Existence of limit for some sequences implies existence of limit Let $f:[0,+\infty)\rightarrow R$ and for every sequence $\{x_n\}$, such that $x_{n+1}=x_n+1$, $f(x_n)\rightarrow 0$. Does it imply that $\lim_{x\rightarrow +\infty}f(x)=0$ ?
I have no idea how to approach this problem. I have been trying to find counterexample, but I can't.
Thanks!
 A: There are counter examples. I'm gonna go with something fancy (just for fun) but there are of course more simple examples. Take the function $f$ for which $f(x) = 0$ except if $x = \pi^n$ for some integer $n \ge 1$, in which case $f(\pi^n) = 1$. Then for every sequence you mentioned, we cannot have $n \ge 1$ such that $f(x) = f(x+n) = 1$ since $\pi^m = \pi^n + n$ implies $\pi$ is algebraic. Therefore $f$ does not converge but all the sequences you wrote above do satisfy $f(x_n) \to 0$.
The idea is just to make $f$ go wrong on some non-regular sequence. I chose a fancy one.
Hope that helps,
A: Suppose 
$$
f(x)=
\begin{cases}
1&x=n+1/n\\
0&\text{otherwise}
\end{cases}
$$
then $\lim_{x\to+\infty}f(x)$ does not exist, but $\lim_{n\to+\infty}f(x_n)=0$ for any such $(x_n)_{n\in\mathbb{N}}$.
A: For any $x\in [0,1)$, let $f(x+k)\stackrel{\rm def}{=}\frac{1}{k(1-x)}$, for $k\in\mathbb{N}$. $f$ is indeed a function defined as required, and satisfies the property; yet on the sequence $(x_n)_{n\geq 0}$ defined by $x_n\stackrel{\rm def}{=}n+1-\frac{1}{n}\xrightarrow[n\to\infty]{}\infty$, $f(x_n)=1$.
