Does there exist any function $f(x)$ such that $(1), (2)$ and $(3)$ can be satisfied:
(1) $f(x)$ is a strictly increasing function.
(2) $f(x)$ is sub-exponential function (i.e. $ \displaystyle \lim_{x\rightarrow\infty} \dfrac{\ln(f(x))}{x}=0$)
(3) $\forall d>0 $ (considered as a constant), $\displaystyle \lim_{x\rightarrow\infty} \dfrac{f(x)}{f(x+d)} \neq 1$ or $\dfrac{f(x)}{f(x+d)}$ does not converge when $x \rightarrow \infty$.
The motivation behind the above problem is that I am trying to find strictly increasing functions or sufficient conditions for strictly increasing functions to satisfy $ \displaystyle \lim_{x\rightarrow\infty} \dfrac{f(x)}{f(x+d)} = 1$. One necessary condition is (2) above since any exponetial function gives $\displaystyle \lim_{x\rightarrow\infty} \dfrac{f(x)}{f(x+d)} < 1$.
What I have tried so far:
The first commonly used non-linear sub-exponential continuous functions are polynomial functions. Given any increasing polynomial functions, we can easily show that $(3)$ is not satisfied since $\displaystyle \lim_{x\rightarrow\infty} \dfrac{f(x)}{f(x+d)} = 1$.
Threfore, then I started to think about using piecewise functions which satisfies condition $(3)$. My current piecewise function is $f(x) = \begin{cases} \dfrac{\beta}{\alpha}x+\alpha & \text{if $0 \leq x \leq d$} \\ f(x-d)\dfrac{f(d)}{f(0)} & \text{if $d <x \leq 2d$} \\ f(x-d)\dfrac{f(x-d)}{f(x-2d)} & \quad \! \! \text{ otherwise} \end{cases}$
where $\alpha>1$, $\beta<1$ (for example, $\alpha = 2$, $\beta = 0.001$, and $d=4$). We can show that $\displaystyle \lim_{x\rightarrow\infty} \dfrac{f(x)}{f(x+d)} = \dfrac{1}{\alpha} \neq 1$ and this function is also increasing. However, this piecewise function can be approximately by exponetial function when $x$ is large, which violates the condition $(2)$.
$$ f(x) = e^{[\ln(x)]} x .$$ Here $[x]$ denotes the integer part of $x$. Note that by considering the sequence $x_n = e^n - d/2 $, we see that $$ \liminf_{x\to\infty} \frac{f(x)}{f(x+d)} \le e^{-1} .$$
Note, if the limit exists and $$ \lim_{x\to\infty} \frac{f(x)}{f(x+d)} < 1 ,$$ then there exists $x_0$ such that for $x \ge x_0$ $$ \sup_{x \ge x_0} \frac{f(x)}{f(x+d)} \le \frac 1L < 1 .$$ Thus $f(x_0+nd) \ge L^n f(x_0)$, or for $x \ge x_0$ $$ f(x) \ge L^{(x-x_0)/d - 1} f(x_0) $$ and it is not possible for $f$ to be sub-exponential.