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If $| x | < 1,$ then $\underset{n \to \infty}{l i m} (1 + x) (1 + x^{2}) (1 + x^{4}) ....... (1 + x^{2 n}) =$

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Correct option is C

limx→∞1+x+x2+.....+x2n=11−x

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If |x|<1,then limn→∞(1+x)(1+x2)(1+x4).......(1+x2n)=

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If $| x | < 1,$ then $\underset{n \to \infty}{l i m} (1 + x) (1 + x^{2}) (1 + x^{4}) ....... (1 + x^{2 n}) =$