Question

The value of $\mathop {\lim }\limits_{n \to \infty } \dfrac{{[r] + [2r] + [3r] + ... + [nr]}}{{{n^2}}}$ where $r$ is a non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$ , is equal to
A. $0$
B. $r$
C. $\dfrac{r}{2}$
D. $2r$

Answer 1

Hint: The greatest integer less than or equal to the number is returned by the greatest integer function. The biggest integer less than or equal to $x$ is denoted by the letter $[x]$. The specified number will be rounded to the nearest integer that is less than or equal to the number. We will use this information to express the greatest integer function of $r$ in terms of the squeeze theorem and thus find the given limit.

Formula used: The squeeze theorem is a theorem in calculus about the limit of a function squeezed between two other functions. According to squeeze (or sandwich) theorem, if $\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } h(x)$
then $\mathop {\lim }\limits_{x \to \infty } g(x) = l$

Complete step by step solution:
It is given that $[r]$ is the greatest integer function, so –
$
  r - 1 < [r] \leqslant r \\
  2r - 1 < [2r] \leqslant 2r \\
  3r - 1 < [3r] \leqslant 3r \\
  . \\
  . \\
  . \\
  nr - 1 < [nr] \leqslant nr \\
 $
Adding all the above equations, we get –
$
  r - 1 + 2r - 1 + 3r - 1...nr - 1 < [r] + [2r] + [3r] + ... + [nr] \leqslant r + 2r + 3r + ... + nr \\
  r(1 + 2 + 3 + ... + n) - n < [r] + [2r] + [3r] + ... + [nr] \leqslant r(1 + 2 + 3 + ... + n) \\
 $
Dividing all the sides by ${n^2}$ and putting $1 + 2 + 3 + .... + n = \dfrac{{n(n + 1)}}{2}$ , we get –
$
  \dfrac{{rn(n + 1)}}{{2{n^2}}} - \dfrac{n}{{{n^2}}} < \dfrac{{[r] + [2r] + [3r] + ... + [nr]}}{{{n^2}}} \leqslant \dfrac{{rn(n + 1)}}{{2{n^2}}} \\
  \dfrac{r}{2}(1 + \dfrac{1}{n}) - \dfrac{1}{n} < \dfrac{{[r] + [2r] + [3r] + ... + [nr]}}{{{n^2}}} \leqslant \dfrac{r}{2}(1 + \dfrac{1}{n}) \\
 $
Now, $\mathop {\lim }\limits_{n \to \infty } \dfrac{r}{2}(1 + \dfrac{1}{n}) = \dfrac{r}{2}$
And $\mathop {\lim }\limits_{n \to \infty } \dfrac{r}{2}(1 + \dfrac{1}{n}) - \dfrac{1}{n} = \dfrac{r}{2}$
Now, according to sandwich theorem, if $\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } h(x)$
Then $\mathop {\lim }\limits_{x \to \infty } g(x) = l$
The squeeze theorem also works if $f(x) < g(x) \leqslant h(x)$
So, we get $\mathop {\lim }\limits_{n \to \infty } \dfrac{{[r] + [2r] + [3r] + .... + [nr]}}{{{n^2}}} = \dfrac{r}{2}$

Thus, the correct option is C.

Note: To understand greatest integer function more clearly, look at this example of a greatest integer function - $[1.9] = 1$ , so we see that $0.9 < [1.9] \leqslant 1.9$ , that’s why we use the relation $r - 1 < [r] \leqslant r$ in the above solution. To find the limit of a given function when the variable tends to infinity, we express the function such that the variable is present only in the denominator of a constant like in this we expressed the function as $\dfrac{1}{n}$.