Question
Let $f:\Bbb{C}\rightarrow \Bbb{C}$ be an entire function such that $$\lim_{|z|\rightarrow\infty} \dfrac{f(z)}z=0.$$ Show that $f$ is constant function.
Attempt:
Since $\lim_{|z|\rightarrow\infty} \dfrac{f(z)}z$ finitely exists, so $\dfrac{f(z)}z$ is bounded in a neighborhood of infinity. Then there exists $R>0$ and a constant $M>0$ such that $\dfrac{|f(z)|}{|z|}\le M$ for $|z|>R$. Thus, $|f(z)|\le M|z|$ for $|z|>R$.
Now, I have proved the following result;
Generalised Liouville's theorem: Let $f$ be an entire function and suppose there is a constant $M$ and $R>0$ and an integer $n\ge 1$ such that $|f(z)|\le M|z|^n$ for $|z|>R$. Show that $f$ is a polynomial of degree $\le n$.
From the above result we get $f$ is a linear polynomial, that is, $f(z)=a+bz$ for some $a,b\in \Bbb{C}$. But $\lim_{|z|\rightarrow\infty} \dfrac{f(z)}z=0$ gives $b=0$. Hence it is done.
Please check whether all the arguments are alright or not. Thank you.
Define a function $g(z):=\frac{f(z)-f(0)}{z}$ if $z\neq 0$ and $g(0)=f'(0).$ Note that $g(z)$ is a entire and $\lim\limits_{|z|\to \infty}|g(z)|= 0$. This shows that $g(z)$ bounded and hence by Lioville's theorem $g(z)\equiv 0.$ This in turn shows that $f(z)\equiv f(0).$
Alternatively, as you have already proved the generalized Liouville's theorem. You know that $f(z)$ is a polynomial of degree $\le 1.$ So take $f(z)=az+b.$ Note that the condition $\lim\limits_{|z|\to \infty}\frac{f(z)}{|z|}=0$ implies that $a=0$ and hence $f(z)\equiv b.$
Alternatively, you can use Cauchy's integral formula to directly show that $f(w_0)=f(w_1)$ for any $w_0, w_1\in \mathbb{C}.$ Choose a disk of radius $R$ large enough so that the disk contains both $w_0, w_1.$ Let $C_R$ be the boundary curve of the disk. Now, Cauchy's formula implies that $$|f(w_0)-f(w_1)|\le \frac{R}{2\pi}\int_{C_{R}}\frac{|f(z)|}{|z|}\left(\frac{|w_1-w_0|}{|(z-w_0)(z-w_1)|}\right)|dz|\to 0,$$ as $R\to \infty.$