Let $a_n>b_n>0$ be sequences tending to infinity and $x=a_n+\delta b_n$ for some constant $\delta$. I am trying to check whether
$$\int_{a_n}^xo\bigg(\dfrac{t-a_n}{b_n^2}\bigg)\mathrm dt\to 0$$
I cannot apply dominated convergence since the limits of integration are varying. But two reasons why I think the sequence of integrals tend to $0$ are because (i) the lower and upper limits are tending to infinity and $o\bigg({\dfrac{x-a_n}{b_n^2}}\bigg)\to 0$ and (ii) $o\bigg({\dfrac{x-a_n}{b_n^2}}\bigg)$ tends to zero so fast that even when we divide it by $\frac{x-a_n}{b_n^2}=\delta/b_n$ (which also tends to 0), the quotient $$\dfrac{o\bigg({\dfrac{x-a_n}{b_n^2}}\bigg)}{\delta/b_n}$$ also tends to $0$.
If this problem is well-formulated, do the integrals tend to $0$?
If your hypothesis is that under the integral sign we have a fonction $f(t)=h_n(t)(\frac{t-a_n}{b_n^2})$ with a function $h_n$, such that $h_n(t)\to 0$ if $n\to +\infty$ independently of $t$, then:
Let $$u_n=\int_{a_n}^{a_n+\delta b_n}h_n(t)(\frac{t-a_n}{b_n^2})dt=\frac{1}{b_n^2}\int_0^{\delta b_n}h_n(u+a_n)udu$$ Let $\varepsilon>0$. As $h_n\to 0$ if $n\to +\infty$, independently in $t$, there exist an $N$ such that for $n\geq N$, we have $|h_n(u+a_n)|\leq \varepsilon$ for $u\geq 0$, and we get immediately that $|u_n|\leq \delta²\varepsilon/2$ for $n\geq N$.
Edit: If your hypothesis is that $h_n(t)=h(t)$ does not depend on $n$, but we have that $h(t)\to 0$ as $t\to +\infty$, then use that as $a_n\to +\infty$, there exist an $N$ such that for $n\geq N$ we have $|h(u+a_n)|\leq \varepsilon$ for $u\geq 0$, and the end of the proof is the same.