If ${a_n}>0$ and strictly increasing, then $\lim_n \int\limits_0^1 \frac{a_nx}{1+a_nx}dx = \int\limits_0^1 \lim\limits_n \frac{a_nx}{1+a_nx}dx$ I don't know what to make of this integral. I know the two limits will be equal provided that both the inner limits exist in $\mathbb{R}$ and the convergence on the right inner limit is uniform, but I'm not sure how to proceed. All help is greatly appreciated.

It is easy to calculate that $x-\log (x+1)$ is the antiderivative of $\frac{x}{1+x}$. Making the change of variables $y=cx$ we arrive at
$$
(\star)\hspace{.7cm}\int_0^1\frac{cx}{1+cx}dx=\frac{1}{c}\int_0^c\frac{y}{1+y}dy=\frac{1}{c}\left[c-\log (c+1)\right]=1-\frac{\log (c+1)}{c},
$$
where $c>0$ is some constant. As $a_n$ is increasing then either $\lim_{n\to\infty}a_n=a\in\mathbb{R}$ or $a_n\nearrow+\infty$.
$\bullet$ If $a_n\to a$ then we obtain from $(\star)$ that
\begin{align}
&\lim_{n\to\infty}\int_0^1\frac{a_nx}{1+a_nx}dx=1-\lim_{n\to\infty}\frac{\log(a_n+1)}{a_n}=1-\frac{\log(a+1)}{a}\\
&\int_0^1\lim_{n\to\infty}\frac{a_nx}{1+a_nx}dx=\int_0^1\frac{ax}{1+ax}dx=1-\frac{\log(a+1)}{a}.\\
\end{align}
$\bullet$ If $a_n\nearrow+\infty$ then, again by $(\star)$,
\begin{align}
&\lim_{n\to\infty}\int_0^1\frac{a_nx}{1+a_nx}dx=1-\lim_{n\to\infty}\frac{\log(a_n+1)}{a_n}=1=\int_0^11dx=\int_0^1\lim_{n\to\infty}\frac{a_nx}{1+a_nx}dx.
\end{align}
