Prove that $\lim_{x \to \infty} f(x) = \infty$ implies that $f$ is strictly increasing on some range $(a, \infty)$, ($a \in \mathbb{R}.$)

Question

Consider the following statement: If $$ \lim_{x \to \infty} f(x) = \infty, $$ exists such an $a \in \mathbb{R}$, that $f$ is strictly increasing on $(a, \infty)$, i.e., for all $a < b < c$, $f(b) < f(c)$.

Proof: Assume on the contrary that there is an $a \in \mathbb{R}$ such that $f$ is monotonically decreasing on $(a, \infty)$, i.e., for all $a < b < c$ $f(b) \geq f(c)$. Let $B = \max_{b > a}\{ f(b) \} \in \mathbb{R}$. We see that $\lim_{x \to \infty} f(x) \leq B \in \mathbb{R}$; a contradiction.

Question: Is my prove correct and rigorous? If not, how can I improve it?

Answer 1

The proof is wrong. It has to be, since the statement is false. Just consider $f(x)=x+2\sin(x)$. You have $\lim_{x\to\infty}f(x)=\infty$, but, for any $a\in\Bbb R$, $f|_{(a,\infty)}$ is not monotonous.

Concerning your proof, you seem to be assuming that if $f|_{(a,\infty)}$ is not increasing, then it is decreasing. That is false.

Answer 2

Monotonically increasing and monotonically decresing are not the only possibilities. If $f(x)=n$ for $2n \leq x <2n+1$ and $2n$ for $2n-1\leq x<2n$ then $f$ is neither monotonically decresing nor monotonically increasing so the assertion is false.