Suppose $f$ is Differentiable on all of ${\Bbb R}$ and that $\lim_{\ x\to\infty} f^{'}(x) = 0.$ Show that $\lim_{\ x\to\infty} (f(x+1)-f(x))=0.$   I tried to prove by this logic,  $\lim_{\ x\to\infty} f^{'}(x) = 0 \Rightarrow \lim_{\ x\to\infty}\left(\lim_{\ h\to\ 0}\frac{f(x+h)-f(x)}{h}\right) = 0 \Rightarrow f^{'}(x) = 0, \forall x\in{\Bbb R}$.  Thus, $f$ is constant which implies $f(x+1)=f(x) \Rightarrow \lim_{\ x\to\infty} (f(x+1)-f(x))=0$.
Having some trouble proving this:     If $f:\mathbb{R}\rightarrow \mathbb{R}$ is differentiable and $\lim_{x\to \infty } f^\prime(x)=0$ show: $\lim _{x\to \infty } (f(x+1)-f(x))=0$   Attempt: from $\int \lim _{x\rightarrow \infty } f^\prime(x) \mathrm{d}x=0$ we can say that $f^\prime(x)=const$ would we then have to show that  $\lim _{x\to \infty } (f(x+1))$ is also a constant? From there if we show that both are equal then the difference is zero. I am on the right track?  Thank you for the help.
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