Find the limit $\lim_{ n\rightarrow \infty} \frac{\left\lfloor\frac{3n}{10}\right\rfloor}{n}$

Question

$\left\lfloor x\right\rfloor$ denotes greatest integer function then $n \ge 1$ and is a positive integer

$$\lim_{ n\rightarrow \infty} \frac{\left\lfloor\frac{3n}{10}\right\rfloor}{n}$$

Answer 1

Using the formula $\displaystyle (x-1)\leq \lfloor x \rfloor \leq x\;,$ Where $\lfloor x \rfloor$ is an floor function of $x$

So Put $\displaystyle x = \frac{3n}{10}\;,$ we get

$\displaystyle\frac{3n}{10}-1\leq \lfloor \frac{3n}{10}\rfloor \leq \frac{3n}{10}$

So $\displaystyle \lim_{n\rightarrow \infty}\frac{\frac{3n}{10}-1}{n}< \lim_{n\rightarrow \infty}\frac{\lfloor \frac{3n}{10}\rfloor}{n} \leq \lim_{n\rightarrow \infty}\frac{\frac{3n}{10}}{n}$

So we get $\displaystyle \frac{3}{10}<\lim_{n\rightarrow \infty}\frac{\lfloor \frac{3n}{10}\rfloor}{n}\leq \frac{3}{10}$

So Using Sandwitch Theorem, We get Limit $\displaystyle = \frac{3}{10}$

Answer 2

Since $x-1<\lfloor x\rfloor\le x$, you get $$ \frac{\frac{3n}{10}-1}{n} < \frac{\left\lfloor\frac{3n}{10}\right\rfloor}{n} \le \frac{3}{10}. $$ Conclude.