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Limit of an Integral

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the function f: R -> R, [tex]f(x) = (x^2 + 1)e^x .[/tex] Find the limit [tex]\mathop {\lim }\limits_{n\rightarrow\infty}n\int_{0}^{1}\left(f\left(\frac{x^2}{n}\right) - 1\right).[/tex]


    2. Relevant equations
    e^x > x + 1 for nonzero real x

    3. The attempt at a solution
    After a bit of algebra, we find that the original limit is
    [tex]\mathop {\lim }\limits_{n\rightarrow\infty}\frac{1}{n}\int_{0}^{1}e^{\frac{x^2}{n}}x^4\,dx + \mathop {\lim }\limits_{n\rightarrow\infty}n\int_{0}^{1}\left(e^{\frac{x^2}{n}}-1\right)\,dx.[/tex]

    In the first limit, the integrand is increasing on [0,1], so we have
    [tex]0 \leq \frac{1}{n}\int_{0}^{1}e^{\frac{x^2}{n}}x^4\,dx \leq \frac{e}{n},[/tex]
    which implies that the first limit is 0 by the squeeze theorem.

    I'm not sure how to compute the second limit though. I can find a lower bound on the integrand via e^x > x + 1, but what is a suitable upper bound on the integrand? Thanks in advance.
     
  2. jcsd
  3. Aug 23, 2009 #2

    tiny-tim

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    Hi snipez90! :smile:

    Have you tried expanding ex2/n - 1 as a function of x2/n ?
     
  4. Aug 23, 2009 #3
    ...

    [tex] \frac{x^2}{n} \leq e^{\frac{x^2}{n}} - 1 &=& \frac{x^2}{n} + \frac{x^4}{n^2}\cdot\frac{1}{2!} + \frac{x^6}{n^3}\cdot\frac{1}{3!} + ... \\
    &=& \frac{x^2}{n} + \frac{x^4}{n^2}\left(\frac{1}{2} + \frac{x^2}{n}\cdot\frac{1}{3!} + \frac{x^4}{n^2}\cdot\frac{1}{4!} + ... \right) \\
    &\leq& \frac{x^2}{n} + \frac{x^4}{n^2}\left(\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...\right) \leq \frac{x^2}{n} + e\cdot\frac{x^4}{n^2} [/tex]

    for x in [0,1]. Hence

    [tex]\frac{1}{3} \leq n\int_{0}^{1} e^{\frac{x^2}{n}}\,dx \leq \frac{1}{3} + \frac{e}{5n},[/tex]

    from which it follows that the limit is 1/3 as n -> infinity by another application of the squeeze theorem, which should be the answer to the original limit.

    Thanks tiny-tim.
     
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