$\lim_{r\to\infty} \int_0^\infty rx/(exp(x) + r^2x)dx = 0$ I've got a feeling I need to use the Dominated Convergence Theorem, but I'm honestly not sure how to even begin tackling this.
I realize the limit of the integrand will ultimately be $0$ as the denominator will be dominated by n, but I'm not sure how to continue.
 A: You can find a dominating function  by differentiating the integrand with respect to $n$. If not clear, here are more details.
Fix $x>0$. For $t\in [1,\infty)$, define $f_t (x) = \frac{t x} {e^x + t^2 x}$.
$$\frac{d}{dt} \frac{1}{f_t (x)} = \frac{d}{dt} (\frac{e^x + t^2x}{tx}) = -\frac{1}{t^2} \frac{e^x}{x}+1.$$
This derivative is increasing in $t$, is negative at $t=1$ ($e^x>x$ for $x>0$) and tends to $1$ as $t\to\infty$.
Therefore the unique minimum  over $t \ge 1$ is attained when $t^2 = e^{x}/x$. As a result, for all $t\ge 1$ we have
$$f_t (x) \le f_{e^{x/2}/\sqrt{x}}(x) = \frac {\frac{e^{x/2}} { \sqrt{x} }x} {e^x + e^x} =\frac{1}{2} \sqrt{x} e^{-x/2}.$$
This upper bound clearly holds for all $t\in{\mathbb N}$. Furthermore the function on the righthand side is integrable. Therefore we can apply the dominated convergence theorem with it as the dominating function.
A: Let $f_n(x)$ be the integrand of your problem, so if we were to differentiate $f_n(x)$, we discover that
$$
f_n'(x)={ne^x+n^3x-nxe^x-n^3\over(e^x+n^2x)^2}
$$
and by some further manipulation we see that this function attains maximum at $x=1$, indicating
$$
0\le f_n(x)\le{n\over e+n^2}
$$
Thus we have
$$
\lim_{n\to\infty}\sup_{x\ge0}|f_n(x)-0|=\lim_{n\to\infty}{n\over e+n^2}=0
$$
Therefore, $f_n(x)$ converges uniformly to zero on $[0,\infty)$, which justified the interchanging:
$$
\lim_{n\to\infty}\int_0^\infty f_n(x)\mathrm dx=\int_0^\infty\lim_{n\to\infty}f_n(x)\mathrm dx=0
$$
