mathop {lim }limits_{x to 0} frac{{sqrt {1 - | vedclass

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(C) આપણી પાસે $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) આપણી પાસે $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin^2(x^2/2)} }}{{2 \sin^2(x/2)}}$ છે.$x 	o 0$ માટે,$\sin(x^2/2) > 0$ હોવાથી:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt{2} \sin(x^2/2)}}{{2 \sin^2(x/2)}} = \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \frac{\sin(x^2/2)}{\sin^2(x/2)}$.$(x^2/2)$ અને $(x/2)^2$ વડે ગુણતા અને ભાગતા:$= \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{(\sin(x/2)/(x/2))^2 \cdot (x^2/4)} ight)$.$= \frac{1}{\sqrt{2}} \cdot \frac{1/2}{1/4} \cdot 1 = \sqrt{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ ની કિંમત શોધો.

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