Discussion Forum

For each t ε R, let [t] be the greatest integer less than or equal to t. Then,

$\lim_{x \rightarrow 0}x([\frac{1}{x}+[\frac{2}{x}]+......+[\frac{15}{x}])$

Answer:

Explanation:

Key Idea Use   property of greatest integer  function [x]= x - {x}

We have,

        $\lim_{x \rightarrow 0}x([\frac{1}{x}+[\frac{2}{x}]+......+[\frac{15}{x}])$

we know,    [x]= x - {x}

$\therefore$       $[\frac{1}{x}]=\frac{1}{x}-\left\{\frac{1}{x}\right\}$

Similarly,      $[\frac{n}{x}]=\frac{n}{x}-\left\{\frac{n}{x}\right\}$

$\therefore$     Given limit  

   $=\lim_{x \rightarrow 0}x \left(\frac{1}{x}-\left\{\frac{1}{x}\right\}+\frac{2}{x}-\left\{\frac{2}{x}\right\}+   ....\frac{15}{x}-\left\{\frac{15}{x}\right\}\right)$

    $=\lim_{x \rightarrow 0}(1+2+3+...+15) -x\left(\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}+.....+\left\{\frac{15}{x}\right\}\right)$

   =120-0=120

$\begin{bmatrix}\because 0\leq\left\{\frac{n}{x}\right\}<1,therefore \\0\leq x \left\{\frac{n}{x}\right\}< x \Rightarrow\lim_{x \rightarrow 0} x \left\{\frac{n}{x}\right\}=0 \end{bmatrix}$