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

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

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

## closed as off-topic by Batominovski, kjetil b halvorsen, Elaqqad, Michael Galuza, user91500Aug 16 '15 at 14:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, kjetil b halvorsen, Elaqqad, Michael Galuza, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

• What do you mean by "N"? is it typo mistake? – Chiranjeev_Kumar Aug 16 '15 at 13:37
• To clarify: $N=n$? Hint: a good way to get a feel for problems like this is to simply try large values. It is easy, for example to evaluate your expression whenever $n$ is an integer. What if $n=100.35$? or $10000.7$? – lulu Aug 16 '15 at 13:40
• @Chiranjeev I think that you don't have the rights to change $N$ to $n$ because this changes the meaning of the question – Elaqqad Aug 16 '15 at 13:43
• yeah that's why i asked to OP @ Elaqqad Sorry for that – Chiranjeev_Kumar Aug 16 '15 at 13:46
• OTOH the question is unanswerable in its original form, and it's highly probable that the OP meant to write $n$ instead of $N$... – PM 2Ring Aug 16 '15 at 13:49

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 $\displaystyle = \frac{3}{10}$

• Wait, something is wrong (maybe it's only notation). But $x<y\leq x$ is tautologically false. When you do the inequalities, you do it without he limits. When you apply limits, then you may get equality on the boundaries, so your $<$ should be a $\leq$. – Zach Stone Aug 16 '15 at 16:15

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.

• please ^give your answer as the largest integer after multiplying by 10^6 – Shivam Chauhan Aug 16 '15 at 13:40
• This is not what you asked in the OP..? – Paolo Leonetti Aug 16 '15 at 13:42