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Let $f(x)=\frac{1}{3} x \sin x-(1-\cos x) .$ The smallest positive interger $k$ such that $\lim _{x \rightarrow 0} \frac{f(x)}{x^{k}} \neq 0$ is
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The correct answers are:
2
Hint:
It $\left.\frac{x \sin x-3(1-\cos x)}{3 x^{k}}=\frac{1}{3} \operatorname{lt}_{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)_{x \rightarrow 0} \mid t_{x \rightarrow 0} \frac{2 x \cos x / 2-6 \sin x / 2}{2 x^{k-1}}\right)$
$k-1=1 \Rightarrow k=2$
It $\left.\frac{x \sin x-3(1-\cos x)}{3 x^{k}}=\frac{1}{3} \operatorname{lt}_{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)_{x \rightarrow 0} \mid t_{x \rightarrow 0} \frac{2 x \cos x / 2-6 \sin x / 2}{2 x^{k-1}}\right)$
$k-1=1 \Rightarrow k=2$
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