Limit of ratio of integrals that go to zero I am working on a statistical problem and trying to understand how to compute
$$
R = \lim_{n\to \infty} \frac{\int_{0}^t x^a (1-x)^b \sqrt{\frac{x}{1+x}}\exp\left(- \frac{n}{v}\left(\frac{x}{1+x}\right)\right)\,dx}{\int_{0}^1 x^a (1-x)^b \sqrt{\frac{x}{1+x}} \exp\left(- \frac{n}{v}\left(\frac{x}{1+x}\right)\right)\,dx},
$$
for $a, b > -1$, $v>0$ and $0 < t < 1$ arbitrary.
I've looked at several resources that discuss when and how one can interchange the limit and integration operations, but here my $f_n$ converge to $0$ and I don't know how to deal with that. I appreciate any insights into how to compute this particular limit, as well as similar limits where the integrand goes to zero as $n$ goes to infinity.
 A: Too long for a comment: not a rigorous proof, but just some heuristic considerations.
$R = \lim_{n\to \infty} \frac{\int_{0}^t x^a (1-x)^b \sqrt{\frac{x}{1+x}}\exp\left(- \frac{n}{v}\left(\frac{x}{1+x}\right)\right)\,dx}{\int_{0}^1 x^a (1-x)^b \sqrt{\frac{x}{1+x}} \exp\left(- \frac{n}{v}\left(\frac{x}{1+x}\right)\right)\,dx}=\lim_{n\to \infty} \frac{(1)}{(2)}$
Let's also denote $m=\frac{n}{v}\to\infty$.
Making change in (1) $x=ts$
$$(1)=t^{a+3/2}\int_0^1s^a(1-st)^b\sqrt{\frac{s}{1+st}}e^{-m\frac{st}{1+st}}ds$$
Making another change $ms=x$
$$(1)=t^{a+3/2}m^{-a-3/2}\int_0^mx^a(1-\frac{t}{m}x)^b\sqrt{\frac{x}{1+xt/m}}e^{-\frac{xt}{1+xt/m}}dx$$
First, we can expand integration to $\infty$ (up to exponentially small terms $\sim e^{-mt}$).
We also note that at $xt\sim \sqrt m \,\,\,\,e^{-xt}\sim e^{-\sqrt m}\ll1 $ and $\frac{xt}{m}\ll1$. Therefore, we can expand the integrand into the Taylor series:
$$(1)\sim t^{a+3/2}m^{-a-3/2}\int_0^\infty x^{a+1/2}\big(1-\frac{bt}{m}x+O(1/m^2)\big)\big(1-\frac{t}{2m}x+O(1/m^2)\big)e^{-xt}e^{(xt)^2/m+O(1/m^2)}dx$$
Expanding $e^{(xt)^2/m+O(1/m^2)}=1+(xt)^2/m+O(1/m^2)$
$$(1)\sim t^{a+3/2}m^{-a-3/2}\int_0^\infty x^{a+1/2}\Big(1-(\frac{bt}{m}+\frac{t}{2m})x+\frac{x^2t^2}{m}+O(1/m^2)\Big)e^{-xt}dx$$
$$=t^{a+3/2}m^{-a-3/2}\int_0^\infty x^{a+1/2}\Big(1-\frac{(2b+t)}{2m}tx+\frac{x^2t^2}{m}+O(1/m^2)\Big)e^{-xt}dx$$
$$=m^{-a-3/2}\Big(\Gamma(a+3/2)+\frac{\Gamma(a+7/2)}{m}-\Gamma(a+5/2)\frac{(2b+3)}{m}+O(1/m^2)\Big)$$
and does not depends on $t$ (up to the exponentially small terms $\sim e^{-tm}$).
As @herb steinberg mentioned above, the integrals are dominated by the value of $x$ near zero, and the limit of the ratio $\frac{(1)}{(2)}$ is equal to 1.
