TITLE: Epsilon Delta proof of a floor function
QUESTION [5 upvotes]: I have to provide a proof for a function but I'm struggling to grasp the main concept.
$$\lim_{x \to 3} \left\lfloor \frac{x}{2}\right\rfloor = 1 $$
Here is what I've come up with:
$$\frac{x}{2} - 1 \lt \left\lfloor \frac{x}{2}\right\rfloor \lt  \frac{x}{2} + 1$$
But I'm stuck from here since I cannot use anything else but $\varepsilon$-$\delta$ (meaning no squeeze theorem.) Each of the limits (for every side) is not helpful. I get: $$ \frac{1}{2} \lt \lim_{x \to 3} \left\lfloor \frac{x}{2}\right\rfloor \lt \frac{5}{2}$$
Any clarifications are welcome!
Thanks!

REPLY [3 votes]: You can just use the fact that $\lfloor \frac{x}{2}\rfloor $ is constant on $[2,4)$, so for any $\varepsilon >0$, $\delta =0.5 $(or any convenient number $<1$) works.