浙江大学 2023年强基第19题

19，解析：当 $x \rightarrow 0$ 时，考虑等价无穷小 $$ \begin{gathered} \ln \left(1+\sin ^{2} x\right)=\sin ^{2} x-\frac{1}{2} \sin ^{4} x+o\left(\sin ^{4} x\right)=\left(x-\frac{1}{6} x^{3}+o\left(x^{4}\right)\right)^{2}-\frac{1}{2}\left(x^{4}+o\left(x^{4}\right)\right)+o\left(x^{4}\right) \# \\ =x^{2}-\frac{5}{6} x^{4}+o\left(x^{4}\right) \\ \sqrt[3]{2-\cos x}-1=\sqrt[3]{1+2 \sin ^{2} \frac{x}{2}}-1=\frac{2}{3} \sin ^{2} \frac{x}{2}-\frac{4}{9} \sin ^{4} \frac{x}{2}=\frac{2}{3}\left(\frac{x}{2}-\frac{1}{6}\left(\frac{x}{2}\right)^{3}+o\left(x^{4}\right)\right)^{2}-\frac{4}{9}\left(\left(\frac{x}{2}\right)^{4}+o\left(x^{4}\right)\right) \\ =\frac{1}{6} x^{2}-\frac{1}{24} x^{4}+o\left(x^{4}\right) \end{gathered} $$ 故原式 $\displaystyle =\lim _{x \rightarrow 0}\left|\frac{\left(x^{2} \frac{5}{6} x^{4}+o\left(x^{4}\right)\right)-6\left(\frac{1}{6} x^{2} \frac{1}{24} x^{4}+o\left(x^{4}\right)\right)}{x^{4}}\right|=-\frac{7}{12}$ ．因此 $p=12, q=7$ ，则 $p+q=19$ 。