mathop {lim }limits_{x to infty } left[ {sqrt | vedclass

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(B) To evaluate the limit $L = \mathop {\lim }\limits_{x 	o \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } ight]$,we rationalize th

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$\frac{1}{2}$

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(B) To evaluate the limit $L = \mathop {\lim }\limits_{x 	o \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } ight]$,we rationalize the expression:$L = \mathop {\lim }\limits_{x 	o \infty } \frac{(\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x)(\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x)}{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x}$$L = \mathop {\lim }\limits_{x 	o \infty } \frac{x + \sqrt {x + \sqrt x } - x}{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x} = \mathop {\lim }\limits_{x 	o \infty } \frac{\sqrt {x + \sqrt x}}{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x}$Divide the numerator and denominator by $\sqrt{x}$:$L = \mathop {\lim }\limits_{x 	o \infty } \frac{\sqrt{1 + \frac{1}{\sqrt{x}}}}{\sqrt{1 + \frac{\sqrt{x+\sqrt{x}}}{x}} + 1} = \mathop {\lim }\limits_{x 	o \infty } \frac{\sqrt{1 + x^{-1/2}}}{\sqrt{1 + \sqrt{x^{-1} + x^{-3/2}}} + 1}$As $x 	o \infty$,$x^{-1/2} 	o 0$,$x^{-1} 	o 0$,and $x^{-3/2} 	o 0$.$L = \frac{\sqrt{1+0}}{\sqrt{1+0} + 1} = \frac{1}{1+1} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right]$ is equal to

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