Question

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

• A. $0$
• B. $r$
• C. $\frac{r}{2}$
• D. $2r$

Answer 1

The correct option is C

$\frac{r}{2}$

Explanation of the correct option.

Compute the required value.

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

We know that,

$$\begin{gathered} r\le \left[r\right]<r+1 \\ 2r\le \left[2r\right]<2r+1 \\ 3r\le \left[3r\right]<3r+1 \\ ... \\ ... \\ nr\le \left[nr\right]<nr+1 \end{gathered}$$

Add all the inequalities,

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

Now,

$\underset{n\to \infty }{\lim }\frac{n(n+1).r}{2.n^2}=\frac{r}{2}$

and,

$\underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{n(n+1).r}{2}}+n}{n^2}=\frac{r}{2}$

So, by sandwich theorem, we can conclude that

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

Hence,option $\left(C\right)$ is the correct option.