Limits at infinity $f(z), g(z)$ are two entire functions, both have no zeros in the closed upper half plane. What does it mean/imply that $$\bigg| \lim_{y\rightarrow \infty}\frac{f(z)}{g(z)}\bigg|=c$$
($z=x+iy$)
i.e. after taking the limit inside the modulus the resulting function -depends on x- have modulus c. (In fact what I have is like: $|\lim_{y\rightarrow \infty} (\dots)|=|..ce^{ix}|=c$)
(I think it implies that $|f(z)|\leq c|g(z)|$ for all $z$  in the upper half plane, is that correct, and if so how to prove it!)
Also, what does it mean/imply that 
$$\bigg|\lim_{y\rightarrow 0}\frac{f(z)}{g(z)}\bigg|=d$$
EDIT: $c$ and $d$ are non zero.
 A: I would take it to mean $\left|\lim_{y\to\infty}\frac{f(x+iy)}{g(x+iy)}\right|=c$, which -- at least a priori -- is something that depends on $x$, but not on $y$, because $y$ is bound by the limit operator. That is, it is just a claim about the magnitude of a real limit $\lim_{y\to\infty}(\cdots y\cdots)$, where $(\cdots y\cdots)$ happens to involve an $x$ and some complex functions.
It is not equivalent to $|f(z)|\le c|g(z)|$. As a simple counterexample, take $f(z)=0$, $g(z)=z$, which satisfies your condition, but not the original, since $\lim\frac{f(z)}{g(z)}=0$ for all $x$.
A counterexample for the other direction is $f(z)=c(1+e^{iz-1})$, $g(z)=1$, where $\lim_{\Im z\to\infty} \frac{f}{g}=c$ but $|f(z)|$ can vary both above and below $c|g(z)|$ over the upper half-plane.
A: For your second question: If $f(z)/g(z) = h(z)$, then $h(z)$ is meromorphic in the complex plane and has no poles or zeros in the closed upper half plane.  Now $\lim_{y \to 0} h(x+iy) = h(x)$, so you're saying $|h(x)| = d$ for $x \in \mathbb R$.  By a version of the Schwarz Reflection Principle, $h(\overline{z}) = d^2/\overline{h(z)}$ for all $z$: in particular $f$ and $g$ have no zeros at all.
That implies $f(z) = e^{F(z)}$ and $g(z) = e^{G(z)}$ where $F$ and $G$ are entire functions, where $\Re(F(x) - G(x)) = \ln(d)$ for $x \in \mathbb R$.  
