I have the following question:
Prove that $\displaystyle\lim_{n \rightarrow \infty} \int_0^{n^2} e^{-x^2}n\sin\left(\frac{x}{n}\right) \ dx = \frac{1}{2}$
When doing this question, I showed that $\displaystyle\lim_{n \rightarrow \infty} e^{-x^2}n\sin\left(\frac{x}{n}\right) = xe^{-x^2}$
and that on the interval $E_n = [0, n^2]$ the function is bounded by an integrable function and that the sequence of functions are continuous (as the product of continuous functions) and hence are measurable - thus DCT can be used. Now, this is where I'm somewhat confused. The rest of my working went like this:
$\displaystyle\lim_{n \rightarrow \infty} \int_0^{n^2} e^{-x^2}n\sin\left(\frac{x}{n}\right) \ dx = \lim_{n \rightarrow \infty} \int_0^{\infty} e^{-x^2}n\sin\left(\frac{x}{n}\right) \ dx$
$ = \displaystyle \int_0^{\infty} xe^{-x^2} \ dx \ \ \ \ \ \ $ (by DCT)
$ = \displaystyle\lim_{x \rightarrow \infty} \left(-\frac{e^{-x^2}}{2}\right) - \left(-\frac{1}{2}\right) = \frac{1}{2}$
Clearly this gives the right answer, but I'm concerned about the justification of the first equality:
$\displaystyle\lim_{n \rightarrow \infty} \int_0^{n^2} e^{-x^2}n\sin\left(\frac{x}{n}\right) \ dx = \lim_{n \rightarrow \infty} \int_0^{\infty} e^{-x^2}n\sin\left(\frac{x}{n}\right) \ dx$
I'm guessing there's some property of $e^{-x^2}n\sin\left(\frac{x}{n}\right)$ that makes this permissible, or is it simply not correct?
Thanks