If [.] denotes the greatest integer function, then find the value of $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left[ x \right] + \left[ {2x} \right] + .......... + \left[ {nx} \right]}}{{{n^2}}}$ is,

Hint: In this particular question use the concept that [x] is greater than or less than equal to x, so [x] is written as, $x - 1 < \left[ x \right] \leqslant x$, and later on use the concept of sandwich theorem, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data
[.] denotes the greatest integer function.
So according to the definition of the greatest integer function we have,
$ \Rightarrow x - 1 < \left[ x \right] \leqslant x$................ (1)
Similarly,
$ \Rightarrow 2x - 1 < \left[ {2x} \right] \leqslant 2x$
$ \Rightarrow 3x - 1 < \left[ {3x} \right] \leqslant 3x$
.
.
.
$ \Rightarrow nx - 1 < \left[ {nx} \right] \leqslant nx$
Now add all these equation we have,
$ \Rightarrow \left( {x + 2x + 3x + .... + nx} \right) - \left( {1 + 1 + 1 + ...... + 1} \right) < \left( {\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]} \right) \leqslant \left( {x + 2x + 3x + .....nx} \right)$
Now as we know that 1 + 1 + 1 +...... + 1 up to n terms is equal to n so we have,
$ \Rightarrow x\left( {1 + 2 + 3 + .... + n} \right) - n < \left( {\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]} \right) \leqslant x\left( {1 + 2 + 3 + .....n} \right)$
Now as we know that 1 + 2 + 3 +...... + n is the summation of first n terms whose sum is given as, $\dfrac{{n\left( {n + 1} \right)}}{2}$ so we have,
$ \Rightarrow x\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) - n < \left( {\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]} \right) \leqslant x\dfrac{{n\left( {n + 1} \right)}}{2}$
Now divide by ${n^2}$ throughout we have,
$ \Rightarrow \dfrac{{x\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) - n}}{{{n^2}}} < \left( {\dfrac{{\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]}}{{{n^2}}}} \right) \leqslant \dfrac{{x\dfrac{{n\left( {n + 1} \right)}}{2}}}{{{n^2}}}$
$ \Rightarrow x\left( {\dfrac{{n\left( {n + 1} \right)}}{{2{n^2}}}} \right) - \dfrac{1}{n} < \left( {\dfrac{{\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]}}{{{n^2}}}} \right) \leqslant x\dfrac{{n\left( {n + 1} \right)}}{{2{n^2}}}$
$ \Rightarrow \dfrac{x}{2}\left( {1 + \dfrac{1}{n}} \right) - \dfrac{1}{n} < \left( {\dfrac{{\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]}}{{{n^2}}}} \right) \leqslant \dfrac{x}{2}\left( {1 + \dfrac{1}{n}} \right)$
Now apply $\mathop {\lim }\limits_{n \to \infty } $in all of the terms we have,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{x}{2}\left( {1 + \dfrac{1}{n}} \right) - \dfrac{1}{n}} \right) < \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]}}{{{n^2}}}} \right) \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{x}{2}\left( {1 + \dfrac{1}{n}} \right)} \right)$
Now as we know that when, $n \to \infty \Rightarrow \dfrac{1}{n} \to 0$ so we have,
$ \Rightarrow \dfrac{x}{2} < \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]}}{{{n^2}}}} \right) \leqslant \dfrac{x}{2}$
Now according to sandwich theorem if, $\mathop {\lim }\limits_{x \to \infty } h\left( x \right) < \mathop {\lim }\limits_{x \to \infty } g\left( x \right) \leqslant \mathop {\lim }\limits_{x \to \infty } f\left( x \right)$ and $\mathop {\lim }\limits_{x \to \infty } h\left( x \right) = \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = P$ then $\mathop {\lim }\limits_{x \to \infty } g\left( x \right) = P$ so we have,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\left[ x \right] + \left[ {2x} \right] + \left[ {3x} \right] + ....... + \left[ {nx} \right]}}{{{n^2}}}} \right) = \dfrac{x}{2}$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the sum of first n terms and always recall the sandwich theorem that if $\mathop {\lim }\limits_{x \to \infty } h\left( x \right) < \mathop {\lim }\limits_{x \to \infty } g\left( x \right) \leqslant \mathop {\lim }\limits_{x \to \infty } f\left( x \right)$, and $\mathop {\lim }\limits_{x \to \infty } h\left( x \right) = \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = P$ then $\mathop {\lim }\limits_{x \to \infty } g\left( x \right) = P$