Question

The value of $lim{x}^2$ [$\frac{1}{x^2}$] where [] denotes G.I.F. is

Answer 1

$\frac{1}{x^2}-1\le \left[\frac{1}{x^2}\right]\le \frac{1}{x^2}$

$(1-x^2)\le x^2\left[\frac{1}{x^2}\right]\le 1$

$\underset{x\to 0}{lim}(1-x^2)\le \underset{x\to 0}{lim}{x}^2\left[\frac{1}{x^2}\right]\le \underset{x\to 0}{lim}1$

$\Rightarrow 1-0\le \underset{x\to 0}{lim}{x}^2\left[\frac{1}{x^2}\right]\le 1$

$\Rightarrow \underset{x\to 0}{lim}{x}^2\left[\frac{1}{x^2}\right]=1$