If |x|<1 then limn→∞ (1+x)1+x21+x4…1+x2n is equal to

If |x|

A
$\frac{1}{x - 1}$
B
$\frac{1}{1 - x}$
C
$1 - x$
D
$x - 1$

Solution:

We have,

$\begin{array}{l} lim_{n \to \infty} \{(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2 n}) \\ = lim_{n \to \infty} \{\frac{(1 - x) (1 + x) (1 + x^{2}) \dots (1 + x^{2 n})}{1 - x} . \\ = lim_{n \to \infty} \frac{1 - x^{4 n}}{1 - x} \\ = \frac{1 - 0}{1 - x} = \frac{1}{1 - x} [∵ lim_{n \to \infty} x^{n} = 0 for - 1 < x < 1] \end{array}$

$If | x | < 1$ then $lim_{n \to \infty} \{(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2 n})\}$ is equal to

Solution:

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