Show f(z)/z is bounded if $\lim\limits_{|z| \to \infty} \frac{f(z)}{z} = 0$ then how do I show that f is bounded. Intuitively, this makes sense to me but I having trouble writing it out formally. I was thing for $|z|>N$, $|f(z)/z|<\epsilon$ and then we know |1/z| will be less than some $\epsilon^2$. Then can I say that these two facts give us that $|f(z)|<$ some number? Does that even help me? 
 A: Since $f(1/z)$ is holomorphic on $\Bbb C \setminus \{0\}$ with $\lim_{z\to 0} zf(1/z) = \lim_{|z| \to \infty} f(z)/z = 0$, $f(1/z)$ has a removable singularity at $z = 0$. So $f(1/z)$ is bounded in a deleted neighborhood of $0$, which implies $f(z)$ is bounded away from $0$, i.e., there exists $R, M > 0$ such that $|f(z)| \le M$ for all $|z| \ge R$. Since $f$ is continuous on the closed disk $|z| \le R$, it bounded by some number $M'$ on that disk. Thus, $f$ is bounded by $\max\{M,M'\}$.
A: Another edited:
Let $a_0 = f(0)$ , and let $g(z)=\frac{f(z)-a_0}{z}$. $g(z)$ is entire, we also have that $g$ is bounded (since the limit is finite). So from Liuoville we have that $g$ is constant.So $g$ must the constant ero function , so $f(z)=f(0)$ .
After the question was edited:
Since it is entire we can write $f$ as a taylor series, since the limit goes to 0 you will be left with only $f(z) = a_0$ , i.e. f is constant (Elseway choose the biggest $m$ such that $a_m\neq 0$, where $a_m$ is the coeffiecent of $x^m$ in the taylor series of $f$ . Then we get that $\lim \limits_{|z| \to \infty} \frac{f(z)}{z} = \lim \limits_{|z| \to \infty}a_m z^{m-1} \neq 0 $. Contradiction) 
Before the question was edited:
Take $f(z) =z^2$ and your argument fails...
You can prove your function is bounded in small enough environment: 
If $f(z)/z$ goes to 0 as $|z|$ goes to zero, then for small enough $r>0$ we have that for every $z \in B_r(0) \ \ \ |f(z)/z| \leq 1$.
Meaning that $|f| \leq |z| <1$ in small enough environment , meaning $f$ is bounded in small enough environment of $0$.
A: Let $g(z)=\dfrac{f(z)}{z}$
Let $\epsilon>o$ then as $|g(z)\rightarrow 0$ as $z\rightarrow \infty$ by definition we will get some $G>0$(large) such that $|g(z)|<\epsilon$ whenever $|z|>G$
Similarly as $|g(z)|\rightarrow 0 $ as $z\rightarrow -\infty$ then we can find a $G_1$ such that as $|z|<-G;|g(z)|<\epsilon $
Choose $M$such that both conditions hold (which is possible)
Then we are only left with the compact set $B[0,\dfrac{M}{2}]$ and since $g$ is entire and hence uniformly continuous on a compact set and hence bounded and thus so is $f$
