TITLE: Does $\lim_{x\to\infty} f(x+1)/f(x) = 1$ imply $\lim_{x\to\infty} f(f(x+1))/f(f(x)) = 1$?
QUESTION [5 upvotes]: As mentioned above, does
\begin{align}
\lim_{x\to\infty} \frac{f(x+1)}{f(x)} = 1
\end{align}
imply
\begin{align}
\lim_{x\to\infty} \frac{f(f(x+1))}{f(f(x))} = 1?
\end{align}
Here, $f:\mathbb{R}\to \mathbb{R}$ is a function such that
(1) $f(x)\neq 0$ for all $x\in\mathbb{R}$ (or for all $x\ge x_0$ for some $x_0$);
(2) $\lim_{x\to\infty} f(x) =\infty$;
(3) $f$ is increasing.
This is not a homework question, but I just wonder if this statement is true (Indeed, I observe that this holds for $f(x)=\log(x)$ so I have this question). I would appreciate any counterexample/proof/hints. Many thanks!

REPLY [4 votes]: No.
Take $f(x)=e^{\sqrt{x}}$.
Then $$f(x+1)=e^{\sqrt{x+1}}=e^{\sqrt{x}\left(1+\frac 1 {2x}+o\left(\frac 1 {x}\right)\right)}=f(x)\left(1+\frac 1 {2\sqrt x}+o\left(\frac 1 {\sqrt x}\right)\right)$$
thus $$\lim_{x\rightarrow+\infty} \frac{f(x+1)}{f(x)}=1$$
But
$$f(f(x+1))=e^{\sqrt{e^{\sqrt{x+1}}}} = e^{e^\frac{\sqrt{x+1}}2}=e^{e^{\frac{\sqrt{x}}2 \left(1+\frac 1 {2x}+o\left(\frac 1 {x}\right)\right)}}\simeq f(f(x))\exp \left [ \frac{e^{\frac {\sqrt{x}}2}}{4\sqrt x}\right]$$
Thus $$\lim_{x\rightarrow+\infty} \frac{f(f(x+1))}{f(f(x))}=+\infty$$