Given a differentiable function $f$ such that $$ \lim_{x \to +\infty}f'(x) = 0 $$ what can we say about $$ \lim_{x\to\infty}(f(x+1)-f(x)) \text{ ?} $$ My first thought was to use mean value theorem on $[x,x+1]$, and I will get that the limit is $0$, is this true? Thank you for your help.
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Yes, you are right. – AndreasT Dec 06 '15 at 17:20
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5:) you answered yourself, why don't you write it down formally..? – AndreasT Dec 06 '15 at 17:23
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Given that $f$ is differentiable consider the interval $[x,x+1]$ and apply the mean value theorem we get $f(x+1)-f(x) = f'(c_x)$ where $c_x\in (x,x+1)$. Now taking limits both sides $$\lim_{x\to\infty}(f(x+1)-f(x)) = \lim_{x\to\infty}(f'(c_x))$$ where $c\in (x,x+1)$ , so as $x$ tends to infinity , $f'(c_x)$ tends to zero as given in the hypothesis . Therefore $$\lim_{x\to\infty}(f(x+1)-f(x)) =0$$ .
Mark Viola
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Butterfly
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3Very minor notational suggestion: since $c$ depends on $x$ perhaps writing it as $c_x$ or $c(x)$ will clarify. Also, in much more complicated problems or proofs, it will keep reminding you that it is not a constant. ("$c$" is quite often a constant, so often, that it can fool you.) – B. S. Thomson Dec 06 '15 at 18:39
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thank you for your advice, i fixed it , and will keep your advice in mind – Butterfly Dec 06 '15 at 18:57