to\infty} \frac{n}{n-1} - \lim_{n\to\infty} \frac{n}{n-1} L

     2L = 1  \implies L = \frac{1}{2}
    Last edited by simplependulum; March 15th 2010 at 03:41 AM.
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  3. #3
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    Fantastic. Thanks a lot for helping me out. Though I understand the simpler solution, I am having trouble parsing the first one. I hope its not too much trouble for you to clarify:

    How do you go from

      = \lim_{n\to\infty}   \int_0^n \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ dt

    to

      =  \lim_{n\to\infty} \int_0^n e^{-t} \left [ \frac{1}{2 - \frac{t}{n} } - \frac{1}{2} + \frac{1}{2} \right ] ~dt   ?

    since  \left( 1 - \frac{t}{n} \right)^{n-1}  tends to and not equals e^{-t}

    Thanks again.
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  4. #4
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    Perhaps I was wrong , the limit sign is outside the integral

    so we cannot put  \left( 1 - \frac{t}{n} \right) ^{n-1} = e^{-t}

    I am here try to make some corrections ,

    Put  \frac{1}{ 2 - \frac{t}{n}} = \frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} + \frac{1}{2}

     = \left( \frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} \right ) + \frac{1}{2}

    By MVT, let  f(x) = \frac{1}{x}

     \frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} = - [ \frac{1}{2} - \frac{ 1 }{ 2 - \frac{t}{n} } ] = -\frac{t}{n} f'(c) = \frac{t}{n} \cdot \frac{1}{c^2} , 2 - \frac{t}{n} \leq c \leq 2

    we have  \frac{1}{c^2} \leq 1

     \frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} \leq \frac{t}{n}

    Therefore ,


    = \lim_{n\to\infty} \int_0^n \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ dt



     \leq \lim_{n\to\infty} ( \frac{1}{2} ) \int_0^n \left( 1 - \frac{t}{n} \right )^{n-1}~dt ~  +~ \lim_{n\to\infty}  \int_0^n \frac{t}{n} \left( 1 - \frac{t}{n} \right )^{n-1} ~dt


    In fact ,we don't need to put \left( 1 - \frac{t}{n} \right )^{n-1} = e^{-t}

    However . after the corrections , I seem to have chosen a terrible detour . Why I thought in this way ...


    BTW , just before the bed , I suddenly think of another method , it is also a simple method :


     \lim_{n\to\infty} \int_0^1 \frac{n x^{n-1}}{1 + x}~dx


    Integration by parts ,

     = \frac{x^n}{1 + x} |_0^1 + \int_0^1 \frac{x^n}{(1 + x)^2}~dx

    we just need to prove that the remaining integral tends to zero .
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  5. #5
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    0\le \frac{x^{n}}{(x+1)^{2}}\le x^{n}\implies 0\le \int_{0}^{1}{\frac{x^{n}}{(x+1)^2}\,dx}\le \frac{1}{n+1}.

    the rest follows by the squeeze theorem.
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  6. #6
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    Thanks a lot simplependulum and Krizalid for your help!

    Also, I realised that the problem can be generalized.

    Notice that \lim_{n \to \infty} n\int_0^1 \left(\frac{1}{1+x} \right)x^ndx = \frac{1}{1+1}
    Also, we observe that \forall polynomial P(x): \lim_{n \to \infty} n\int_0^1 P(x)x^ndx = P(1)

    Now, let f be continuous on [0,1]. Since, the conditions for Weierstrauss Approximation theorem
    are satisfied, we can now establish that \lim_{n \to \infty} n\int_0^1 f(x)x^ndx = f(1)

    Proof :
    Given f and for \epsilon < 0 we have some polynomial P_{\epsilon} such that
    P_{\epsilon}(x)-\epsilon < f(x) < P_{\epsilon}(x)+\epsilon \quad \forall x \in [0,1]
    So,  \int_0^1 nP_{\epsilon}(x)x^{n}dx -\epsilon\frac{n}{n+1} < \int_0^1 nf(x)x^{n}dx < \int_0^1 nP_{\epsilon}(x)x^{n}dx +\epsilon\frac{n}{n+1}
    Taking limits we see  P_{\epsilon}(1)-\epsilon < \lim_{n \to \infty} n\int_0^1 f(x)x^ndx < P_{\epsilon}(1)+\epsilon
    That is lim_{n \to \infty}\int_0^1 nf(x)x^{n}dx is \epsilon near P_{\epsilon}(1) and consequently 2\epsilon near f(1).
    Since \epsilon was arbitrary, we are done.
    Last edited by doxian; March 16th 2010 at 01:25 AM.
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