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Q.

If $x > 0$ , then $\sqrt{x \sqrt{x \sqrt{x \dots \infty}}} =$

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a

$x$

b

$x^{2}$

c

$2 x$

d

$\frac{x}{2}$

answer is A.

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Detailed Solution

Suppose, $k = \sqrt{x \sqrt{x \sqrt{x \dots \infty}}}$

$k = \sqrt{x . k}$

$k^{2} = x k$

$k (k - x) = 0$

$k = 0 (or) k = x$

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