Let f(x) = frac{1}{3} x sin x - (1 - cos x). | vedclass

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(C) We are given $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$.Using the Taylor series expansions for $\sin x$ and $\cos x$ near $x = 0$:$\sin x = x -

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$2$

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(C) We are given $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$.Using the Taylor series expansions for $\sin x$ and $\cos x$ near $x = 0$:$\sin x = x - \frac{x^3}{6} + O(x^5)$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$Substituting these into $f(x)$:$f(x) = \frac{1}{3} x \left(x - \frac{x^3}{6} + O(x^5)ight) - \left(1 - (1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6))ight)$$f(x) = \frac{1}{3} x^2 - \frac{x^4}{18} - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$f(x) = (\frac{1}{3} - \frac{1}{2}) x^2 + (\frac{1}{24} - \frac{1}{18}) x^4 + O(x^6)$$f(x) = -\frac{1}{6} x^2 - \frac{1}{72} x^4 + O(x^6)$Now,consider the limit $\lim_{x ightarrow 0} \frac{f(x)}{x^k} = \lim_{x ightarrow 0} \frac{-\frac{1}{6} x^2 - \frac{1}{72} x^4 + O(x^6)}{x^k}$.If $k If $k = 2$,the limit is $-\frac{1}{6} 
eq 0$.If $k > 2$,the limit is $0$.Thus,the smallest positive integer $k$ such that the limit is non-zero is $k = 2$.

Let $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$. The smallest positive integer $k$ such that $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is

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