1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook