Question

Let [.] denotes the greatest integer function, then

$lim_{n \: \to \: \infty } \: \: \frac{[x] + [2x] + [3x] + - - - + [nx]}{ {n}^{2} }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{n \to \infty }\rm \frac{[x] + [2x] + [3x] + - - - + [nx]}{ {n}^{2} }$

We know, by definition of Greatest Integer function,

$\rm \: x - 1 < [x] \leqslant x$

$\rm \: 2x - 1 < [2x] \leqslant 2x$

$\rm \: 3x - 1 < [3x] \leqslant 3x$

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.

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$\rm \: nx - 1 < [nx] \leqslant nx$

On adding above all inequalities, we get

$\rm \: (x - 1) + (2x - 1) + \cdots + (nx - 1) < [x] + [2x] + \cdots + [nx] \leqslant x + 2x + \cdots + nx$

$\rm \: x(1 + 2 + \cdots + n) - n < [x] + [2x] + \cdots + [nx] \leqslant x(1 + 2 + \cdots + n)$

We know,

$\color{green}\boxed{ \rm{ \:1 + 2 + 3 + \cdots + n \: = \: \frac{n(n + 1)}{2} \: }}$

So, on using this result, we get

$\rm \: x\dfrac{n(n + 1)}{2} - n< [x] + [2x] + \cdots + [nx] \leqslant x\dfrac{n(n + 1)}{2}$

$\rm \: \dfrac{xn(n + 1)}{2} - n< [x] + [2x] + \cdots + [nx] \leqslant \dfrac{xn(n + 1)}{2}$

$\rm \: \frac{x}{2}( {n}^{2} + n) - n< [x] + [2x] + \cdots + [nx] \leqslant \frac{x}{2}( {n}^{2} + n)$

can be further rewritten as by dividing each term by n^2.

$\rm \: \dfrac{x}{2} \bigg(1 + \dfrac{1}{n} \bigg) - \dfrac{1}{n} < \dfrac{[x] + [2x] + \cdots + [nx]}{ {n}^{2} } \leqslant \dfrac{x}{2} \bigg(1 + \dfrac{1}{n} \bigg)$

On applying limits, we get

$\rm \:\displaystyle\lim_{n \to \infty }\rm \bigg(\dfrac{x}{2} \bigg(1 + \dfrac{1}{n} \bigg) - \dfrac{1}{n} \bigg) < \displaystyle\lim_{n \to \infty }\rm \dfrac{[x] + [2x] + \cdots + [nx]}{ {n}^{2} } \leqslant \displaystyle\lim_{n \to \infty }\rm \dfrac{x}{2} \bigg(1 + \dfrac{1}{n} \bigg)$

Now,

$\rm \:\displaystyle\lim_{n \to \infty }\rm \bigg(\dfrac{x}{2} \bigg(1 + \dfrac{1}{n} \bigg) - \dfrac{1}{n} \bigg) = \frac{x}{2}$

and

$\rm \:\displaystyle\lim_{n \to \infty }\rm \dfrac{x}{2} \bigg(1 + \dfrac{1}{n} \bigg) = \frac{x}{2}$

So, By squeeze theorem, we have

$\color{green}\boxed{ \rm{ \:\rm\implies \:\displaystyle\lim_{n \to \infty }\rm \frac{[x] + [2x] + [3x] +\cdots + [nx]}{ {n}^{2} } = \dfrac{x}{2}}}$

$\rule{190pt}{2pt}$

Remark :- Squeeze Theorem or Sandwich Theorem :-

$\rm \: If \: f(x) \: \leqslant \: g(x) \: \leqslant \: h(x) \: such \: that \\ \rm \: \displaystyle \lim_{x \: \to \: a}h(x) = l \\ and \: \\ \displaystyle \lim_{x \: \to \: a}f(x) = l \\ \rm \: then \\ \rm \: \displaystyle \lim_{x \: \to \: a}g(x) = l$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$