2023年浙江大学强基计划数学试题（回忆版）#20
已知 $\left|\lim\limits_{x \rightarrow 0} \dfrac{\left.\ln \left(1+\sin ^2 x\right)-66 \sqrt[3]{2-\cos x}-1\right)}{x^4}\right|=\dfrac{q}{p}$，$p $ 和 $ q $ 是互素的正整数，则 $ p+q=$_____．
答案 $19$．
解析 根据题意，有\[\begin{split} \dfrac qp&=\left|\lim _{x \rightarrow 0} \frac{\ln \left(1+\sin ^2 x\right)-6(\sqrt[3]{2-\cos x}-1)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{\sin ^2 x-\frac{1}{2} \sin ^4 x+o\left(\sin ^4 x\right)-6\left(\sqrt[3]{1+2 \sin ^2 \frac{x}{2}}-1\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{\left(x-\frac{1}{6} x^3+o\left(x^4\right)\right)^2-\frac{1}{2} x^4+o\left(x^4\right)-6\left(1+\frac{2}{3} \sin ^2 \frac{x}{2}-\frac{4}{9} \sin ^4 \frac{x}{2}+o\left(\sin ^4 \frac{x}{2}\right)-1\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{\left(x-\frac{1}{6} x^3+o\left(x^4\right)\right)^2-\frac{1}{2} x^4+o\left(x^4\right)-6\left(\frac{2}{3} \sin ^2 \frac{x}{2}-\frac{4}{9} \sin ^4 \frac{x}{2}+o\left(\sin ^4 \frac{x}{2}\right)\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{x^2-\frac{1}{3} x^4+o\left(x^4\right)-\frac{1}{2} x^4+o\left(x^4\right)-6\left(\frac{2}{3}\left(\frac{1}{2} x-\frac{1}{48} x^3+o\left(\frac{1}{16} x^4\right)\right)^2-\frac{1}{36} x^4+o\left(x^4\right)\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{x^2-\frac{5}{6} x^4+o\left(x^4\right)-6\left(\frac{1}{6} x^2-\frac{1}{72} x^4+o\left(x^4\right)-\frac{1}{36} x^4+o\left(x^4\right)\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{x^2-\frac{5}{6} x^4+o\left(x^4\right)-6\left(\frac{1}{6} x^2-\frac{1}{24} x^4+o\left(x^4\right)\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{x^2-\frac{5}{6} x^4+o\left(x^4\right)-x^2+\frac{1}{4} x^4+o\left(x^4\right)}{x^4}\right| \\ & =\left|\lim _{x \rightarrow 0} \frac{-\frac{7}{12} x^4+o\left(x^4\right)}{x^4}\right| \\ & =\left|-\frac{7}{12}+\lim _{x \rightarrow 0} \frac{o\left(x^4\right)}{x^4}\right| \\ & =\left|-\frac{7}{12}\right| \\ & =\frac{7}{12} ,\end{split}\]于是 $p+q=7+12=19$．