Hey Paze.
If you allow discontinuities then the fastest growing function at a particular point is the delta function at x = 0.
Dirac delta function - Wikipedia, the free encyclopedia
Sure, it grows faster at first. But what happens for $\displaystyle x = 1000$? $\displaystyle e^{1000}>e^{900}=(e^3)^{300} >10^{300}=(10^3)^{100}=1000^{100}$. Then what happens when we increase x by 1? $\displaystyle e^{x+1}/e^x=e$ for all x. In contrast, $\displaystyle (x+1)^{100}/x^{100}\to 1$ as $\displaystyle x\to\infty$ because the numerator and the denominator are polynomials of the same degree with the same leading coefficient 1. For example, $\displaystyle 1001^{100}/1000^{100}\approx 1.1<e\approx 2.7$. Thus, when x is increased by 1, $\displaystyle e^x$ is always multiplied by $\displaystyle e$, while $\displaystyle x^{100}$ is multiplied by smaller and smaller numbers that tend to 1.
This is clever, though delta function is not really a function. Even the piecewise function $\displaystyle \begin{cases}0&x<0\\ 1&x\ge0\end{cases}$ grows infinitely fast at 0.
If we restrict ourselves to continuous functions or to functions on natural numbers, suppose we have a candidate for the fastest-growing function $\displaystyle f(x)$. Then what about $\displaystyle 2^{f(x)}$?
I don't believe there is such a thing as "the fastest growing function," as you could always multiply whatever you think is the fastest function by 2 to get one that has a greater slope. However, in thinking about "normal" functions that grow very fast $\displaystyle x^x$ is a pretty good one. Once you get past about x=2.1 it grows much faster than $\displaystyle e^x$. Of course the next logical extension of this is to consider $\displaystyle x^{(x^x)}$, which grows so fast that it exceeds one googol around x= 3.84, then $\displaystyle x^{(x^{(x^x)})}$, etc, etc.