\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(x-1)}^{2n}}-1}{{{(x-1)}^{2n}}+1}\]is discontinuous at

\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(x-1)}^{2n}}-1

1.	\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(x-1)}^{2n}}-1}{{{(x-1)}^{2n}}+1}\]is discontinuous at
A.	\[x=0\]only
B.	\[x=2\]only
C.	\[x=0\]and 2
D.	none of these
Answer» D. none of these

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\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(x-1)}^{2n}}-1}{{{(x-1)}^{2n}}+1}\]is discontinuous at

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