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I understand that if a function is monotonic then the limit at infinity is either $\infty$,a finite number or $-\infty$. If I know the derivative is bigger than $0$ for every $x$ in $[0, \infty)$ then I know that $f$ is monotonically increasing but I don't know whether the limit is finite or infinite.

If $f'(x) \geq c$ and $c \gt 0$ then I know the limit at infinity is infinity and not finite, but why? How do I say that if the limit of the derivative at infinity is greater than zero, then the limit is infinite?

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If $$C=\lim_{x\rightarrow\infty} f'(x) > 0$$ then pick some $c\in(0,C)$. Then, by the definition of the limit, there is an $N$, such that $\forall x>N: f'(x) > c$. (You don't need that $f'(x)>c$ for all $x$ for any of these arguments to work, just that this is true for all $x\in(N,\infty)$ for some $N$.) – Thomas Andrews May 25 '11 at 16:54
up vote 6 down vote accepted

You can also prove it directly by the Mean Value Theorem:

$$f(x)-f(0)=f'(\alpha)(x-0) \geq cx \,.$$

Thus $f(x) \geq cx + f(0)$.

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In each interval $[n,n+1)$ the function increases by at least $c$.

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In addition to Ross's answer, note that if the derivative is always positive, it might be positive and yet convergent to zero. For example, consider $\dfrac{-1}{1+x^2}$ on the positive x axis. The derivative is always positive, and yet the limit is finite.

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But this function doesn't satisfy the condition on the derivative specified in the original question. – wckronholm May 25 '11 at 15:36
@wckronholm: I see this. I see him referring to two possibilities: when the derivative is greater than zero, and when it is greater than a positive number. I believe he is confused about them. Ross addressed the latter, so I gave an example of how that was a sharp answer. – mixedmath May 25 '11 at 15:41

For some $C>0$,

$$f(x) = \int_0^x f^\prime (t)\,dt \geq \int_0^x C \, dt = C\cdot x \longrightarrow \infty \;(x\rightarrow \infty)$$

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