Prove that if $f(z)$ is analytic at infinity, then $\lim_{z \to \infty}{ f'(z)}= 0$

Question

I don't really know how can i prove that. I know a function $f(z)$ is analytic at $z=\infty$ means $f(\frac{1}{\xi})$ is analytic at $\xi = 0$.

In particular:

$$\lim_{z \to \infty}{f(z)} = \lim_{\xi \to 0}{f(\frac{1}{\xi})} = f(\infty)$$

Always exists and is finite. Also, I think it follows that:

$$\lim_{z \to \infty}{f'(z)}=\lim_{\xi \to 0}{f'(\frac{1}{\xi})}$$

But I do not know how to continue.

Answer 1

Expanding a suggestion by Daniel Fischer:

Let $g(\xi) = f(1/\xi)$. Then $g'(\xi)=-f'(1/\xi)/\xi^2$. Since $g$ is holomorphic at $0$, the derivative $g'$ is bounded in a neighborhood of $0$: $|g'|\le M$. It follows that $$|f'(1/\xi)|\le M|\xi|^2 $$ or, to put it another way, $$|f'(z)|\le M/|z|^2 $$ when $|z|$ is sufficiently large.