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Let $f:[0, \infty)\to\mathbb{R}$ be a function such that for any positive $a$ sequence $\{f(a+n)\}$ converges to zero. Does the limit $\lim\limits_{x\to\infty}f(x)$ exist?

Thanks in advance.

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    You need to clarify the question. E.g. you ask about the limit of $f(x)$ as $n\rightarrow \infty$. – 2017-02-02
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    @I guess the answer is no but I don't prove it – 2017-02-02
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    If $f$ has no restrictions there will be for sure nonconverging functions. If you give an enumeration to the rationals in the unit interval and make a function that satisfies easy rules like: $f(q_i+i)=1$ and then start decreasing to 0 it will satisfy your property but have a subsequence going to 1. Is this agreeable and vaguely clear? – 2017-02-02
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    What do you mean by $\lim\limits_{n\to\infty}f(x)$? – 2017-02-02
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    @Thanks your comment, I'm sorry I mean $x\to \infty$ – 2017-02-02
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    @ZenoCozeno You should post that as an answer! – 2017-02-02

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I will provide a function satisfying the hypothesis and not having a limit for $x \rightarrow \infty$.

Since $\mathbb{Q} \cap [0,1)$ is countable we can fix a sequence of rationals $q_{i \in \mathbb{N}}$.

Define $f:[0,\infty) \rightarrow \mathbb{R}$ as $0$ everywhere, except for the points of the form $q_i+i$ where it is defined as $1$. We notice that definitely every sequence of the form $\{f(a+n)\}_n$ is null, so $f$ satifies the hypothesis. (I can meet at most one $q_i+i$ in the sequence)

Since $\{q_i+i\}_i$ is a sequence of real numbers going to infinity with value constant to $1$ there is no $\lim_{x\to\infty} f(x)$ because we have sequences going to different limits.