Let $f: \mathbb{R} \to \mathbb{R}$ such that $\lim\limits_{x \to \infty} \frac{f(x)}{x} = c$ where $c\neq 0$ or $\infty$.
Is it true that for all such functions $f$, if $\lim\limits_{x \to \infty}(f(x+1)-f(x))$ exists, then it is equal to $c$?
Note : Say if $f(x) = x+\sin(x)$, then the condition is clearly false because the second limit does not exist. But say, we impose another condition that the second limit should exist, then is it necessary that it goes to $c$ as well?
Yes! Without loss of generality assume $c=1$ and assume that $\lim_{x\to\infty}f(x+1)-f(x)=d<1$. Then for any $\epsilon>0$, there exists some $T_\epsilon$ such that for $x>T_\epsilon$, $f(x+1)-f(x)<d+\epsilon$. Fix $x_0>t_\epsilon$. Then, we have $f(x_0+k)-f(x_0)<k(d+\epsilon)$, and by dividing both sides by $k$: $$\frac{f(x_0+k)}{k}-\frac{f(x_0)}{k}<(d+\epsilon)\leq \frac{d+1}{2},$$ where the last inequality holds for $\epsilon<\frac{1-d}{2}$. Letting $k\to\infty$, we have $\frac{f(x_0)}{k}\to 0$ as $f(x_0)$ is a constant, and hence: $$1=\lim_{k\to\infty}\frac{f(x_0+k)}{x_0+k}=\lim_{k\to\infty}\frac{f(x_0+k)}{k}\frac{k}{x_0+k}=\lim_{k\to\infty}\frac{f(x_0+k)}{k}\leq \frac{d+1}{2}<1.$$ Which is a contradiction. Similar argument holds for $d>1$, and letting $T_\epsilon$ be such that $f(x+1)-f(x)>d-\epsilon$ for $x>T_\epsilon$ and again, considering $\epsilon<\frac{d-1}{2}$.
If $f(x+1)-f(x) \to l$ as $x \to \infty$ then $f(n+1)-f(n) \to l$ as $n \to \infty$. By Cesaro's theorem $\frac 1 n \sum\limits_{k=1}^{n} [f(k+1)-f(k)] \to l$. This means $\frac {f(n+1)-f(1)} n \to l$. Hence $\frac {f(n+1)} n \to l$. Since $\frac {f(n+1)} {n+1}=\frac {n+1} n \frac {f(n+1)} n $ we get $c=l$.
