The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots .+[n r]}{n^{2}}$, where $r$ is non-zero real number and $[r]$ denotes the greatest integer less than or equal to $\mathrm{r}$, is equal to :
Correct Option: 1
We know that
$\leq[\mathrm{r}]+[2 \mathrm{r}]+\ldots . .+[\mathrm{nr}]<(\mathrm{r}+2 \mathrm{r}+\ldots .+\mathrm{nr})+\mathrm{n}$
Now, $\lim _{n \rightarrow \infty} \frac{n(n+1) \cdot r}{2 \cdot n^{2}}=\frac{r}{2}$
and $\lim _{n \rightarrow \infty} \frac{\frac{n(n+1) r}{2}+n}{n^{2}}=\frac{r}{2}$
So, by Sandwich Theorem, we can conclude that
$\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^{2}}=\frac{r}{2}$
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