$\lim _{x \rightarrow \infty} x^2\left(2-x \sin \frac{1}{x}-\cos \frac{1}{x}\right)=$
【答案】 $\frac{2}{3}$

【解析】 $\lim _{x \rightarrow \infty} x^2\left(2-x \sin \frac{1}{x}-\cos \frac{1}{x}\right)=x^2\left[2-x\left(\frac{1}{x}-\frac{1}{6 x^3}+o\left(\frac{1}{x^3}\right)\right)-\left(1-\frac{1}{2 x^2}+o\left(\frac{1}{x^2}\right)\right)\right]$ $=x^2\left[\frac{1}{6 x^2}+\frac{1}{2 x^2}+o\left(\frac{1}{x^2}\right)\right]=\frac{2}{3}$.
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