JEE Main 2021 — Limits Question with Solution

We know that $r\le \left[r\right]<r+1$

$2r\le \left[2r\right]<2r+1$

$3r\le \left[3r\right]<3r+1$

$⋮ ⋮ ⋮$

$nr\le \left[nr\right]<nr+1$

On adding all the above inequalities, we get

$r+2r+\dots +nr\le \left[r\right]+\left[2r\right]+...+\left[nr\right]<r+2r+...nr+n$

Using $1+2+3+...+n=\frac{n\left(n+1\right)}{2}, n\in N,$

$\frac{n\left(n+1\right)}{2}·r\le \left[r\right]+\left[2r\right]+...+\left[nr\right]<\frac{n\left(n+1\right)}{2}·r+n$

$\Rightarrow \frac{{\displaystyle \frac{n\left(n+1\right)}{2}}·r}{n^2}\le \frac{\left[r\right]+\left[2r\right]+...+\left[nr\right]}{n^2}<\frac{{\displaystyle \frac{n\left(n+1\right)}{2}}·r+n}{n^2}$

Now, $\underset{n\to \infty }{\lim }\frac{n(n+1)·r}{2·n^2}=\underset{n\to \infty }{\lim }\frac{n^2\left(1+{\displaystyle \frac{1}{n}}\right)·r}{2·n^2}=\frac{r}{2}$

and $\underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{n\left(n+1\right)}{2}}·r+n}{n^2}=\underset{n\to \infty }{\lim }\frac{n^2\left\{\left(1+{\displaystyle \frac{1}{n}}\right)·r+{\displaystyle \frac{1}{n}}\right\}}{2·n^2}=\frac{r}{2}$

So, by Sandwich Theorem, we can conclude that

$\underset{n\to \infty }{\lim }\frac{\left[r\right]+\left[2r\right]+...+\left[nr\right]}{n^2}=\frac{r}{2}.$