The value of mathop {lim }limits_{x to 0} fra | vedclass

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Question

(C) We have $\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

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$\sqrt{2}$

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(C) We have $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin^2(x^2/2)} }}{{2 \sin^2(x/2)}}$.Since $\sqrt{\sin^2 	heta} = |\sin 	heta|$,and for $x 	o 0$,$\sin(x^2/2) > 0$,we have:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt{2} \sin(x^2/2)}}{{2 \sin^2(x/2)}} = \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \frac{\sin(x^2/2)}{\sin^2(x/2)}$.Multiply and divide by $(x^2/2)$ and $(x/2)^2$:$= \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{(\sin(x/2)/(x/2))^2 \cdot (x^2/4)} ight)$.$= \frac{1}{\sqrt{2}} \cdot \frac{1/2}{1/4} \cdot \mathop {\lim }\limits_{x 	o 0} \frac{1}{1^2} = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ is

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