MathematicsIf x>0, then xxx⋯∞=

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

A
$x$
B
$x^{2}$
C
$2 x$
D
$\frac{x}{2}$

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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