Question: Does the function $f(x)=\{1$ if $x\in \Bbb Z $, $0$ if $x \notin \Bbb Z$, tend to zero as $x$ tends to infinity?

Not sure how to do the piece-wise function, feel free to edit or tell me how to. I've used the comma instead to illustrate a new line.

Defintion for a function $f(x)$ which tends to $0$ as $x$ tends to infinity:

$\forall \varepsilon > 0 \exists K \in \Bbb R \forall x>K :|f(x)|<\varepsilon$

My claim: Does not tend to zero as $x$ tends to infinity. The logic behind my claim is that if I were to imagine a graph of this, there would be two dotted lines, one at $y=0$ and another at $y=1$.

I find proving questions like these quiet difficult. I usually start off by writing what I need to prove, in this case I need to prove the negation. Then I would proceed with finding my choices for $\varepsilon$ and $x$, this is where I am struggling at the moment.

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    $\begingroup$ Calculate $\varliminf$ and $\varlimsup$ of $f$ at infinity. These are different so $f$ does not converge. $\endgroup$ – zwim May 12 at 12:56
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    $\begingroup$ Take $\epsilon=1/2$ and for each $K$ take $x=\lceil K\rceil+1$. That proves that $\exists \epsilon >0$, the $1/2$ value, such that $\forall K$ there exists $x>K$, the value $x=\lceil K\rceil + 1$, such that $|f(x)|=1>1/2=\epsilon$. This is $|f(x)|<\epsilon$ is false. $\endgroup$ – logarithm May 12 at 12:57
  • $\begingroup$ @zwim Do you mean take the limit of the top, and the bottom and then conclude since they are not the same then $f$ does not converge? I get the concept behind this but what theorem are you using to get to this conclusion? $\endgroup$ – ViB May 12 at 13:00
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    $\begingroup$ The value $1$ is taken for arbitrarily large values of $x$. This value is at distance $|1-0|=1$ from the supposed limit $0$. The idea is to take $\epsilon$ to be any number smaller than that distance. All we needed was the inequality $1>1/2$. You could have taken $e/\pi$, or $\pi/4$, or $0.yourbirthday$, anything less than $1$ and larger than $0$. The choice of $x$ is more relaxed, all you need is that when you are given a $K$ you take some integer $x>K$ at which $f(x)=1$. $\endgroup$ – logarithm May 12 at 13:02
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    $\begingroup$ $\varliminf$ and $\varlimsup$ are alternate notations for $\liminf$ and $\limsup$ $\endgroup$ – saulspatz May 12 at 13:09

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