lim _{x rightarrow 0} frac{sqrt{1-cos x^2}}{1 | vedclass

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(A) We know that $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$.Applying this to the numerator and denominator:$\lim _{x ightarrow 0} \f

Vedclass · Accepted Answer

$\sqrt{2}$

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(A) We know that $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$.Applying this to the numerator and denominator:$\lim _{x ightarrow 0} \frac{\sqrt{2 \sin^2(x^2/2)}}{2 \sin^2(x/2)} = \lim _{x ightarrow 0} \frac{\sqrt{2} |\sin(x^2/2)|}{2 \sin^2(x/2)}$.Since $x ightarrow 0$,$\sin(x^2/2) > 0$,so we can remove the absolute value.$\lim _{x ightarrow 0} \frac{\sqrt{2} \sin(x^2/2)}{2 \sin^2(x/2)} = \frac{1}{\sqrt{2}} \lim _{x ightarrow 0} \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{\sin^2(x/2)}$.Using $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$:$= \frac{1}{\sqrt{2}} \cdot 1 \cdot \lim _{x ightarrow 0} \frac{x^2/2}{(\sin(x/2))^2} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot \lim _{x ightarrow 0} \left( \frac{x/2}{\sin(x/2)} ight)^2 \cdot 4 = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot 1^2 \cdot 4 = \frac{2}{\sqrt{2}} = \sqrt{2}$.

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} = $

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