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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?

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    $\begingroup$ For $n>1$, these functions are constant! $\endgroup$ – Bernard Oct 18 '17 at 17:48
  • $\begingroup$ @Bernard. How to prove it? $\endgroup$ – Almot1960 Oct 18 '17 at 17:52
  • $\begingroup$ Simply with the definition of the floor function. Is there any difficulty? $\endgroup$ – Bernard Oct 18 '17 at 17:55
  • $\begingroup$ @Bernard. this way? if $ \ \ : \to \ \ 0<a<b$ then $0=\frac{0}{b}<\frac{a}{b}<\frac{b}{b}=1 \ \to \lfloor \frac{a}{b} \rfloor=0$ $\endgroup$ – Almot1960 Oct 18 '17 at 17:59
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    $\begingroup$ 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. $\endgroup$ – Lubin Oct 18 '17 at 19:10

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