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

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Question

(C) લક્ષ $\mathop {\lim }\limits_{x 	o \infty } \sqrt x (\sqrt {x + 5} - \sqrt x )$ ની ગણતરી કરવા માટે,આપણે પદનું સંમેયીકરણ કરીએ:$\mathop {\lim }\lim

Vedclass · Accepted Answer

$5/2$

Vedclass · Answer

(C) લક્ષ $\mathop {\lim }\limits_{x 	o \infty } \sqrt x (\sqrt {x + 5} - \sqrt x )$ ની ગણતરી કરવા માટે,આપણે પદનું સંમેયીકરણ કરીએ:$\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}$જેમ $x 	o \infty$,તેમ $5/x 	o 0$,તેથી પદ $\frac{5}{\sqrt{1+0} + 1} = \frac{5}{2}$ થાય છે.

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

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$\mathop {\lim }\limits_{x \to 0} \cos \frac{1}{x}$

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