微积分(A)随缘一题[23]

求：\(\lim_{x \to 0^+}\frac{x \ln \sin x-\sin x \ln x}{x^3\ln x}\)

\[\lim_{x \to 0^+}\frac{x \ln \sin x-\sin x \ln x}{x^3\ln x} \\ =\lim_{x \to 0^+}\frac{x \ln \sin x-x \ln x +x \ln x -\sin x \ln x}{x^3\ln x} \\ =\lim_{x \to 0^+} \left( \frac{\ln \frac{\sin x}{x}}{x^2\ln x}+\frac{x-\sin x}{x^3} \right) \\ =\left(\lim_{x \to 0^+} \frac{\ln (1+\frac{\sin x}{x}-1)}{x^2\ln x}\right)+\frac{1}{6} \\ =\left(\lim_{x \to 0^+} \frac{\frac{\sin x}{x}-1}{x^2\ln x}\right)+\frac{1}{6} \\ =\left(\lim_{x \to 0^+} \frac{-\frac{x^3}{6}}{x^3\ln x}\right)+\frac{1}{6}=\frac{1}{6} \]

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由拉格朗日中值定理可得：

\[\lim_{x \to 0^+}\frac{x \ln \sin x-\sin x \ln x}{x^3\ln x} \\ =\lim_{x \to 0^+} \frac{\sin x}{x^2\ln x}\left( \frac{\ln \sin x}{\sin x} -\frac{\ln x}{x} \right) \\ =\lim_{x \to 0^+,\sin x < \zeta < x} \frac{\sin x}{x^2\ln x}(\sin x-x) \frac{1-\zeta}{\zeta^2} \\ \]

又由夹逼定理可得：

\[\lim_{x \to 0^+,\sin x < \zeta < x} \frac{\sin x}{x^2\ln x}(\sin x-x) \frac{1-\ln \zeta}{\zeta^2} \\ \ge \lim_{x \to 0^+,\sin x < \zeta < x} \frac{\sin x}{x^2\ln x}(\sin x-x) \frac{1-\ln x}{x^2} \\ = \lim_{x \to 0^+} \frac{x}{x^2\ln x}\left(-\frac{x^3}{6} \right) \frac{1-\ln x}{x^2} \\ = \lim_{x \to 0^+} \frac{x}{x^2\ln x}\left(-\frac{x^3}{6} \right) \frac{1-\ln x}{x^2} \\ =\frac{1}{6}\lim_{x \to 0^+}\frac{\ln x-1}{\ln x} = \frac{1}{6} \]

\[\lim_{x \to 0^+,\sin x < \zeta < x} \frac{\sin x}{x^2\ln x}(\sin x-x) \frac{1-\ln \zeta}{\zeta^2} \\ \le \lim_{x \to 0^+} \frac{\sin x}{x^2\ln x}(\sin x-x)\frac{1-\ln \sin x}{\sin^2x} \\ =\lim_{x\to0^+}\frac{-\frac{x^3}{6}(1-\ln\sin x)}{x^3\ln x} \\ =\frac{1}{6}\lim_{x \to 0^+} \frac{\ln \sin x-1}{\ln x}=\frac{1}{6}\lim_{x \to 0^+} \frac{\cot x}{\frac{1}{x}}=\frac{1}{6} \]

所以：\(\lim_{x \to 0^+}\frac{x\ln \sin x-\sin x\ln x}{x^3\ln x}=\frac{1}{6}\)