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Let $f:[0, + \infty] \rightarrow \mathbb R$ be a function bounded in each bounded interval. Prove that if $\lim_{x\rightarrow +\infty}[ f(x+1) - f(x)] = L$, then $\lim_{x \rightarrow + \infty} f(x) = L$.

I know that $f:[0, + \infty] \rightarrow \mathbb R$ bounded in each interval $\Rightarrow$ $\forall \epsilon >0$, $\forall a \in [0, +\infty]$, $\exists M>0$, such that $|x-a|< \epsilon \Rightarrow |f(x)|<M$.

I also know that $\lim_{x\rightarrow +\infty}[ f(x+1) - f(x)] = L$ $\Rightarrow$ $\forall \epsilon >0$, $\exists N \ge 0$ such that $\forall x>N$, $|f(x+1)-f(x)-L|< \epsilon$.

What are the next steps?

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The function $f(x) = \sin 2 \pi x$ satisfies $f(x+1)-f(x) = 0$ for all $x$, and is bounded, but clearly has no limit.

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I don't think it is true. $f(x)=x$ is bounded in every interval, $f(x+1)-f(x)=1$, so we have a limit $L=1$, but $\lim_{x \rightarrow + \infty} f(x) \neq 1$. In fact, unless $L=0$, I believe you will always go off to $\pm \infty$

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