Let $g$ and $h$ be two entire functions such that $\lim_{|z|\to\infty}\frac{|g(z)|}{|z|^5} = 0$ and $|h(z)|\le |z|^4$. What can we say about $g$ and $h$?
I got to both of them being polynomials of at most degree 4. I'll give you my reasoning:
Since $g$ is entire, we can write it as $g(z) = \sum_{n=0}^\infty a_nz^n$, so \begin{equation} 0 = \lim_{z\to\infty}\frac{|g(z)|}{|z|^5} = \lim_{z\to\infty}\frac{|a_0| + |a_1||z| + |a_2||z|^2 + \cdots}{|z|^5} \end{equation} But this can only happen if $|a_n| = 0$ for $n\ge5$ (and then $a_n = 0$ for $n\ge5$). Hence, $g(z) = a_0 + a_1z + a_2z^2+a_3z^3+a_4z^4$, with $a_i\in\Bbb C$ for $i\in\{0,1,2,3,4\}$.
Since $h$ is entire, we can write it as $h(z) = \sum_{n=0}^\infty b_nz^n$, with $b_n = \int_{|z|=R}\frac{h(z)}{z^{n+1}}dz$^with $R>0$. Then, \begin{equation} |b_n|\le \frac1{2\pi}2\pi R \frac{R^4}{R^{n+1}} = R^{4-n}\text{ for }n\in\Bbb N\cup\{0\} \end{equation} If $n>4$, $|b_n|\le \lim_{R\to\infty}R^{4-n} = 0$, so $b_n = 0$ for $n\ge 5$. Hence, $h(z) = b_0 + b_1z + b_2z^2+b_3z^3+b_4z^4$, with $b_i\in\Bbb C$ for $i\in\{0,1,2,3,4\}$.
Is this alright, or am I mistaken?
What you wrote about the function $g$ is not correct. Why would it follow from what you wrote that $a_n=0$ when $n\geqslant5$. For instance, if you consider the sine function from $\Bbb R$ to $\Bbb R$, you also have $\lim_{x\to\infty}\frac{|\sin(x)|}{x^5}=0$, but it is not true that the coefficients of its Taylor series centered at $0$ are all equal to $0$ when $n>5$.
You can use that fact that $|a_n|\leqslant\frac{\sup_{|z|=r}|f(z)|}{r^n}$ to prove that $n\geqslant5\implies|a_n|=0$.
What you wrote about $h$ looks fine.