USTC2025秋数学分析B1midterm

一、计算下列各题.

(1)用极限定义计算$\lim\limits_{ x \to 1 }x^{2}$.
解：$\forall\varepsilon>0$,取$\delta=\min\{1,\frac{1}{3}\varepsilon\}$，则当$0<|x-1|<\delta$时,$|x^{2}-1|=|x-1|\cdot|x+1|<3\delta\leq\varepsilon$.故$\lim\limits_{ x \to 1 }x^{2}=1$.
(2)求$\lim\limits_{ n \to \infty }\left(1 +\frac{ 2}{n}+\frac{3}{n^{2}} \right)^n$.
解：原式$=e^{ \lim\limits_{ n \to \infty }n\ln(1 +\frac{ 2}{n}+\frac{3}{n^{2}} ) }=e^{ \lim\limits_{ n \to \infty }n(\frac{2}{n}+\frac{3}{n^{2}}) }=e^{ 2 }$.
(3)$\lim\limits_{ n \to \infty }(2^{n}+3^{n}+5^{n})^{\frac{1}{n}}$.
解：原式$=\lim\limits_{ n \to \infty }5\sqrt[n]{\left( \frac{2}{5} \right)^{n}+\left( \frac{3}{5} \right)^{n}+1}= 5$.
(4)$\lim\limits_{ x \to 0 }\left( \frac{1}{x^{2}}-\frac{1}{\sin ^{2}x} \right)$.
解：原式$=\lim\limits_{ x \to 0 }\frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}=\lim\limits_{ x \to 0 }\frac{\left( x-\frac{1}{6}x^{3}+o(x^{3}) \right)^{2}-x^{2}}{x^{4}}=\lim\limits_{ x \to 0 }\frac{-\frac{1}{3}x^{4}+o(x^{4})}{x^{4}}=-\frac{1}{3}$.
(5)$\lim\limits_{ x \to 1 }\frac{x-x^{x}}{1-x+\ln x}$.
解：原式$=\lim\limits_{ x \to 1 }\frac{1-x^{x}(\ln x+1)}{-1+\frac{1}{x}}=\lim\limits_{ x \to 1 }\frac{x^{x} \cdot \frac{1}{x}+(\ln x+1)^{2} \cdot x^{x}}{\frac{1}{x^{2}}}=2$.
(6)求参数曲线$\begin{cases}x=t^{2} \\ y=2t^{3}\end{cases}$上点$\left( \frac{1}{4},\frac{1}{4} \right)$处切线的直角坐标方程.
解：即$t=\frac{1}{2}$处,切线斜率$\frac{\mathrm{d}y}{\mathrm{d}t}|_{t=\frac{1}{2}}=\frac{3}{2}$.故切线方程为$y=\frac{3}{2}x-\frac{1}{8}$.
(7)求$(x^{2}e^{ x })^{(20)}$.
解：原式$=e^{ x }(C_{20}^{0}x^{2}+C_{20}^{1}\cdot 2x+C_{20}^{2}\cdot 2)=e^{ x }(x^{2}+40x+380)$.
(8)$f(x)=\begin{cases} \frac{\pi}{2} +\ln(1+\sin x), & x\leq 0 \\ 2\arctan(e^{ x }), & x>0\end{cases},$求$f'(x)$.
解：$f'_{-}(0)=\lim\limits_{ x \to 0^{-} }\frac{\ln(1+\sin x)}{x}=1;f'_{+}(0)=\lim\limits_{ x \to 0^{+} }\frac{2\arctan(e^{ x })- \frac{\pi}{2}}{x}=1$,故$f'(0)=1$.
则$f'(x)=\begin{cases}\frac{\cos x}{1+\sin x}, & x\leq 0 \\ \frac{2e^{ x }}{1+e^{ 2x }}, & x>0\end{cases}$.

二、求$f(x)=\sqrt[3]{ \sin(x^{3}) }$的带Peano余项的7阶Maclaurin公式，并求$f^{(7)}(0)$.

解：$x\to 0$时,$\sin(x^{3})=x^{3}-\frac{x^{9}}{6}+o(x^{9})$,
$f(x)=(x^{3})^{\frac{1}{3}}\cdot\left( 1-\frac{x^{6}}{6}+o(x^{6}) \right)^{\frac{1}{3}}=x\cdot\left( 1-\frac{x^{6}}{18}+o(x^{6}) \right)=x-\frac{x^{7}}{18}+o(x^{7})$.
$f^{(7)}(0)=7!\cdot\left( -\frac{1}{18} \right)=-280$.

三、设函数$f(x)=\begin{cases}x^{2}\sin \frac{1}{x}, & x<0 \\ ax+b, & x\geq 0\end{cases}.$

(1)确定$a,b,$使得$f(x)$在$\mathbb{R}$上可导；
(2)当$f(x)$在$\mathbb{R}$上可导时，分析其导函数的连续性，如有间断点指明其类型.

解：(1)由于$f(x)$在$x>0$和$x<0$时均为初等函数，都是可导的，故只需讨论$x=0$处的可导性.由$f(x)$在$x=0$处连续得$f(0)=b=\lim\limits_{ x \to 0^{-} }f(x)=\lim\limits_{ x \to 0^{-} }x^{2}\sin \frac{1}{x}=0$;
再由$f(x)$在$x=0$处可导知$f'_{+}(0)=f'_{-}(0)$,即

$$\lim\limits_{ x \to 0^{+} } \frac{f(x)-f(0)}{x-0}=a=\lim\limits_{ x \to 0^{-} }\frac{x^{2}\sin \frac{1}{x}-0}{x-0} =\lim\limits_{ x \to 0^{-} } x\sin \frac{1}{x}=0.$$

故当$a=0,b=0$时,$f(x)$在$\mathbb{R}$上可导.
(2)由(1)知$f'(0)=0$,故$f'(x)=\begin{cases}2x\sin \frac{1}{x}-\cos \frac{1}{x}, & x<0 \\ 0, & x\geq 0\end{cases}$.
由于$\lim\limits_{ x \to 0^{-} }f'(x)$不存在，故$x=0$是$f'(x)$的第二类间断点，$f'(x)$在其它点处连续。

四、已知函数$f(x)=2x^{3}-3x^{2}(x \in [-2,2])$.

