How can I prove that if $f$ is differential on $(0, \infty)$ and $f'$ is strictly growing on $(0,\infty)$ and, starting at some point $x_0 \in (0,\infty)$, $f'>0$, than $\lim_{x\to \infty} f(x) = \infty$?
It makes common sense yet I don't know how to approach this.
If $f'(x_0) = M > 0$ and $f'$ is growing then $$ f'(x) \ge M \text{ for } x \ge x_0 $$ and therefore, using the mean-value theorem, $$ f(x) \ge f(x_0) + (x-x_0)M \text{ for } x \ge x_0 \, . $$
You could also argue that $f$ is convex and therefore its graph lies above the tangent line at $x=x_0$, which leads to the same result.