I am trying to prove that the $$\lim_{x\to0}\left\lfloor\frac1x\right\rfloor$$ does not exist with the $\epsilon-\delta$ definition. I am struggling with how to structure the proof. I graphed the function and when we approach the function from $x\le0$ and $x\ge0$ we get $-\inf$ and $\inf$ respectively, implying the function is not continuous at $x=0$.
In order to prove limit does not exist we can use contradiction. Say the limit exists and it is $L$. From the epsilon delta definition, say we fix epsilon to be 1. Then there is a delta such that $\forall |x| < \delta$, $\left|\left\lfloor\frac1x\right\rfloor- L\right| < 1$. If we look at only $x > 0$, we can choose a $x = 1/n < \delta$ for $n\in \mathbb Z^+ $. So, $\left\lfloor\frac1x\right\rfloor= n$. We can always pick a $1/n$ such that the epsilon inequality does not hold as $\left\lfloor\frac1x\right\rfloor$ is unbounded near $0$ but I can't complete the formal argument from here.