The value of lim _{x rightarrow 0} frac{cos ( | vedclass

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Question

(B) Let $L = \lim _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1 - \frac{	heta^2}{2!}

Vedclass · Accepted Answer

$1/6$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!} - O(	heta^6)$,we have:$\sin x = x - \frac{x^3}{6} + O(x^5)$$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24}$.Substituting these into the limit:$L = \lim _{x ightarrow 0} \frac{(1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24})}{x^4}$.$L = \lim _{x ightarrow 0} \frac{\frac{5x^4}{24} - \frac{x^4}{24}}{x^4} = \lim _{x ightarrow 0} \frac{4x^4/24}{x^4} = \frac{4}{24} = \frac{1}{6}$.

The value of $\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to

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