Prove these functions $F_n$ are bounded by a function $G$ 
Hi,
I know that to apply the Dominated Convergence Theorem to these functions, they must be bounded by another function $G$ which does not depend on $n$, for all $x$.
However I'm really struggling to find a function.
In the first case, $F_n$ behaves differently depending on whether $x$ is in $(0,1)$ or larger than $1$, but I can't analyse the $(0,1)$ case, the maximum value keeps changing depending on $n$ and $a$.
In the second, I'm completely stuck as to how to approach it.
Any help would be greatly appreciated!
Thanks

For second, note that
$$
\lim_{n\to\infty}\int_0^{n^2}xe^{-x^2}dx=\lim_{n\to\infty}\frac{1-e^{-n^4}}{2}=\frac1{2}
$$
Since the Taylor series of $\sin x$ is alternating, for $x>0$ we have
$$
|\sin x-x|\leqslant \frac{x^3}{6}
$$
Thus
\begin{align}
\left|\int_0^{n^2}e^{-x^2}n\sin{\frac{x}{n}}dx-\int_0^{n^2}xe^{-x^2}dx\right|&=\left|\int_0^{n^2}\left(e^{-x^2}n\sin{\frac{x}{n}}-xe^{-x^2}\right)dx\right|
\\
&\leqslant\int_0^{n^2}\left|\sin{\frac{x}{n}}-\frac{x}{n}\right|ne^{-x^2}dx
\\
&\leqslant\int_0^{n^2}\frac{x^3}{6n^2}e^{-x^2}dx
\\
&=\frac{1}{12n^2}\left(-x^2e^{-x^2}-e^{-x^2}\right)\bigg |_0^{n^2}
\\
&=\frac{1}{12n^2}(1-e^{-n^4}-n^4e^{-n^4})
\\
&\to0\quad\text{as}\quad n\to \infty
\end{align}
So 
$$
\lim_{n\to\infty}\left|\int_0^{n^2}e^{-x^2}n\sin{\frac{x}{n}}dx-\int_0^{n^2}xe^{-x^2}dx\right|=0
$$
And so
$$
\lim_{n\to\infty}\int_0^{n^2}e^{-x^2}n\sin{\frac{x}{n}}dx=\lim_{n\to\infty}\int_0^{n^2}xe^{-x^2}dx=\frac1{2}
$$
