2021春数学分析期中(回忆版)-答案

一

1-(1)

$\underset{(x,y)\to (0,0)}{lim}\frac{e^x+e^y}{\cos x-\sin y}=2$

1-(2)

$\underset{\begin{array}{c}x\to +\mathrm{\infty },y\to +\mathrm{\infty }\end{array}}{lim}{(1+\frac{1}{x})}^{\frac{x}{x+y}}$ = $e^{\underset{x\to +\mathrm{\infty },y\to +\mathrm{\infty }}{lim}\frac{x}{x+y}\ln (1+\frac{1}{x})}$ = $e^{\underset{x+\mathrm{\infty },y+\mathrm{\infty }}{lim}\frac{1}{x+y}}=e^0=1$

1-(3)

$\frac{\partial f}{\partial x}(x,y)=2x\ln (x^2+y^2)+\frac{2x^3}{x^2+y^2}((x,y)\ne (0,0))$

$\frac{\partial f}{\partial x}(0,0)=\underset{t\to 0}{lim}2t|n|t|=0$

$\frac{\partial f}{\partial y}(x,y)=\frac{2x^{2}y}{x^2+y^2}((x,y)\ne (0,0))$

$\frac{\partial f}{\partial y}(0,0)=0$

$\frac{\partial f}{\partial x}(x,y)=\{\begin{array}{ll}2x\ln (x^2+y^2)+\frac{2x^3}{x^2+y^2},& (x,y)\ne (0,0),\\ 0,& (x,y)=(0,0)\end{array}$

$\frac{\partial f}{\partial y}(x,y)=\{\begin{array}{ll}\frac{2x^{2}y}{x^2+y^2},& (x,y)\ne (0,0),\\ 0& (x,y)=(0,0)\end{array}$

1-(4)

$\frac{\partial u}{\partial x}=-\frac{2xz}{{(x^2+y^2)}^2},\frac{\partial u}{\partial y}=-\frac{2yz}{{(x^2+y^2)}^2},\frac{\partial u}{\partial z}=\frac{1}{x^2+y^2}$

$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy+\frac{\partial u}{\partial z}dz=-\frac{2x}{{(x^2+y^2)}^2}dx-\frac{2y}{{(x^2+y^2)}^2}dy+\frac{1}{x^2+y^2}dz$

二

$\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0,M>0$, 使得当 $0<|x-a|<\delta ,y>M$ 时, $|f(x,y)-A|<\epsilon$

证明: $\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta =min(\frac{\epsilon }{6},a),M=2a\sqrt{\frac{2max(|16a^3-3|,3)}{\epsilon }}$

$0<|x-a|<\delta ,y>M$ 时, $|\frac{2x^5+3x{y}^2}{x+y^2}-3a|=|3(x-a)+\frac{x^2(2x^3-3)}{x+y^2}|$

$⩽3|x-a|+\frac{x^2|2x^3-3|}{x+y^2}<\frac{\epsilon }{2}+\frac{x^2|2x^3-3|}{y^2}<\frac{\epsilon }{2}+\frac{4a^{2}max(|16a^3-3|,3)}{y^2}$

$<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$

三

$V=\pi {\int }_{-1}^{15}(x+1)dx-\pi {\int }_{11}^{15}(x-11{)}^{2}dx=\frac{320}{3}\pi$

四

4-(1)

$E^0=E,\partial E=[0,1)\times \{0\}\cup \{(x,y)\in {\mathbb{R}}^2\mid x^2+y^2=1\},\overline{E}=\{(x,y)\in {\mathbb{R}}^2\mid x^2+y^2\le 1\}$

4-(2)

$E^0=\varphi ,\partial E=[-1,1]\times [-1,1],\overline{E}=[-1,1]\times [-1,1]$

五

证明

(1)

$F\mathrm{\setminus }G=F\cap G^C$

由于G是开集, 故 $G^C$ 是闭集, $F\mathrm{\setminus }G$ 是闭集

(2)

$G\mathrm{\setminus }F=G\cap F^C$

由于F是闭集, 故 $F^C$ 是开集, $G\mathrm{\setminus }F$ 是开集

六

6-(1)

证明 $=$ : $u=(\cos \theta ,sin\theta ),\theta \in [0,2\pi )$ 为所有方向,

$$\mathrm{则}\frac{\partial f}{\partial n}(0,0)=\underset{t\to 0}{lim}\frac{f(tn)-f(0,0)}{t}={\cos }^3\theta$$

6-(2)

不可微. 证明如下:

$\underset{(x,y)\to (0,0)}{lim}\frac{f(x,y)-f(0,0)-\frac{\partial f}{\partial x}(0,0)x-\frac{\partial f}{\partial y}(0,0)y}{\sqrt{x^2+y^2}}=\underset{(x,y)\to (0,0)}{lim}\frac{\frac{x^3}{x^2+y^2}-x}{\sqrt{x^2+y^2}}$

