$\left\lfloor x\right\rfloor$ 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, user91500 Aug 16 '15 at 14:44

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  • 1
    $\begingroup$ What do you mean by "N"? is it typo mistake? $\endgroup$ – Chiranjeev_Kumar Aug 16 '15 at 13:37
  • $\begingroup$ 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 $10000.7$? $\endgroup$ – lulu Aug 16 '15 at 13:40
  • $\begingroup$ @Chiranjeev I think that you don't have the rights to change $N$ to $n$ because this changes the meaning of the question $\endgroup$ – Elaqqad Aug 16 '15 at 13:43
  • $\begingroup$ yeah that's why i asked to OP @ Elaqqad Sorry for that $\endgroup$ – Chiranjeev_Kumar Aug 16 '15 at 13:46
  • $\begingroup$ OTOH the question is unanswerable in its original form, and it's highly probable that the OP meant to write $n$ instead of $N$... $\endgroup$ – 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 $\displaystyle = \frac{3}{10}$

  • $\begingroup$ 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 $\leq$. $\endgroup$ – 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.

  • $\begingroup$ please ^give your answer as the largest integer after multiplying by 10^6 $\endgroup$ – Shivam Chauhan Aug 16 '15 at 13:40
  • 2
    $\begingroup$ This is not what you asked in the OP..? $\endgroup$ – Paolo Leonetti Aug 16 '15 at 13:42

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