If we take your definition of little-O, taking $g(\zeta) = \zeta -z$, taking $f$ to be the little-o function of $g$ in the numerator, and $D$ to be the domain you have:
$$\lim_{\zeta \to z}\frac{f(\zeta)}{g(\zeta)} = 0$$ implies
$$\forall \epsilon>0, \exists \delta>0: \forall \zeta \in D: 0<|\zeta - z| < \delta \implies |\frac{f(\zeta)}{g(\zeta)}|<\epsilon$$
$$\implies |\frac{f(\zeta)}{\zeta-z}|<\epsilon$$
By the definition of limits.
So your however small ($\epsilon$) you want the magnitude of your quotient ($|\frac{f(\zeta)}{\zeta-z}|$) to be, you can bound it below that just by finding the appropriate $\delta$.
So your integrand can be as small as you wish. If you're asking about the integral itself, look up the estimation lemma.
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