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(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Using the Taylor series expansion for $\sin x = x -

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$\frac{1}{6}$

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(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ and $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$,we substitute $\sin x$ into the cosine series:$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.Now,$f(x) = \frac{(1 - x^2/2 + 5x^4/24) - (1 - x^2/2 + x^4/24)}{x^4} = \frac{4x^4/24}{x^4} = \frac{1}{6}$.Thus,$k = \lim_{x 	o 0} f(x) = \frac{1}{6}$.

Given,$\sin x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}$. If the function $f(x)$ given by $f(x) = \frac{\cos(\sin x) - \cos x}{x^4}$ for $x \neq 0$ and $f(0) = k$ is continuous at $x = 0$,then $k =$

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