Equality of limits or counterexample

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 $c=l$.