\[\underset{x\to 1}{\mathop{\lim }}\,\frac{(1-x)(1-{{x}^{2}})...(1-{{x}^{2n}})}{{{\{(1-x)(1-{{x}^{2}})...(1-{{x}^{n}})\}}^{2}}},n\in N,\] equals

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{(1-x)(1-{{x}^{2}})...(1-{{x

1.	\[\underset{x\to 1}{\mathop{\lim }}\,\frac{(1-x)(1-{{x}^{2}})...(1-{{x}^{2n}})}{{{\{(1-x)(1-{{x}^{2}})...(1-{{x}^{n}})\}}^{2}}},n\in N,\] equals
A.	\[^{2n}{{P}_{n}}\]
B.	\[^{2n}{{C}_{n}}\]
C.	\[(2n)!\]
D.	None of these
Answer» C. \[(2n)!\]

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\[\underset{x\to 1}{\mathop{\lim }}\,\frac{(1-x)(1-{{x}^{2}})...(1-{{x}^{2n}})}{{{\{(1-x)(1-{{x}^{2}})...(1-{{x}^{n}})\}}^{2}}},n\in N,\] equals

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