mathop {lim }limits_{x to infty } sqrt x (sqr | vedclass

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(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \sqrt x (\sqrt {x + 5} - \sqrt x )$,we rationalize the expression:$\mathop {\lim }\l

Vedclass · Accepted Answer

$5/2$

Vedclass · Answer

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \sqrt x (\sqrt {x + 5} - \sqrt x )$,we rationalize the expression:$\mathop {\lim }\limits_{x 	o \infty } \sqrt x (\sqrt {x + 5} - \sqrt x ) 	imes \frac{\sqrt {x + 5} + \sqrt x}{\sqrt {x + 5} + \sqrt x}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{\sqrt x (x + 5 - x)}{\sqrt {x + 5} + \sqrt x}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{5\sqrt x}{\sqrt x(\sqrt {1 + 5/x} + 1)}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{5}{\sqrt {1 + 5/x} + 1}$As $x 	o \infty$,$5/x 	o 0$,so the expression becomes $\frac{5}{\sqrt{1+0} + 1} = \frac{5}{2}$.

$\mathop {\lim }\limits_{x \to \infty } \sqrt x (\sqrt {x + 5} - \sqrt x ) = $

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