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$\lim_{n \in N,n \to \infty} \frac{ \lfloor rn \rfloor }{n} = r$

I know that for every r, real number there exists a sequence of rational numbers. But in above how left side limit = r. Is it tends to o? i did not understand how it tends to r. Kindly elaborate

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    $\begingroup$ Essentially this works because the fractional amount removed by the floor function makes less and less difference with larger numbers, i.e., when $r\mapsto rn$. So for large $n$, $\lfloor rn \lfloor \approx rn$. $\endgroup$
    – Jam
    Feb 21, 2020 at 3:51
  • $\begingroup$ nice........... $\endgroup$ Feb 21, 2020 at 3:55

2 Answers 2

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Note that $$ \lfloor x \rfloor \leq x \leq \lfloor x \rfloor+1, \quad \mbox{when } x \geq 0 $$ yields $$ rn - 1 \leq \lfloor rn \rfloor \leq rn $$ Dividing all sides by $n$ will give you the limit.

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  • $\begingroup$ oh yes! understood. thank you $\endgroup$ Feb 21, 2020 at 3:07
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    $\begingroup$ You can make the upper part of the inequality strict: $\lfloor x\rfloor \le x < \lfloor x \rfloor +1$ but in this case it doesn't really matter. $\endgroup$
    – Jam
    Feb 21, 2020 at 3:48
  • $\begingroup$ Sandwich theorm! $\endgroup$
    – sai-kartik
    Feb 21, 2020 at 4:12
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Let $rn=I+f,\; 0\le f<1$

As $n\to\infty, I\to\pm\infty$ according as the sign of $r$

$$\dfrac{[rn]}n=\dfrac I {\dfrac{I+f}r}=r\cdot\dfrac1{1+\dfrac fI}$$

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