Can we get quantitative information from the Dominated (resp. Monotone) Convergence Theorem? It's a common exercise in measure theory courses to solve problems like the following:

Compute $$\lim_{n \to \infty} \int_0^\infty (1+ (x/n))^{-n} \sin(x/n)\,\mathrm{d}x.$$

(cf. Folland Ch 2.4, exercise 28a).
The dominated convergence theorem makes short work of things like this, since (for large $n$) we get roughly exponential decay. A little bit of work turns this intuition into a pointwise bound, and then we swap $\lim \int = \int \lim$ to solve the problem.
Often in computer science, combinatorics, etc. this is not enough. We're interested in the rate of convergence. That is, it's not enough to know
$$
\lim_{n \to \infty} \int_0^\infty (1+ (x/n))^{-n} \sin(x/n)\ dx 
= \int_0^\infty 0\  dx
= 0
$$
We want to say that
$$\int_0^\infty (1+ (x/n))^{-n} \sin(x/n)\ dx = O(1/n),$$
or something similar.
I'm not well versed with computing integrals while using estimates of this kind, but I'm sure that people have thought about this. Is there a way to get these kinds of error estimates from the dominated (resp. monotone) convergence theorem? I would be happy to use this problem (or any others that you may feel are more emblematic of the technique) as a case study.
I suspect one could argue by integrating some error bounds between $f_n$ and $f$, and it may be helpful to look at, say $f_n \cdot \chi_{[0,n]}$ in order to make the error bounds more useful. This feels like it is kind of sidestepping the convergence theorems entirely, though, and I'm curious if there's a more elegant approach.

Thanks in advance! ^_^
 A: Asymptotic analysis- especially of integrals with infinite bounds- is somewhat of an art. There exists a proliferation of techniques whose goal is to estimate the leading behavior of an integral with respect to some interesting limit.  A good place to start diving into the field is the book "Introduction to asymptotics" by D.S. Jones (and of course there are other readily googlable choices).
The convergence theorems you mention unfortunately cannot provide any information on how to perform asymptotic estimates on integrals. Sometimes, procedures and lemmata used to prove that DCT or MCT are applicable may hint at the right asymptotic behavior (this was hinted in the comment section).
In this particular example, it is very simple to obtain arbitrarily high orders of an asymptotic formula involving only powers of $n$. Simply rewrite $(1+x/n)^{-n}=e^{-n\ln(1+x/n)}$ and Taylor expand the integrand in powers of $1/n$:
$$\sin(x/n)e^{-n\ln(1+x/n)}=\frac{xe^{-x}}{n} + \frac{x^3e^{-x} }{2 n^2} + \frac{x^3 e^{-x}  (-4 - 8 x + 3 x^2)}{ 24 n^3}+\mathcal{O}(n^{-4})$$
Due to the presence of $e^{-x}$ on every single one of the coefficients, the integrals can be computed in this case and we end up with the asymptotic estimate
$$I(n)=\int_{0}^{\infty}\frac{\sin(x/n)}{(1+x/n)^n}dx\sim\frac{1}{n}+\frac{3}{n^2}+\frac{6}{n^3}+\mathcal{O}\left(\frac{1}{n^4}\right)$$
You can verify that you can obtain asymptotic estimates of any order, and that order by order, the limit $n^{m+1}(I(n)-\sum_{m=1}^N c_m/n^m)$ exists and is finite. However, you may complain that this series does not converge, and you would be right, but that's the nature of most asymptotic series. As this list comprised of exactly one technique for solving  asymptotics is far from being conclusive, I really do hope this helps!
A: I think the answer in total generality has to be no. For any sequence $a_n$ with a limit of $0$, Let $f_n$ be the function which is equal to $a_n$ on the closed unit interval $[0, 1]$ and 0 elsewhere.
Then you could apply the dominated convergence theorem to get $\lim_{n \to \infty} \int f_n = 0$, but of course you can also just compute the integral, which is $a_n$. But $a_n$ can decay as slowly as you like, e.g. take $a_n = 1/\log(\log(n))$ (for $n$ large).
