Prove by the definition of the limit
$\lim_{x\to-\infty} \frac{5}{\lfloor{x}\rfloor} = 0$
(If you use that for every $x \in R, x-1 \leq \lfloor{x}\rfloor \leq x$, prove it).
I did the following:
$ |f(x) - L| = | \frac{5}{\lfloor{x}\rfloor} | = - \frac{5}{\lfloor{x}\rfloor} \leq - \frac{5}{x} < - \frac{5}{M} = \epsilon $
So $M= - \frac{5}{\epsilon}$
But it seems too obvious. Also I didn't know how to prove the definition of the floor function.
Is it correct?