The value of lim _{n rightarrow infty} frac{[ | vedclass

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Question

(A) We know that for any real number $x$,$x-1 < [x] \leq x$. Applying this to the terms $[kr]$ for $k=1, 2, \ldots, n$,we have: $kr-1 < [kr] \leq kr$.

Vedclass · Accepted Answer

$\frac{r}{2}$

Vedclass · Answer

(A) We know that for any real number $x$,$x-1 Applying this to the terms $[kr]$ for $k=1, 2, \ldots, n$,we have: $kr-1 Summing these inequalities from $k=1$ to $n$: $\sum_{k=1}^{n} (kr-1) $r \frac{n(n+1)}{2} - n Dividing the entire inequality by $n^2$: $\frac{r \frac{n(n+1)}{2} - n}{n^2} Taking the limit as $n \rightarrow \infty$: $\lim_{n \rightarrow \infty} \left( \frac{r(n^2+n)}{2n^2} - \frac{n}{n^2} \right) = \frac{r}{2}$ and $\lim_{n \rightarrow \infty} \frac{r(n^2+n)}{2n^2} = \frac{r}{2}$. By the Sandwich Theorem,the limit is $\frac{r}{2}$.

The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$,where $r$ is a non-zero real number and $[x]$ denotes the greatest integer less than or equal to $x$,is equal to:

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