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+ B. 低阶无穷小 + C. 等价无穷小 + D. 高阶无穷小 +4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$,则当$n\rightarrow\infty$时,$x_n$是 + A. 无穷大量 + B. 无穷小量 + C. 有界变量 + D. 无界变量 +5. 下列四个级数,**发散**的是 + A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ + B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ + C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ + D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ +#### 二、填空题(共5小题,每小题3分,共15分) +6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. +7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$,则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. +8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛,则常数$a$的取值范围是_____. +9. 已知函数$f(x)$在$x=1$处可导,且$f(1)=0,f'(1)=2$,则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. +10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切,则常数$c$的值为\_\_\_\_\_\_. +#### 三、解答与证明题(11~19小题,共70分) +11. (6分)求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. +12. (6分)求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. +13. (6分)已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导,求常数$a,b$的值. +14. (6分)设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数,求曲线$y=y(x)$在$x=0$对应点处的切线方程. +15. (8分)设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$,求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$,$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. +16. (8分)设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数,证明对任一实数$x_0$,方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. +17. (10分)求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. +18. (10分)百米跑道上正在举行百米赛跑,摄影师站在50米处为4号跑道夺冠种子选手A录像,摄像头始终对准选手A,摄影师距离4号跑道5米,若选手A以10m/s的速度从摄影师正前方经过,问此时摄影镜头的角速度是多少? ![[期中试卷-18.png]] +19. (10分)设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ + (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在,并求其极限值;(5分) + (2) 证明:对于任意实数$p$,级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. (5分) \ No newline at end of file