mathop {lim }limits_{x to infty } frac{{sqrt | vedclass

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(C) Let $f(x) = \sqrt{x^{2} - \sqrt{x^{2} - \sqrt{x^{2} - \dots}}}$Then $f(x) = \sqrt{x^{2} - f(x)}$.Squaring both sides,we get $f(x)^{2} = x^{2} - f(

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(C) Let $f(x) = \sqrt{x^{2} - \sqrt{x^{2} - \sqrt{x^{2} - \dots}}}$Then $f(x) = \sqrt{x^{2} - f(x)}$.Squaring both sides,we get $f(x)^{2} = x^{2} - f(x)$,which implies $f(x)^{2} + f(x) - x^{2} = 0$.Using the quadratic formula for $f(x)$,we have $f(x) = \frac{-1 \pm \sqrt{1 + 4x^{2}}}{2}$.Since $f(x) > 0$,we take the positive root: $f(x) = \frac{-1 + \sqrt{1 + 4x^{2}}}{2}$.Now,we evaluate the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{f(x)}{x} = \mathop {\lim }\limits_{x 	o \infty } \frac{-1 + \sqrt{1 + 4x^{2}}}{2x}$.Dividing numerator and denominator by $x$,we get $\mathop {\lim }\limits_{x 	o \infty } \frac{-\frac{1}{x} + \sqrt{\frac{1}{x^{2}} + 4}}{2} = \frac{0 + \sqrt{0 + 4}}{2} = \frac{2}{2} = 1$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - \dots} } } }}{x}$ is equal to-

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