来源 [尊重原有作者劳动成果]
一、 计算题
1：解：由于
$underset{x o {{0}^{+}}}{mathop{lim }}\,frac{ln (1-x)}{-x}=underset{x o {{0}^{+}}}{mathop{lim }}\,frac{1}{1-x}=1,underset{x o {{0}^{+}}}{mathop{lim }}\,frac{{{e}^{{{x}^{2}}}}-1}{{{x}^{2}}}=underset{x o {{0}^{+}}}{mathop{lim }}\,frac{2x{{e}^{{{x}^{2}}}}}{2x}=1$
于是
$underset{x o {{0}^{+}}}{mathop{lim }}\,frac{{{ln }^{3}}(1-x)sin [sin (ln frac{1}{x})]}{{{e}^{{{x}^{2}}}}-1}=underset{x o {{0}^{+}}}{mathop{lim }}\,frac{{{(-x)}^{3}}}{{{x}^{2}}}sin [sin (ln frac{1}{x})]=0$
2：不妨设${{t}_{1}}={{x}^{x}},{{t}_{2}}={{x}^{ln x}}$，于是
$ln {{t}_{1}}=xln x,ln {{t}_{2}}={{(ln x)}^{2}}$
则$frac{t_{1}^{}}{{{t}_{1}}}=ln x+1,frac{t_{2}^{}}{{{t}_{2}}}=frac{2ln x}{x}$
即$t_{1}^{}=(ln x+1){{x}^{x}},t_{2}^{}=frac{2ln x}{2}{{x}^{ln x}}$
于是$y=frac{2ln x}{x}{{x}^{ln x}}+(ln x+1){{x}^{x}}$
3：解：令$t={{x}^{4}}$，于是
$int_{0}^{+infty }{{{e}^{-{{x}^{4}}}}}dxcdot int_{0}^{+infty }{{{x}^{2}}{{e}^{-{{x}^{4}}}}dx=frac{1}{16}}int_{0}^{+infty }{{{t}^{-frac{3}{4}}}}{{e}^{-t}}cdot int_{0}^{+infty }{{{t}^{-frac{1}{4}}}}{{e}^{-t}}dt$
$=frac{1}{16}Gamma (frac{1}{4})Gamma (frac{3}{4})overset{}{mathop{=}}\,frac{1}{16}cdot frac{pi }{sin frac{pi }{4}}=frac{pi }{8sqrt{2}}$
此处介绍余元公式：
$forall pin (0,1)$，有$Gamma (p)Gamma (1-p)=frac{pi }{sin ppi }$
4：解：
$F(x)={{f}_{x}}[f(x,x),f(x,x)]cdot [{{f}_{x}}(x,x)+{{f}_{y}}(x,x)]+{{f}_{y}}[f(x,x),f(x,x)]cdot [{{f}_{x}}(x,x)+{{f}_{y}}(x,x)]$
于是$F(1)={{(a+b)}^{2}}$
5：解：不妨设$x=rcos heta ,y=rsin heta ,0le rle 1,0le heta le 2pi $
于是
$iint_{D}{({{x}^{3}}+{{y}^{3}}){{e}^{{{x}^{2}}+{{y}^{2}}}}}dxdy=int_{0}^{2pi }{d heta int_{0}^{1}{{{r}^{4}}}({{sin }^{3}} heta +{{cos }^{3}} heta ){{e}^{{{r}^{2}}}}dr=}int_{0}^{2pi }{({{sin }^{3}} heta +{{cos }^{3}} heta )}d heta cdot int_{0}^{1}{{{r}^{4}}{{e}^{r}}dr}$
而
$int_{0}^{2pi }{({{sin }^{3}} heta +{{cos }^{3}} heta )}d heta =int_{0}^{2pi }{(sin heta +cos heta )(1+sin heta cos heta )d heta }$
$=int_{0}^{2pi }{(sin heta +cos heta )d heta +int_{0}^{2pi }{{{sin }^{2}}}} heta cos heta d heta +int_{0}^{2pi }{sin heta {{cos }^{2}}} heta d heta $
$=int_{0}^{2pi }{{{sin }^{2}}} heta d(sin heta )-int_{0}^{2pi }{{{cos }^{2}}} heta d(cos heta )$
$=frac{1}{3}{{sin }^{3}} heta |_{0}^{2pi }-frac{1}{3}{{cos }^{3}} heta |_{0}^{2pi }$
$=0$
则$iint_{D}{({{x}^{3}}+{{y}^{3}}){{e}^{{{x}^{2}}+{{y}^{2}}}}}dxdy=0$
6：解：记${{L}_{1}}:0le xle 2,y=0$，$D$是由$L$和${{L}_{1}}$所围成的封闭曲线
而
$int_{{{L}_{1}}}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=int_{0}^{2}{{{e}^{x}}}}dx={{e}^{x}}|_{0}^{2}={{e}^{2}}-1$
另一方面，由格林公式可知：
$int_{L+{{L}_{1}}}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=iint_{D}{({{e}^{x}}sin y-{{e}^{x}}sin y)dxdy=0}}$
于是
$int_{L}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=-}int_{{{L}_{1}}}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=}1-{{e}^{2}}$
二、
（1）证明：由于$sqrt{x}f(x)$在$(0,+infty )$上有界，则存在$M0$，使得
$left| sqrt{x}f(x) ight|le M$
对任意的$varepsilon 0$，对任意的${{x}_{1}},{{x}_{2}}in (0,+infty )$，不妨设${{x}_{1}}{{x}_{2}}$
