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

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(C) We know that the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$. Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$,th

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) We know that the limit is of the form $\frac{0}{0}$ as $x ightarrow 0$. Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$,the expression becomes: $\lim _{x ightarrow 0} \frac{2 \sin^2(\frac{x}{2})}{x^2}$ $= 2 \lim _{x ightarrow 0} \left( \frac{\sin(\frac{x}{2})}{x} ight)^2$ $= 2 \lim _{x ightarrow 0} \left( \frac{\sin(\frac{x}{2})}{2 \cdot \frac{x}{2}} ight)^2$ $= 2 \cdot \frac{1}{4} \lim _{x ightarrow 0} \left( \frac{\sin(\frac{x}{2})}{\frac{x}{2}} ight)^2$ Since $\lim _{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we get: $= 2 \cdot \frac{1}{4} \cdot (1)^2 = \frac{1}{2}$.

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

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