It is true that busy beaver functions are noncomputable, but that does not mean that they have an infinite value do they? Does some f(x)=infinity for some beaver function? Doesn't noncomputable just mean that the exact value can not be mathematically determined? Therefore, is it not possible for computable functions, such as the ones I came up with, to grow faster than noncomputable functions, such as busy beaver functions?
No! Infinity is not a number!
(or partial function ...) is computable means there is a finite state Turing machine (or equivalently a register machine which is usually easier to demonstrate) that will accept any natural number as input and compute f for that input (and in the case of partial functions, need machine halting for those input where f is defined and doesn't halt if f is not defined at the input).
Even if f is noncomputable, any truncation
While you can cook up computable functions that beats the busy beavers function for the first N inputs, eventually the busy beavers function will win (see the construction in the other thread). That is why the answer to your initial question (does your function grow faster than the beavers function in the long run) is negative.