$=\underset{(x,y)\to 0,0)}{lim}-\frac{x{y}^2}{{(x^2+y^2)}^{\frac{3}{2}}}$

设 $g(x,y)=-\frac{x{y}^2}{{(x^2+y^2)}^{\frac{3}{2}}}$

取 $s_i=(\frac{1}{i},0),t_i=(\frac{1}{i},\frac{1}{i})$, 则 $s_i\to 0(i\to \mathrm{\infty }),t_i\to 0(i\to \mathrm{\infty }),g\left(s_i\right)=0.g\left(t_i\right)=-\frac{\sqrt{2}}{4}$

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}g(x,y)$ 不存在

$\Rightarrow f(x,y)$ 于 $(0,0)$ 处不可微

七

7-(1)

证明: $\overline{D}=\{(x,y)\mid x^2+y^2\le a\}$ 是有界闭集, 因此是紧致集

又由于f是 $\overline{D}$ 上的连续函数, 故f在 $\overline{D}$ 可上一致造续, 则 $f$ 在 $D$ 上也一致连续

7-(2)

不一致连续. 证明如下:

取 $S_i=(i,\frac{\pi }{2i}),t_i=(i,\frac{\pi }{6i})$ .则 $\Vert S_i-t_i\Vert =\frac{\pi }{3i}\to 0(i\to \mathrm{\infty }),f\left(S_i\right)=1,f\left(t_i\right)=\frac{1}{2}$, $|f\left(S_i\right)-f\left(t_i\right)|=\frac{1}{2}$

$\Rightarrow f$ 在 ${\mathbb{R}}^2$ 上不一致连续

八

证明

8-(1)

$S\subseteq B_{2a}(0)=\{x\in {\mathbb{R}}^n\mid \parallel x\parallel <2a\}$

$\Rightarrow S$ 具有界集

$S^c=\{x\in {\mathbb{R}}^n\mid \parallel x\parallel <a\mathrm{或}\parallel x\parallel >a\}$

对任意 $x\in S^C$,

若 $\parallel x\parallel <a$, 令 $r=a-\parallel x\parallel$, 对任意 $y\in B_r(x)$,

$\parallel y\parallel ⩽\parallel y-x\parallel +\parallel x\parallel <a-\parallel x\parallel +\parallel x\parallel =a$

$\Rightarrow B_r(x)\subseteq S^c$

若 $\parallel x\parallel >a$, 令 $r=\parallel x\parallel -a$, 对任意 $y\in Br(x)$

$\parallel y\parallel \ge \parallel x\parallel -\parallel x-y\parallel >\parallel x\parallel -(\parallel x\parallel -a)=a$

$\Rightarrow B_r(x)\subseteq S^C$

$\Rightarrow S^c$ 是开集

$\Rightarrow$ S是闭集

又由于S是有界集, 故S是紧致集

又由于f是连续的, 故 $f$ 在S上可以取到最大值和最小值

8-(2)

设 $f(x)=M,f(y)=m,S=\{(x,y)\mid x=a\cos \theta ,y=a\sin \theta \}$, $x=(a\cos {\theta }_1,a\sin {\theta }_1),y=(a\cos {\theta }_2,a\sin {\theta }_2)$, 不妨设 ${\theta }_1>{\theta }_2\mathrm{且}|{\theta }_1-{\theta }_2\mid <2\pi$

令 $S_1=\{(x,y)\mid x=a\cos \theta ,y=a\sin 0,\theta \in [{\theta }_2,{\theta }_1]]$,

$S_2=\{(x,y)\mid x=a\cos \theta ,y=a\sin \theta ,\theta \in [{\theta }_1,{\theta }_2+2\pi ]\},$

对 $\mathrm{\forall }x,y\in S_1,$ 设 $x=(a\cos {\phi }_1,a\sin {\phi }_1),y=(a\cos {\phi }_2,a\sin {\phi }_2)$, 不妨设 ${\phi }_2>{\phi }_1$

则存在连续映射 $x=\phi (t)=(a\cos t,a\sin t),t\in [{\phi }_1,{\phi }_2]$ 并且 $\phi \left({\phi }_1\right)=x.\phi \left({\phi }_2\right)=y$

$\Rightarrow S$ 是连通集

考虑$S_1$上的连续函数 $f$, 则 $\mathrm{\exists }{\xi }_1\in S_1$, 使得 $f\left({\xi }_1\right)=\eta \in (f(y),f(x))=(m,M)$

同理, $\mathrm{\exists }{\xi }_2\in S_2$ 使得 $f\left({\xi }_2\right)=\eta \in (m,M)$

$\Rightarrow S$ 中至少存在两个互异的点使得$f$在这两点处函数值等于$\eta$

$\mathrm{QED}$