# If $\lim_{x\rightarrow\infty}f(f(x))=\infty$, $f:(-\infty,\infty)\rightarrow\mathbb{R}$ is continuous, prove $\lim_{x\rightarrow\infty}|f(x)|=\infty$

I'm working on another homework problem that I could use some help with. The question is posed in the title, and given here (exactly):

Problem: Let $f:(-\infty,+\infty)\rightarrow\mathbb{R}$ be continuous and the $\!\!\lim\limits_{x\rightarrow+\infty}\!\!f(f(x))=+\infty$. Prove that $\!\!\lim\limits_{x\rightarrow+\infty}\!\!|f(x)|=+\infty$.

It is a former, prelim problem from earlier in the year, and that was why I was requesting help - please note, I'm finishing up an honors, undergraduate degree in Mathematics, so this is a regular homework problem. As we speak, I'm trying to prove this by contradiction. If I'm correct, we have $\exists\varepsilon>0:\forall M\in\mathbb{N},\exists x\in(-\infty,+\infty):x>M~\text{and}~|f(x)|>\varepsilon$ negating the sufficient part, or we can suppose that $\lim\limits_{x\rightarrow\infty}|f(x)|\neq +\infty$ so that $\lim\limits_{x\rightarrow\infty}|f(x)|=L$. I've been working with both, going back and forth, using the precise definitions overall in order to get a contradiction against one of the two given assumptions using the other. I'm not sure if this is the way to go, and any help is GREATLY appreciated!

The negation of $\lim\limits_{x\rightarrow\infty}{|f(x)|}=\infty$ is: There exists $M$ such that for every $N\in\mathbb{R}$, there exists $x>N$ such that $|f(x)|<M$.

Assuming this negation, we can obtain a sequence $\{x_n\}$ such that $x_n\rightarrow\infty$ and $f(x_n)$ converges to some finite limit $L$ (why can we do this?). We then look at what $\lim\limits_{n\rightarrow\infty}{f(f(x_n))}$ must be.

• Thank you Joey! I see what you mean - by continuity we have that $f(x_{n})$ converges to $L$. I will start working this out! – Procore Oct 20 '15 at 4:05
• Sorry for the huge delay, but I should've said $f(x_{n})\rightarrow L$, as $x_{n}\rightarrow+\infty$ since $f$ is a continuous function. – Procore Apr 18 '16 at 1:31