Hurwitz's theorem:
"Let $\{ f_n\}$ be a sequence of analytic functions on an open set $D$, which converges uniformly to a function $f$ on compact subsets of $D$. If the functions $f_n$ are nowhere zero, then $f$ is either identically zero or nowhere zero".
The following proof by contradiction is taken from my textbook (modified a little):
Suppose $f (z) = 0$ for some $z ∈ D$ and that $f(z) \not \equiv 0$. Take a small curve $\gamma \subset D $, centered at $z$, then $f(z) \neq 0$ on $\gamma$. Thus, $$ \frac{f'_n}{f_n} \to \frac{f'}{f} \text{ on } \gamma $$ and therefore, $$ \frac{1}{2\pi i}\oint_{\gamma}\frac{f'(z)}{f(z)}dz=\lim_{n\to\infty}\frac{1}{2\pi i}\oint_{\gamma}\frac{f_n'(z)}{f_n(z)}dz=0 \tag{*} $$ However, $$ \frac{1}{2\pi i}\oint_{\gamma}\frac{f'(z)}{f(z)}dz \geq 1 $$ according to the argument principle. Thus, we have a contradiction and therefore $f(z) \equiv 0$ or nowhere zero.
My question is:
In (*), why is $\lim_{n\to\infty}\frac{1}{2\pi i}\oint_{\gamma}\frac{f_n'(z)}{f_n(z)}dz=0$? Based on my understanding, if $f_n \to f$, then $\lim_{n\to\infty}\frac{1}{2\pi i}\oint_{\gamma}\frac{f_n'(z)}{f_n(z)}dz \text{ shoud}=1$ instead.
Each function $f_n$ is nowhere $0$, and therefore, by the argument principle$$\oint\frac{f_n'(z)}{f_n(z)}\,\mathrm dz=0.$$So$$\lim_{n\to\infty}\oint\frac{f_n'(z)}{f_n(z)}\,\mathrm dz=0$$