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Obviously, if $f(x)$, $x \to \infty$, has a finite limit, the sequence $a_n = f(n)$ converges. The converse does not hold ($\sin \pi x)$. A sufficient condition for the converse to hold is the monotonicity of $f$ (I think).

What's a condition which is both necessary and sufficient?

In other words, how do we characterize those functions for which

$(1) $ The limit of the sequence and function both exist.

$(2) \lim\ f(x) = \lim \ f(n)$

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  • $\begingroup$ Other than the fact that the natural numbers are equally spaced, I don't see why you can't ask the same question for any unbounded increasing subsequence of real numbers. I don't know the answer, but tackling it in this generality may give some insight into the problem. $\endgroup$ – астон вілла олоф мэллбэрг Sep 29 '16 at 5:40
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    $\begingroup$ The oscillations must be controlled in one way or another, hence define $$\mathrm{osc}(f,x,r)=\sup\{|f(y)-f(z)|\mid|y-x|\leqslant r,|z-x|\leqslant r\}$$ Then the function $f$ converges if and only if the sequence $(f(n))$ converges and $\mathrm{osc}(f,x,\frac12)\to0$ when $x\to\infty$ ($x$ real), if and only if the sequence $(f(n))$ converges and $\mathrm{osc}(f,n,\frac12)\to0$ when $n\to\infty$ ($n$ integer). $\endgroup$ – Did Sep 29 '16 at 6:34
  • $\begingroup$ As currently written, condition $(2)$ is redundant, and $(1)$ is half-redundant since if the function has a limit, then the sequence will have the same limit (as noted in the first paragraph of the question). $\endgroup$ – dxiv Sep 29 '16 at 6:34
  • $\begingroup$ @Did Maybe post that as an answer? $\endgroup$ – MathematicsStudent1122 Sep 30 '16 at 1:18

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