Question
Let $$f\left( x \right) = x - \left[ x \right],$$ where $$\left[ x \right]$$ denotes the greatest integer $$ \leqslant x$$ and $$g\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left\{ {f\left( x \right)} \right\}}^{2n}} - 1}}{{{{\left\{ {f\left( x \right)} \right\}}^{2n}} + 1}},$$ then $$g\left( x \right)$$ is equal to :
A.
$$0$$
B.
$$1$$
C.
$$ - 1$$
D.
none of these
Answer :
$$ - 1$$
Solution :
$$\eqalign{
& {\text{As }}0 \leqslant x - \left[ x \right] < 1\,\forall \,x\, \in \,R,\,0 \leqslant f\left( x \right) < 1 \cr
& \therefore \,\mathop {\lim }\limits_{n \to \infty } {\left\{ {f\left( x \right)} \right\}^{2n}} = 0 \cr
& {\text{Thus, for}}\,x\, \in \,R, \cr
& g\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left\{ {f\left( x \right)} \right\}}^{2n}} - 1}}{{{{\left\{ {f\left( x \right)} \right\}}^{2n}} + 1}} = \frac{{0 - 1}}{{0 + 1}} = - 1{\text{ }} \cr} $$