Proving a piece-wise function does not converge

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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.