Question

Let $f\left(x\right)=x-\left[x\right]$ and $g\left(x\right)=\underset{n\to \infty }{lim}\frac{\left[f\right(x){]}^{2n}-1}{\left[f\right(x){]}^{2n}+1},$ then the absolute value of $g\left(x\right)$ is
(where $[.]$ denotes the greatest integer function)

Answer 1

We know, fractional part function $\left\{x\right\}=x-\left[x\right]$ and $0\le \left\{x\right\}<1$
$\Rightarrow f\left(x\right)=\left\{x\right\}$ and $0\le f\left(x\right)<1$
$\therefore \left[f\right(x\left)\right]=0$
Now,
$g\left(x\right)=\underset{n\to \infty }{lim}\frac{\left[f\right(x){]}^{2n}-1}{\left[f\right(x){]}^{2n}+1}$
$\Rightarrow g\left(x\right)=\frac{0-1}{0+1}=-1$
So, the absolute value of $g\left(x\right)$ is $1.$