On Stein and Shakarchi's Complex Analysis book, in the course of proving that for $f$ a holomorphic (i.e. differentiable) function on an open set, and $C$ a circle whose interior is also contained in the domain of $f$, $$f^{(n)} = \frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta$$ by induction on $n$, the author makes the following assertion: if we let $A=\frac{1}{\zeta-z-h}, B = \frac{1}{\zeta-z}$ we have that $$\frac{f^{(n-1)}(z+h)-f^{(n-1)}(z)}{h} = \frac{(n-1)!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta-z-h)(\zeta-z)}(A^{n-1}+A^{n-2}B+...+B^{n-1}) d\zeta.$$
I get everything until this point. However, I do not get why the following claim by the author implies that I can interchange limit and integral: "but observe that if $h$ is small, then $z+h$ and $z$ stay at a finite distance from the boundary circle $C$, so in the limit as $h \to 0$, we find that the quotient converges to $$\frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta.$$
I know that we can interchange the limit and integral if the limit is uniform. However, it seems very complicated to determine that the integrand $g_m :=\frac{f(\zeta)}{(\zeta-z-h_m)(\zeta-z)}(A^{n-1}+A^{n-2}B+...+B^{n-1})$ converges uniformly to $g :=\frac{f(\zeta)}{(\zeta-z)^{n+1}}$ as $m \to \infty$ directly from the definition of uniform convergence. (Here, $\{h_m\}$ is an arbitrary sequence approaching $0$. The $h$'s in the definition of $A$ is also subtituted by $h_m$ in the definition of $g_m$.) How can the author so quickly justify the switching of the integral and the limit?
Define $g : C \times C^c \to \mathbb{C}$ by $$g(\zeta, z) = \frac{f(\zeta)}{(\zeta - z)^{n}}.$$ Let $g_z : C \times C^c \to \mathbb{C}$ denote the derivative of $g$ wrt $z$. Fix $z \in C^c$. You want to prove that $$\sup_{\zeta \in C}|\frac{g(\zeta, z + h) - g(\zeta, z)}{h} - g_z(\zeta, z)| \to 0 \text{ as } h \to 0.$$ By the FTC, for $\zeta \in C$, and $|h| < r := \text{dist}(z, C)$, $$\frac{g(\zeta, z + h) - g(\zeta, z)}{h} = \int_{0}^{1}g_z(\zeta, z + th)\,dt.$$ Hence for $\zeta \in C$ and $|h| < r$, \begin{align} |\frac{g(\zeta, z + h) - g(\zeta, z)}{h} - g_z(\zeta, z)| &= |\int_{0}^{1}(g_z(\zeta, z + th) - g_z(\zeta, z))\,dt| \\ &\leq \int_{0}^{1}|g_z(\zeta, z + th) - g_z(\zeta, z)|\,dt \\ &\leq \sup_{z_1 \in D_{|h|}(z)}|g_z(\zeta, z_1) - g_z(\zeta, z)| \\ &\leq \omega_{g_z, C \times \overline{D_{|h|}(z)}}(|h|), \end{align} where $\omega_{g_z, C \times \overline{D_{|h|}(z)}}$ is a modulus of continuity for the uniformly continuous function $g_z$ on $C \times \overline{D_{|h|}(z)}$. $g_z$ is indeed uniformly continuous on $C \times \overline{D_{|h|}(z)}$ because $g_z$ is continuous on $C \times C^c$, and $C \times \overline{D_{|h|}(z)}$ is compact. For $|h| < r/2$, $$\omega_{g_z, C \times \overline{D_{|h|}(z)}}(|h|) \leq \omega_{g_z, C \times \overline{D_{r/2}(z)}}(|h|) \to 0 \text{ as } h \to 0.$$ This completes the proof.
Edit: Here are some details on "modulus of continuity". I like to think of the "canonical" modulus of continuity of a function $f : X \to Y$ between two metric spaces as the map $\omega : [0, \infty) \to [0, \infty]$ defined by $$\omega(\delta) = \sup_{d(x_1, x_2) \leq \delta}d(f(x_1), f(x_2)).$$ Note that $d(x_1, x_2) \leq \delta \implies d(f(x_1), f(x_2)) \leq \omega(\delta)$. It is easy to show that $f$ is uniformly continuous if and only if $\omega(\delta) \searrow 0$ as $\delta \searrow 0$.
For this problem, we don't need to compute the modulus of continuity explicitly, i.e. we don't need hard quantitative estimates on it. For this problem, all was needed was that $g_z$ was continuous so that we could conclude that it's modulus of continuity has the limit $0$ at $0$.