I'm looking at the proof of the Cauchy integral formula given in Stein and Shakarchi's Complex Analysis.
Corollary 4.2 If $f$ is holomorphic in an open set $\Omega$, then $f$ has infinitely many complex derivatives in $\Omega.$ Moreover, if $C\subset \Omega$ is a circle whose interior is also contained in $\Omega$, then $$f^{(n)}(z)=\frac{n!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta -z)^{n+1}} d\zeta$$ for all $z$ in the interior of $C$.
I don't understand the last paragraph in the below proof. Why does he say that because $z+h$ and $z$ stay at a finite distance from the boundary circle $C$, so the limit as $h$ tends to $-$, we have the final convergence? I think he's switching integral and the limit here, but to do that, don't we need uniform convergence? I'm not sure how this fact is exactly established. I would greatly appreciate any help to understand this completely.
