$f'(x) > 0$, does that imply the limit as $x\rightarrow \infty$ is $\infty$?

Question

Given the following situation: $f'(x) > 0$ and $f: \mathbb{R} \rightarrow \mathbb{R}$.

I am trying to find if there is a flaw in my understanding of the properties of this situation. I assert that in the situation where $x$ approaches $\infty$ that the limit is not always $\infty$.

If I were for example to construction a piecewise function like:

$$ f(x) = \begin{cases} \frac{|x|}{|x|+ 1} & x\geq 0 \\ \frac{-|x|}{|x|+ 1} & x\lt 0 \end{cases} $$

We can see that the limit is 1. Am I correct in my assertion and understanding?

Answer 1

You are correct, another example would be $f(x) = \arctan(x)$.
The derivative is stricly positive and the function is bounded.

Some more details: The derivative is $f'(x) = \frac {1}{1+x^2}\leq1$, but $ |f(x)| < \frac {\pi}{2}$.

We can further investigate a somewhat reverse scenario, take a function $g(x)$ such that $g'(x) \xrightarrow {x \rightarrow \infty} 0$.
Does that mean the limit $\lim_{x\rightarrow\infty} g(x) \in \mathbb R$?
No. Take, for example $g(x) = \log(x)$

I hope that gives you some intuition.

Answer 2

If you just want the limit as $x \to \infty$, then $-1/x$ works.