(1)求$f(x)$的单调区间、凹凸区间；
(2)求$f(x)$的最值.

解：(1)$f'(x)=6x(x-1)$,故$f(x)$在$[-2,0]$和$[1,2]$上单调递增,在$[0,1]$上单调递减.
又$f''(x)=6(2x-1)$,故$f(x)$在$\left[ -2, \frac{1}{2} \right]$上凹,在$\left[ \frac{1}{2},2 \right]$上凸.
(2)由(1)中分析知$x=0$为极大值点,$x=1$为极小值点;最值在$\{f(-2)=-28,f(0)=0,f(1)=-1,f(2)=4\}$中取到，故函数最大值为$4$,最小值为$-28$.

五、函数$f(x)$在$[-1,1]$上连续,$(-1,1)$内可导,且$f(-1)=f(1)$.证明：存在$\xi \in (-1,1)$使得$\xi f(\xi)+f'(\xi)=0$.

证明：令$F(x)=f(x)\cdot e^{ \frac{1}{2}x^{2} }$,则$F(-1)=F(1)$.
由罗尔定理,存在$\xi \in (-1,1)$使得$F'(\xi)=0$,代入即证.

六、设$f(x)$在$[0,1]$上连续,在$(0,1)$内可导,$f(0)=f(1)=0$且$f(x)$不恒为0.记$M=\max\limits_{0\leq x\leq 1 }{|f(x)|}$.

(1)证明：存在$\xi \in (0,1)$,使得$|f'(\xi)|\geq 2M$;
(2)进一步证明,存在$\eta \in (0,1)$,使得$|f'(\eta)|> 2M$.

证明：(1)设$|f(c)|=M,c\in(0,1)$.若$c\in(0, \frac{1}{2}]$,则$\exists \xi \in(0,c) s.t. |f'(\xi)|=|\frac{f(c)-0}{c-0}|\geq 2M$;若$c\in\left( \frac{1}{2},1 \right)$,则$\exists \xi \in(c,1)s.t.|f'(\xi)|=|\frac{0-f(c)}{1-c}|>2M$.
(2)由(1),当$c\ne \frac{1}{2}$时$\xi$满足题意.当$c= \frac{1}{2}$时,设$f\left( \frac{1}{2} \right)=M$,假设$|f'(x)|\leq 2M,x\in(0,1)$,则$M=f\left( \frac{1}{2} \right)-f(0)=\int_{0}^{\frac{1}{2}}f'(x)dx\leq \frac{1}{2}\cdot 2M$中等号取等,故$f'(x)=2M,x\in\left( 0, \frac{1}{2} \right)$,同理有$f'(x)=-2M,x\in\left( \frac{1}{2},1 \right)$.此时$f'_{+}\left( \frac{1}{2} \right)\ne f'_{-}\left( \frac{1}{2} \right)$,与$f(x)$在$(0,1)$内可导矛盾,故假设不成立.
$f\left( \frac{1}{2} \right)=-M$时同理.

七、设$f(x)$在$[0,+\infty)$上二阶可导,$\lim\limits_{ x \to +\infty }f(x)$和$\lim\limits_{ x \to +\infty }f''(x)$存在且有限.证明：

(1)$\lim\limits_{ x \to +\infty }f''(x)=0$.
(2)$\lim\limits_{ x \to +\infty }f'(x)=0$.

证明：(1)设$\lim\limits_{ x \to +\infty }f''(x)=A$.假设$A>0$,则$\exists x_{1}>0$,当$x>x_{1}$时,$f''(x)> \frac{A}{2}$,记$f'(x_{1})=t$,若$t\leq \frac{A}{2}$,取$x_{2}-x_{1}=\frac{\frac{A}{2}-t}{\frac{A}{2}}$,即$x_{2}=x_{1}+1- \frac{2t}{A}$;若$t> \frac{A}{2}$,取$x_{2}=x_{1}$.则当$x>x_{2}$时,$f'(x)> \frac{A}{2}$,此时不可能有$\lim\limits_{ x \to +\infty }f(x)$有限,因为$\forall G>0$,当$x>x_{3}=2\frac{G-f(x_{2})}{A}+x_{2}$时,$f(x)>G$,故假设不成立.$A<0$时同理，所以$A=0$.
(2)对$\forall x,h>0$,有$Taylor$展开

$$f(x+h)=f(x)+f'(x)h+ \frac{f''(\xi)}{2}h^{2}.$$

分别取$h=1$和$h=2$,有

$$\begin{cases} f(x+1)=f(x)+f'(x)+\frac{1}{2}f''(\xi_{1}(x)) \\ f(x+2)=f(x)+2f'(x)+2f''(\xi_{2}(x)) \end{cases}$$

两式相减可得

$$f'(x)=f(x+2)-f(x+1)-2f''(\xi_{2}(x))+\frac{1}{2}f''(\xi_{1}(x))$$

令$x\to +\infty$,此时$x_{i}(x)\to +\infty(i = 1,2)$,有

$$\lim\limits_{ x \to +\infty } f'(x)=\lim\limits_{ x \to +\infty } f(x+2)-\lim\limits_{ x \to +\infty } f(x+1)-2\cdot 0+0=0$$