limlimits _{x rightarrow 0} frac{cos (sin x)- | vedclass

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Question

(C) We evaluate the limit: $\lim\limits _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(C) We evaluate the limit: $\lim\limits _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!} - \dots$:$\sin x = x - \frac{x^3}{6} + O(x^5)$$\cos(\sin x) = 1 - \frac{(x - \frac{x^3}{6})^2}{2} + \frac{(x)^4}{24} + O(x^6) = 1 - \frac{x^2 - \frac{x^4}{3}}{2} + \frac{x^4}{24} + O(x^6) = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} + O(x^6) = 1 - \frac{x^2}{2} + \frac{5x^4}{24} + O(x^6)$.$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$.Subtracting these:$\cos(\sin x) - \cos x = (1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24}) + O(x^6) = \frac{4x^4}{24} = \frac{x^4}{6}$.Therefore,$\lim\limits _{x ightarrow 0} \frac{\frac{x^4}{6}}{x^4} = \frac{1}{6}$.

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

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