Let \[f(x)=x-[x],\] where [x] denotes the greatest integer \[\le x\] and \[g(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{{{\{f(x)\}}^{2n}}-1}{{{\{f(x)\}}^{2n}}+1},\] then g(x) is equal to

Let \[f(x)=x-[x],\] where [x] denotes the greatest integer \[\le x\] a

1.	Let \[f(x)=x-[x],\] where [x] denotes the greatest integer \[\le x\] and \[g(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{{{\{f(x)\}}^{2n}}-1}{{{\{f(x)\}}^{2n}}+1},\] then g(x) is equal to
A.	0
B.	1
C.	-1
D.	None of these
Answer» D. None of these

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Let \[f(x)=x-[x],\] where [x] denotes the greatest integer \[\le x\] and \[g(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{{{\{f(x)\}}^{2n}}-1}{{{\{f(x)\}}^{2n}}+1},\] then g(x) is equal to

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