Let $f(z) : X \to \mathbb{C}$ be an (analytic) complex function, $X$ is $\mathbb{C}$ except finite sets. If $\lim_{z\to \infty}f(z)=0$, then $f$ is bounded?
This question originates from following proof ( Joseph Bak, Complex analysis, p.125 ) :
Why $A_n$ is bounded function? Can anyone help?
Since $A_n$ approaches $0$ when $z \to \infty$, there is $M>0$ such that $|A_n(z)| \le 1$ if $|z|>M$.
By the continuity of $A_n$, there is some $N>0$ such that $|A(z)|\le N$ on the compact disk centered on $0$ and with radius $M$.
Therefore, $$|A(z)| \le\max\{1,N\}.$$