Prove/disprove: if $\lim\limits_{ n\to\infty} f(n)=\infty$ then $\lim\limits_{ n\to\infty}f(f(n))=\infty$ 
Let $f(x)$ a continuous function on $\Bbb{R}$.
Prove/disprove: If $\lim\limits_{n\to\infty} f(n)=\infty$, then $\lim\limits_{n\to\infty}f(f(n))=\infty,$ where the limits are taken over $n \in \mathbb N$. 

I have an hunch this statement isn't always true, but couldn't find a proper function to show this.
 A: Let $M$ be given. There exists $N$ with the property that $x > N$ implies $f(x) > M$. There exists $K$ with the property that $x > K$ implies $f(x) > N$. Thus $$x > K \implies f(x) > N \implies f(f(x)) > M.$$
A: Counterexample:
$
 f(x)=
 \begin{cases}
  \dfrac{1}{2}+x-2x(x-\lfloor{x}\rfloor)
  &
  \text{$0\leq{x}-\lfloor{x}\rfloor<\dfrac{1}{2}$}
  \\
  \dfrac{1}{2}+2x(x-\lfloor{x}\rfloor-\dfrac{1}{2})
  &
  \text{$\dfrac{1}{2}\leq{x}-\lfloor{x}\rfloor<1$}
 \end{cases}
$

For $n\in\mathbb{N}$:


*

*$f(n)=n+\dfrac{1}{2}\implies\lim\limits_{n\to\infty}f(n)=\infty$

*$f(f(n))=\dfrac{1}{2}\implies\lim\limits_{n\to\infty}f(f(n))\neq\infty$


A: Consider the function
$$
f(x):=x\cos(2\pi x)+\frac{1}{4}.
$$
Then $f$ is continuous in $\mathbb R$ such that $f(n)=n+\frac{1}{4}$ for $n\in\mathbb N$, which tends to $\infty$ as $n\to\infty$. But $f(f(n))=(n+\frac{1}{4})\cos(2\pi n+\frac{\pi}{2})+\frac{1}{4}=\frac{1}{4}$ for $n\in\mathbb N$.
A: When you say $f(n) \to \infty$ as $n \to \infty$, the usage of symbol $n$ usually means that $n$ is a positive integer (you have mentioned this explicitly also). Thus when the argument of $f$ tends to $\infty$ through positive integers then $f$ tends to $\infty$. In case of $f(f(n)) = f(t)$ the argument of $f$ is $t = f(n)$ and this itself may not be an integer. So even though the argument $t \to \infty$ it is not guaranteed to be an integer and hence $f(t)$ may not tend to $\infty$. If for all $n \in \mathbb{N}$ we ensure that $f(n) \in \mathbb{N}$ then $f(f(n)) \to \infty$ as $n \to \infty$.
