How do I prove that the floor function has no limit?

Question

I don't really know where to start. I had this approach in mind: take any integer, let's name it $k$, and show that the limit on the left side of $k$ equals to $k-1$, and that the limit on the right side of $k$ equals to $k$, thus the limits are different, and so the floor function has no limit.

Answer 1

As this is a limit, we can restrict ourselves to $\;x\in\left(k-\frac12\,,\,\,k+\frac12\right)\;,\;\;k\in\Bbb Z\;$ , and then

$$\begin{cases}\lim\limits_{x\to k^-}\lfloor x\rfloor=\lim\limits_{x\to k^-}(k-1)=k-1\\{}\\\lim\limits_{x\to k^+}\lfloor x\rfloor=\lim\limits_{x\to k^+}\,k=k\end{cases}$$