# If $\lim\limits_{x \to \infty} f(f(x))= \infty$, $\lim\limits_{x \to \infty} f(x)=\pm \infty$

Let $f: \mathbb{R} \to \mathbb{R}$ such that $\displaystyle \lim_{x \to \infty} f(f(x))= \infty$ and $\displaystyle \lim_{x \to -\infty} f(f(x))= -\infty$ and $f$ has the intermediate value property (it is not necessarily continuous).
Prove that $\displaystyle\lim_{x \to \infty}f(x)$ and $\displaystyle\lim_{x \to -\infty}f(x)$ exist and are infinite.

I managed to prove that those two limits exist. But I have a really hard time proving that they are infinite. I supposed that $\displaystyle \lim_{x \to \infty} f(x)=a \in \mathbb{R}$ and tried to get a contradiction from here, but since $f$ is not continuous, I must somehow use the intermediate value property and. I see that, for all $\epsilon>0$, there is $\delta>0$ such that when $x>\delta$, $f(x)$ is getting bounded in $(a-\epsilon, a+\epsilon)$ and so $(f\circ f)((\delta,\infty))\subset f((a-\epsilon,a+\epsilon))$, but I couldn't proceed further.

• If the limit is $a$, then $f(x)$ must tend to infinite when $x\to a$ in some sense, so you can inquire what is the value $f(a)$ and get an absurd – Exodd Mar 18 '18 at 11:26
• You also have to consider what happens if the limit doesn't exist. – Wojowu Mar 18 '18 at 11:36

## 1 Answer

The limit must exist or be infinite.

Assume that $\liminf_{x\to+\infty}f(x)=a$ and $\limsup_{x\to+\infty}f(x)=b$, where $a,b\in[-\infty,+\infty]$.

By the intermediate value property it follows that all values in $(a,b)$ are limit points of $f$ as $x\to+\infty$. Therefore $f$ must be unbounded from above in every neighborhood around every point in $(a,b)$ and also all values in $(a,b)$ are attained in every neighborhood of $+\infty$.

Assume that on $[a,b]$ there are infinitely many points such that $|f(x)|<n$. Then pick a convergent sequence among them $x_k$. Then we can pick $y_k\to+\infty$ such $f(y_k)=x_k$. It follows that $f(f(y_k))$ doesn't tend to infinity. This is a contradiction.

Therefore, for all $n$ the number of points in $[a,b]$ such that $|f(x)|<n$ must be finite. But then $[a,b]$ is a countable union of finite sets. But $[a,b]$ is uncountable if $a\neq b$, while a countable union of finite sets is countable. Therefore $a=b$.

The limit must be infinite.

Assume now that $a$ is finite. Then, by the intermediate value property, $f$ takes all values between $f(a)$ and $+\infty$ in any neighborhood of $a$. To be organized, pick a side of $a$, either $>a$ or $<a$ such that $f([n,+\infty))$ intersect that side for all $n$. Without loss of generality let's say it is $>a$. The first claim of this paragraph can be made for these one-sided neighborhoods. Let $x_k$ be a sequence of points $x_k\to a$ such that $f(x_k)\to f(a)$ and $x_k>a$.

Therefore, by the intermediate value property, we can take $y_k\to\infty$ such that $f(y_k)=x_k$. This is a contradiction because $f(f(x_k))\to f(a)\neq +\infty$. Therefore, $a$ is infinite.

Nice exercise. I will throw it to my graduate students.