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

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Given, $\begin{aligned} f(x) & =\frac{\cos (\sin x)-\cos x}{x^4} \ & =\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{\sin x+x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^4} \ & =\lim _{x \rightarrow 0} \frac{2 \sin (x) \sin \left(\frac{x^3}{12}\right)}{x \times \frac{x^3}{12} \times 12}\left[\text { Applying } \sin x=x-\frac{x^3}{3 !}+\ldots\right] \ & =\lim _{x \rightarrow 0} 2\left(\frac{\sin x}{x}\right) \times \frac{\sin \left(\frac{x^3}{12}\right)}{\left(\frac{x^3}{12}\right)} \times \frac{1}{12}=2 \times 1 \times 1 \times \frac{1}{12}=\frac{1}{6} \end{aligned}$

Given, \(\sin x=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2 n-1}}{(2 n-1) !}\). If the function \(f(x)\) given…

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