Find Limit of Integral: Homework Statement

[snipez90]

Homework Statement

Consider the function f: R -> R, $$f(x) = (x^2 + 1)e^x .$$ Find the limit $$\mathop {\lim }\limits_{n\rightarrow\infty}n\int_{0}^{1}\left(f\left(\frac{x^2}{n}\right) - 1\right).$$

Homework Equations

e^x > x + 1 for nonzero real x

The Attempt at a Solution

After a bit of algebra, we find that the original limit is
$$\mathop {\lim }\limits_{n\rightarrow\infty}\frac{1}{n}\int_{0}^{1}e^{\frac{x^2}{n}}x^4\,dx + \mathop {\lim }\limits_{n\rightarrow\infty}n\int_{0}^{1}\left(e^{\frac{x^2}{n}}-1\right)\,dx.$$

In the first limit, the integrand is increasing on [0,1], so we have
$$0 \leq \frac{1}{n}\int_{0}^{1}e^{\frac{x^2}{n}}x^4\,dx \leq \frac{e}{n},$$
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.

 

[tiny-tim]

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.

Hi snipez90!

Have you tried expanding e^x2/n - 1 as a function of x²/n ?

 

[snipez90]

...

$$\frac{x^2}{n} \leq e^{\frac{x^2}{n}} - 1 &=& \frac{x^2}{n} + \frac{x^4}{n^2}\cdot\frac{1}{2!} + \frac{x^6}{n^3}\cdot\frac{1}{3!} + ... \\ <br /> &=& \frac{x^2}{n} + \frac{x^4}{n^2}\left(\frac{1}{2} + \frac{x^2}{n}\cdot\frac{1}{3!} + \frac{x^4}{n^2}\cdot\frac{1}{4!} + ... \right) \\ <br /> &\leq& \frac{x^2}{n} + \frac{x^4}{n^2}\left(\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...\right) \leq \frac{x^2}{n} + e\cdot\frac{x^4}{n^2}$$

for x in [0,1]. Hence

$$\frac{1}{3} \leq n\int_{0}^{1} e^{\frac{x^2}{n}}\,dx \leq \frac{1}{3} + \frac{e}{5n},$$

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.