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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 $0$?

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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.

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    By $o(\frac{t-a_n}{b_n^2})$ I mean that $\frac{o(\frac{t-a_n}{b_n^2})}{\frac{t-a_n}{b_n^2}}\to 0$ as $n\to \infty.$ – 2017-01-18
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    $\to 0$ for fixed $t$ ? Independently of $t\in [a_n,a_n+\delta b_n]$ ? – 2017-01-18
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    Independently of $t$ – 2017-01-18
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    What if $h(t) \to 0$ but $h$ doesn't depend on $n$? – 2017-01-18
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    $h \to 0$ with respect to what variable ? $n$ or $t$ ? – 2017-01-18
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    $h$ depends on $t$. However, as $n$ increases, so do the limits of integration (therefore, $t$ increases with $n$), so I think your work might settle it. – 2017-01-18
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    If you have a function $h(t)$, that doesn't depend on $n$, such that $h(t)\to 0$ if $t\to +\infty$, this is my previous answer... – 2017-01-18
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    Sorry about that! Can you re-post your other answer separately? I wasn't finished looking at it before you deleted it. – 2017-01-18
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    See the edit at the end of the answer. Sorry, I must go. Have a good day. – 2017-01-18