Proving function's (un)boundeness via MVT

Question

I posted a problem yesterday and someone solved it and explained to me that function is unbounded because $f(x)=f(a)+(x-a)f'(\xi)>f(a)+(x-a)c$. But I can't see why this inequality means that function is unbounded. For example if $g(x)=f(a)+c(x-a)$ is maybe bounded (which we don't know it is until we prove it but if I knew how I wouldn't post this)it wouldn't necessarily mean that $f$ is unbounded simply because it's greater than $g$. $f:(a,+\infty)\rightarrow\mathbb{R}$ is differentiable, $c>0$, $f'(x)>c$, $\forall x\in(a,+\infty)$

Answer 1

Assuming that the domain of $f$ contains all large enough real numbers and that $c>0$, then$$\lim_{x\to\infty}f(a)+(x-a)c=\infty,$$and therefore, since $f(x)>f(a)+(x-a)c$,$$\lim_{x\to\infty}f(x)=\infty$$too. Therefore, $f$ is unbounded.