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Thread: Help finding limit

  1. #1
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    Help finding limit

    Hi

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

    Thanks a lot.
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  2. #2
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    Attempt:

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

    $\displaystyle n dx = - dt $

    the integral becomes :

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

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

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

    $\displaystyle = \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 $

    $\displaystyle = \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} $

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

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

    Another simplier method that i've just thought of :

    Suppose the limit exists ,

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

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

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


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

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

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

    $\displaystyle 2L = 1 \implies L = \frac{1}{2}$
    Last edited by simplependulum; Mar 15th 2010 at 02:41 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

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

    to

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

    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 $\displaystyle \left( 1 - \frac{t}{n} \right) ^{n-1} = e^{-t} $

    I am here try to make some corrections ,

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

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

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

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

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

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

    Therefore ,


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



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


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


    Integration by parts ,

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

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

    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 $\displaystyle \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): $\displaystyle \lim_{n \to \infty} n\int_0^1 P(x)x^ndx = P(1)$

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

    Proof :
    Given $\displaystyle f$ and for $\displaystyle \epsilon < 0$ we have some polynomial $\displaystyle P_{\epsilon}$ such that
    $\displaystyle P_{\epsilon}(x)-\epsilon < f(x) < P_{\epsilon}(x)+\epsilon \quad \forall x \in [0,1] $
    So, $\displaystyle \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 $\displaystyle P_{\epsilon}(1)-\epsilon < \lim_{n \to \infty} n\int_0^1 f(x)x^ndx < P_{\epsilon}(1)+\epsilon $
    That is $\displaystyle lim_{n \to \infty}\int_0^1 nf(x)x^{n}dx$ is $\displaystyle \epsilon$ near $\displaystyle P_{\epsilon}(1)$ and consequently $\displaystyle 2\epsilon$ near $\displaystyle f(1)$.
    Since $\displaystyle \epsilon$ was arbitrary, we are done.
    Last edited by doxian; Mar 16th 2010 at 12:25 AM.
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