Let assume $f(z)$ is defined and analytic at $|z|&gt;1$ and such that $\lim_{z\to \infty}f(z)=L$ when $L&lt;\infty$ Prove: $\lim_{z\to \infty}f'(z)=0$  $f(z)$ is defined and analytic at $|z|&gt;1$ so it can be expanded to a power series $f(z)=\sum_{n=0}^{\infty}\frac{a_n}{z^n}$ at $|z|&gt;1$ now we know that $lim_{z\to \infty}f(z)=\lim_{z\to \infty}\sum_{n=0}^{\infty}\frac{a_n}{z^n}=L$ Taking the derivative we get: $\lim_{z\to \infty}f'(z)=\lim_{z\to \infty}\sum_{n=1}^{\infty}-n*\frac{a_n}{z^{n-1}}=0$ Or am I missing something?
Prove that if $ f\mathrm{(}z\mathrm{)} $ is analytic at the infinity  ,then :  $ {\lim}_{{z}\mathrm{\rightarrow}\mathrm{\infty}}{f}\prime{\mathrm{(}}{z}{\mathrm{)}}\mathrm{{=}}{0} $
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