How to find: $\lim\limits_{n\rightarrow\infty} \left\lfloor \frac{-1}{n}\right\rfloor$

How to find:

$$\lim\limits_{n\rightarrow\infty} \left\lfloor \frac{-1}{n}\right\rfloor =?$$

And :

$$\lim\limits_{n\rightarrow\infty} \left\lfloor \frac{1}{n}\right\rfloor =?$$

My Try :

we know that :

$$\frac{-1}{n}-1 < \left\lfloor \frac{-1}{n}\right\rfloor \leq \frac{-1}{n}$$

$$\frac{-(1+n)}{n} < \left\lfloor \frac{1}{n}\right\rfloor \leq \frac{-1}{n}$$

NoW :

$$\lim\limits_{n\rightarrow\infty} \frac{-1}{n} =0$$

$$\lim\limits_{n\rightarrow\infty} \frac{-(1+n)}{n} =-1$$

So we can not use from Sandwich theorem . So what can we do?

• For $n>1$, these functions are constant! – Bernard Oct 18 '17 at 17:48
• @Bernard. How to prove it? – Almot1960 Oct 18 '17 at 17:52
• Simply with the definition of the floor function. Is there any difficulty? – Bernard Oct 18 '17 at 17:55
• @Bernard. this way? if $\ \ : \to \ \ 0<a<b$ then $0=\frac{0}{b}<\frac{a}{b}<\frac{b}{b}=1 \ \to \lfloor \frac{a}{b} \rfloor=0$ – Almot1960 Oct 18 '17 at 17:59
• I would insist that you try to evaluate $\lfloor1/n\rfloor$ for $n=1,2,3,4,5$ (that should be enough). From your question and your approach to it, it really looks as if you did not have a practical friendship with the function. – Lubin Oct 18 '17 at 19:10