The value of $\lim _{n \rightarrow \infty} \frac{[\mathrm{r}]+[2 \mathrm{r}]+\ldots \ldots+[\mathrm{nr}]}{\mathrm{n}^{2}}$, where $\mathrm{r}$ is non-zero real number and
$[r]$ denotes the greatest integer less than or equal to $r$, is equal to :
Correct Option: 1
We know that
$\mathrm{r} \leq[\mathrm{r}]<\mathrm{r}+1$
and
$2 r \leq[2 r]<2 r+1$
$3 r \leq[3 r]<3 r+1$
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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