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

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

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

• 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 $n$, $\lfloor rn \lfloor \approx rn$. – Jam Feb 21 '20 at 3:51
• nice........... – Nascimento de Cos Feb 21 '20 at 3:55

## 2 Answers

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.

• oh yes! understood. thank you – Nascimento de Cos Feb 21 '20 at 3:07
• 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. – Jam Feb 21 '20 at 3:48
• Sandwich theorm! – sai-kartik Feb 21 '20 at 4:12

Let $$rn=I+f,\; 0\le f<1$$

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

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