User blog comment:Mh314159/Question about growth rate/@comment-35470197-20191107005001

> what is the growth rate of h(x) for sufficiently large x?

Is this a typo of g(x)? Or are you considering h(x) = g^{g(x-1)}(x)? The iteration roughly contributes to ordinals as "+1". Therefore the growth rate of g(x) corresponds to ω^2+1, and the growth rate of g^{g(x-1)}(x) corresponds to ω^2+2, if you set reasonable initial values at x=0.