微积分(A)随缘一题[1]

求：\(\lim_{x \to 0}\frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}\)

\[\begin{aligned} &\lim_{x \to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2} \\ =&\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2})\sqrt{1-2x^2}\sqrt[3]{1-\frac{9x^2}{2}}}{x^2} \\ =&\lim_{x \to 0}\frac{1-\sqrt{1-2x^2}\sqrt[3]{1-\frac{9x^2}{2}}}{x^2} +\lim_{x \to 0}\frac{\sqrt{1-2x^2}\sqrt[3]{1-\frac{9x^2}{2}}}{2}\\ =&?+\frac{1}{2} \end{aligned} \]

\[\begin{aligned} &\lim_{x \to 0} \frac{ 1-\sqrt{1-2x^2}\sqrt[3]{1-\frac{9x^2}{2}} }{x^2} \\ =&\lim_{x \to 0} \frac{ 1-(1-2x^2)(1-\frac{9x^2}{2})^\frac{2}{3} }{(1+\sqrt{1-2x^2}\sqrt[3]{1-\frac{9x^2}{2}})x^2} \\ =&\lim_{x \to 0} \frac{ 1-(1-2x^2)(1-\frac{9x^2}{2})^\frac{2}{3} }{2x^2} \\ =&\lim_{x \to 0} \frac{ 1-(1-2x^2)(1-\frac{9x^2}{2})^\frac{2}{3} }{2x^2} \end{aligned} \]

\[\begin{aligned} &\lim_{x \to 0} \frac{ 1-(1-2x^2)(1-\frac{9x^2}{2})^\frac{2}{3} }{2x^2} \\ =&\lim_{x \to 0} \frac{ 1-(1-\frac{9x^2}{2})^\frac{2}{3} }{2x^2}+\lim_{x \to 0}\frac{2x^2(1-\frac{9x^2}{2})^{\frac{2}{3}}}{2x^2} \\ =& (\lim_{x \to 0}\frac{1-(1-\frac{9x^2}{2})^{\frac{2}{3}}}{2x^2})+1 \end{aligned} \]

令 \(t=\sqrt[3]{1-\frac{9x^2}{2}},x^2=\frac{2}{9}(1-t^3)\)

\[\begin{aligned} &\lim_{x \to 0} \frac{1-(1-\frac{9x^2}{2})^{\frac{2}{3}}}{2x^2} \\ =&\lim_{t \to 1}\frac{1-t^2}{\frac{4}{9}(1-t^3)} \\ =&\frac{9}{4}\lim_{t \to 1}\frac{(1+t)(1-t)}{(1-t)(1+t+t^2)} \\ =&\frac{9}{4}\frac{2}{3}=\frac{3}{2} \end{aligned} \]

所以 \(?=\frac{3}{2}+1=\frac{5}{2}，LHS=\frac{5}{2}+\frac{1}{2}=3\)