If - x - - 1- then lim x- 1-x 1- x 2 1- x 4 - 1- x 2n -

The correct option is D 0 x →∞lim (1 + x) ( 1 + x^2) (1 + x^4) ..... (1 + x^2x) x →∞lim(1 x) (1 + x) (1 + x^2) ( 1 + x^4) .... (1 + x^2x)(1 ...

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Solving Linear Differential Equations of First Order

If | x | < ...

Question

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

\frac{1}{1 + x}

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\frac{1}{1 - x}

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1

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Similar questions

| x | < 1

lim n \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}

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

\frac{d y}{d x}

x = 0

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

\frac{d y}{d x}

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

\frac{d y}{d x}

x = 0

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