I know this is a classical exercise and that I can find many answers around here, but I would like for someone to check if my solution is right.
Suppose that $f:\mathbb{C}\longrightarrow \mathbb{C}$ is a holomorphic function such that $\displaystyle{\lim_{z \to \infty} \frac{f(z)}{z} = 0}$. Prove that $f$ is constant.
My solution is the following:
Let $\varepsilon >0$. Then, $\exists R>0$ such that $\forall z \in \mathbb{C}: |z|>R \implies \frac{|f(z)|}{|z|} < \varepsilon$. Equivalently, $\forall z \in \mathbb{C}: |z|>R \implies |f(z)|<\varepsilon |z|$. Now, the set $\overline{D(0,R)}$ is a compact subset of $\mathbb{C}$, so $f$, which is a continuous function -since it's holomorphic-, takes a maximum value $M$ on $\overline{D(0,R)}$. Note that $M$ depends on $R$ and $R$ depends on $\varepsilon$, so we'll write $M(\varepsilon)$. Now $\forall z \in \overline{D(0,R)}: |f(z)| \leq M(\varepsilon)$ and $\forall z \in \mathbb{C} \setminus \overline{D(0,R)}: |f(z)| < \varepsilon |z|$. We can conclude that $\forall z \in \mathbb{C}: |f(z)| \leq M(\varepsilon) + \varepsilon |z|$. Let $z \in \mathbb{C}$. $\forall r>0$, by Cauchy's integral theorem we get $f'(z) = \displaystyle{\frac{1}{2\pi i}\int_{\gamma_{r}}\frac{f(\zeta)}{(\zeta - z)^2}d\zeta}$, where $\gamma_r(t)=z+re^{it}, 0\leq t\leq 2\pi$. Now $|f'(z)| \leq \displaystyle{\frac{1}{2\pi}\max\{|f(\zeta)|: \zeta \in \overline{\gamma_r}\}\frac{1}{r^2}L(\gamma_r)}$ $\displaystyle{\leq \frac{1}{r}\max\{M(\varepsilon) + \varepsilon|\zeta|: \zeta \in \overline{\gamma_r}\}}\leq \frac{M(\varepsilon)+\varepsilon(r+|z|)}{r} = \varepsilon + \frac{M(\varepsilon)+|z|}{r}$. This holds for every $r>0$, so we can take the limit as $r$ tends to $+\infty$ and we get $|f'(z)|\leq\varepsilon$. But $\varepsilon$ was arbitary small, so $f'(z)=0$, hence $f$ is constant.