Limn-1 - x1 - x21 - x4 - 1 - x2n- -x- - 1- is equal to

The correct option is B 1/1 x lim_n→∞(1 + x)(1 + x^2)(1 + x^4) ... (1 + x^2n) =1/1 xlim_n→∞(1 x)(1 + x)(1 + x^2)(1 + x^4) ... (1 + x^2n) =1 ...

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Algebra of Limits

limn→∞1 + x1 ...

Question

$lim n \to \infty (1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2 n}), | x | < 1$ , is equal to

\frac{1}{x - 1}

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\frac{1}{1 - x}

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1 - x

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1

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Solution

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Similar questions

| x | < 1

lim n \to \infty {(1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2 n})}

| x | < 1

lim x \to \infty (1 + x) (1 + x^{2}) (1 + x^{4}) + . . . (1 + x^{2 n}) =

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})

\frac{d y}{d x}

x = 0

1

y = \frac{1 - x^{2^{n + 1}}}{1 - x}

x^{2^{n + 1}}

\frac{1}{(1 + x) (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2^{n}})}

| x | < 1

x^{2^{n + 1}}

\frac{1}{(1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})}

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