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

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Question

(C) Given the limit $ \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} $. Substituting $ x = 0 $,we get the indeterminate form $ \left(\frac{0}{0}\right

Vedclass · Accepted Answer

$ \frac{1}{2} $

Vedclass · Answer

(C) Given the limit $ \lim _{x ightarrow 0} \frac{1-\cos x}{x^{2}} $. Substituting $ x = 0 $,we get the indeterminate form $ \left(\frac{0}{0}ight) $. Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $ x $: $ \lim _{x ightarrow 0} \frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x^{2})} = \lim _{x ightarrow 0} \frac{\sin x}{2x} $. Using the standard limit $ \lim _{x ightarrow 0} \frac{\sin x}{x} = 1 $,we get: $ \frac{1}{2} 	imes \lim _{x ightarrow 0} \frac{\sin x}{x} = \frac{1}{2} 	imes 1 = \frac{1}{2} $. Thus,the value of the limit is $ \frac{1}{2} $.

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

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