For entire funtion $h$, $zh(z)$ cannot have a nonzero limit as $z\to \infty$ This is a question I encounter in learning complex variables.

Show that for any entire funtion $h$, $zh(z)$ cannot have a nonzero limit as $z\to \infty$.

Furthermore, use this statement to give an alternate proof of Liouville's theorem.

I am really having trouble proving the statement, and can't see any obvious connection between this and the Liouville's theorem. Any hints or suggestions are appreciated.
Let us assume the opposite, i.e. that $\lim_{z\to \infty} zh(z)=k;k\neq 0$

Assuming Liouville theorem:
$$\lim_{z\to \infty}zh(z)=k\neq 0\\ \lim_{z\to \infty}h(z)=0\\
h(z)=a (\text{const})\\
\lim_{z\to \infty}zh(z)=\infty$$


This contradiction proves that the limit, if it exists, must be either $0$ or $\infty$

Without Liouville theorem:
By the Cauchy integral formula:

$$|h^{(n)}(0)|=\left|\frac{n!}{2\pi i}\oint_{c_r}\frac{h(z)}{z^{n+1}}dz\right|\le \frac{n!}{r^n}\underset{|z|=r}{\text{Max}}|h(z)|$$
Letting $r\to \infty$ we have, thanks to the assumption, that $\forall_{n&gt;1}h^{(n)}(0)=0$. This implies that $h(z)=az+b$, which contradicts
$$\lim_{z\to \infty}zh(z)=k$$
Proving Liouville with this statement:
Given a bounded entire function $h(z)$, let us define $$f(z):=f\left(\frac{1}{z}\right)$$ This function is  holomorphic on $\mathbb{C}-\{0\}$, and is bounded in a neighbourhood of $0$; By  Riemann's theorem on removable singularities $\lim_{z\to 0}f(z)=l\in \mathbb{C}$ exists, and thus $\lim_{z\to \infty}h(z)=l\in \mathbb{C}$ too.
Defining $g(z)=\frac{h(z)-h(a)}{z-a}$, we obtain an entire function, for which $$\lim_{z\to \infty}g(z)z=l-h(a)$$
Thanks to our statement, $l-h(a)=0$. Since $a$ was generic, $h$ is constant
Your statement is wrong. What about $h(z)=1$ ? Then, for example, for $t&gt;0$, $$\lim_{t\to \infty }t h(t)=+\infty.$$