So I have $\displaystyle\lim_{n \to \infty} \int^{n^2}_0 e^{-x^2} n \sin\left(\frac{x}{n}\right) dx$. I'd like to apply the MCT but the trouble is there is a limit which also depends on $n$
So I thought let's do :$\displaystyle\lim_{m \to \infty}\displaystyle\lim_{n \to \infty} \int^{m^2}_0 e^{-x^2} n \sin\left(\frac{x}{n}\right) \,dx$.
First deal with $\displaystyle\lim_{n \to \infty} \int^{m^2}_0 e^{-x^2} n \sin\left(\frac{x}{n}\right)\, dx$ (This is all in my Lebesgue Integration course by the way.)
For this I want to let $f_n=e^{-x^2} n \sin\left(\frac{x}{n}\right)$ and $\displaystyle\lim_{n \to \infty} f_n= x e^{-x^2}=f$ if you think about limit of $\frac{\sin(x)}{x}$ being $1$ as $x$ tends to $0$- that limit should be simple to get. I want to find an upper bound for all the $|f_n|$ -$g(x)$ which is INDEPENDENT of $n$ and apply the Dominated Convergence Theorem so I realised $ne^{-x^2}$ would not work. $f_n$ is not montonically increasing in $n$ so Monotone Convergence theorem I doubt will work. However I plan to use MCT on another sequence $h_m=h\chi_{[0,m^2]}$ where h is the limit $\displaystyle\lim_{n \to \infty} \int^{m^2}_0 e^{-x^2} n \sin\left(\frac{x}{n}\right) \,dx$. I can see none rigorously that $\frac{1}{2}(1-e^{-m^2})$ will come out of first integral which is positive for large enough $m$ so we MCT which only works for non negative functions.
EDIT:Actually wait after that we dont get any integral or anything to do with $x$ we just consider $\displaystyle\lim_{m \to \infty}$ $\frac{1}{2}(1-e^{-m^2})$ and the limit of that is $\frac{1}{2}$ so yup the only problem is DCT at the start then... Could anybody please help?
Edit: Or am I going about this the completely wrong way...
Let $f_n(x) = e^{-x^2} n \sin\left(\dfrac{x}{n}\right)\cdot \chi_{[0,n^2]}(x)$.
On the interval $[0,1]$, this is a monotone increasing sequence of functions, and MCT can be applied.
On the interval $[1,\infty)$, you have
$$ \int^{n^2}_1 e^{-x^2} n \sin\left(\frac{x}{n}\right) \, dx = \int\limits_{[1,\infty)} f_n(x)\,dx $$
and
$$ |f_n(x)| \le xe^{-x^2} $$ and $$ \int\limits_{[1,\infty)} xe^{-x^2} \, dx < \infty $$ so the dominated convergence theorem can be applied.
I'm not at all sure this is the simplest way, but it should work.