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

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(B) લક્ષ $L = \mathop {\lim }\limits_{x 	o \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } ight]$ ની ગણતરી કરવા માટે,આપણે પદનું સંમે

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) લક્ષ $L = \mathop {\lim }\limits_{x 	o \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } ight]$ ની ગણતરી કરવા માટે,આપણે પદનું સંમેયીકરણ કરીએ:$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}$અંશ અને છેદને $\sqrt{x}$ વડે ભાગતા:$L = \mathop {\lim }\limits_{x 	o \infty } \frac{\sqrt{1 + x^{-1/2}}}{\sqrt{1 + \sqrt{x^{-1} + x^{-3/2}}} + 1}$જેમ $x 	o \infty$,તેમ $x^{-1/2} 	o 0$,$x^{-1} 	o 0$,અને $x^{-3/2} 	o 0$.$L = \frac{\sqrt{1+0}}{\sqrt{1+0} + 1} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right]$ ની કિંમત શોધો.

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