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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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up vote 5 down vote accepted

Use the mean value theorem. Since $f$ is differentiable on the interval $[x,x+1]$, there exists $c_x \in [x,x+1]$ such that $f'(c_x) = f(x+1)-f(x)$. Thus $$ \lim_{x \to \infty} f(x+1) - f(x) = \lim_{x \to \infty} f'(c_x) = \lim_{c_x \to \infty} f'(c_x) = 0. $$ The last limit substitution is justified because $c_x \to \infty \quad \Longleftrightarrow \quad x \to \infty$ since $x \le c_x \le x+1$.

Hope that helps,

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Thanks for the help! – jake Nov 20 '11 at 5:10
No problem! I see some people liked this question. =) – Patrick Da Silva Nov 20 '11 at 7:42

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