于是
$left| f({{x}_{1}})-f({{x}_{2}}) ight|=left| int_{{{x}_{1}}}^{{{x}_{2}}}{f(t)dt} ight|le int_{{{x}_{1}}}^{{{x}_{2}}}{left| f(t) ight|}dt=int_{{{x}_{1}}}^{{{x}_{2}}}{left| sqrt{t}f(t) ight|}cdot frac{1}{sqrt{t}}dt$
$le 2Mint_{{{x}_{1}}}^{{{x}_{2}}}{frac{1}{sqrt{t}}}dt=2M(sqrt{{{x}_{2}}}-sqrt{{{x}_{1}}}le 2Msqrt{{{x}_{2}}-{{x}_{1}}}$
令$delta =frac{{{varepsilon }^{2}}}{4{{M}^{2}}}$，当$left| {{x}_{1}}-{{x}_{2}} ight|delta $时，$left| f({{x}_{1}})-f({{x}_{2}}) ight|varepsilon $
即$f(x)$在$(0,+infty )$上一致收敛
（或利用柯西中值定理，令$g(x)=sqrt{x}$）
（2）由（1）可知，对任意的$varepsilon 0$，令${{x}_{1}} o {{0}^{+}}$，当$0xdelta $时，有
$left| f(x)-f(0) ight|varepsilon $
由极限定义可知：
$underset{x o {{0}^{+}}}{mathop{lim }}\,f(x)=f({{0}^{+}})$存在
（3）能，理由如下：
证明：由于$underset{x o {{0}^{+}}}{mathop{lim }}\,sqrt{x}f(x)$存在，则存在$M0$，存在${{X}_{1}}0$，当$x{{X}_{1}}$时，$left| sqrt{x}f(x) ight|le M$
由（1）可知：$f(x)$在$(0,{{X}_{1}})$上一致收敛
而$underset{x o +infty }{mathop{lim }}\,sqrt{x}f(x)$存在，同理可证$f(x)$在$({{X}_{2}},+infty )$上一致收敛
而$f(x)$在$[{{X}_{1}},{{X}_{2}}]$上连续，则$f(x)$在$[{{X}_{1}},{{X}_{2}}]$上一致连续
于是$f(x)$在$(0,+infty )$上一致连续
三、
（1）证明：对任意的$xin (a,+infty )$，则$f(x)$在$[x,x+1]$上可导
记$underset{x o +infty }{mathop{lim }}\,f(x)=A$
$underset{x o +infty }{mathop{lim }}\,f(x+1)=A$，则
$underset{x o +infty }{mathop{lim }}\,[f(x+1)-f(x)]=0$
由拉格朗日中值定理可知：存在$xxi x+1$，使得
$f(x+1)-f(x)=f(xi )[(x+1)-x]=f(xi )$
且当$x o +infty $时，$xi o +infty $，于是
$underset{x o +infty }{mathop{lim }}\,f(xi )=underset{x o +infty }{mathop{lim }}\,[f(x+1)-f(x)]=1$
由归结原则可知：
$underset{x o +infty }{mathop{lim }}\,f(xi )=underset{x o +infty }{mathop{lim }}\,f(x)=0$
（2）可以推出，理由如下：
证明：应用$Taylor$公式可知：对任意的$m=1,2,3,4$
$f(x+m)=f(x)+mf(x)+frac{{{m}^{2}}}{2!}f(x)+frac{{{m}^{3}}}{3!}f(x)+frac{{{m}^{4}}}{4!}f({{xi }_{m}}(x)),x{{xi }_{m}}(x)x+m$
由于$f(x),f(x),f(x)$均可用$f(x+m)-f(x)$及$f({{xi }_{m}}(x))$表达
而$underset{x o +infty }{mathop{lim }}\,[f(x+m)-f(x)]$与$underset{x o +infty }{mathop{lim }}\,f({{xi }_{m}}(x))=underset{x o +infty }{mathop{lim }}\,f(x)$均存在有限
故$underset{x o +infty }{mathop{lim }}\,{{f}^{(k)}}(x)$均存在有限，不妨记
$underset{x o +infty }{mathop{lim }}\,{{f}^{(k)}}(x)={{A}_{k}},k=0,1,2,3,4$
在上面的$Taylor$公式中，令$x o +infty $，则${{xi }_{m}}(x) o +infty $，于是有
$0=underset{x o +infty }{mathop{lim }}\,[f(x+m)-f(x)]=munderset{x o +infty }{mathop{lim }}\,f(x)+frac{{{m}^{2}}}{2!}underset{x o +infty }{mathop{lim }}\,f(x)+frac{{{m}^{3}}}{3!}underset{x o +infty }{mathop{lim }}\,f(x)+frac{{{m}^{4}}}{4!}underset{x o +infty }{mathop{lim }}\,f({{xi }_{m}}(x))$
即$0={{A}_{0}}-{{A}_{0}}=m{{A}_{1}}+frac{{{m}^{2}}}{2!}{{A}_{2}}+frac{{{m}^{3}}}{3!}{{A}_{3}}+frac{{{m}^{4}}}{4!}{{A}^{4}},m=1,2,3,4$
列举出来为
$left{egin{array}{ll} {A_1} + frac{1}{{2!}}{A_2} + frac{1}{{3!}}{A_3} + frac{1}{{4!}}{A_4} = 0, \ 2{A_1} + frac{{{2^2}}}{{2!}}{A_2} + frac{{{2^3}}}{{3!}}{A_3} + frac{{{2^4}}}{{4!}}{A_4} = 0, \ 3{A_1} + frac{{{3^2}}}{{2!}}{A_2} + frac{{{3^3}}}{{3!}}{A_3} + frac{{{3^4}}}{{4!}}{A_4} = 0, \ 4{A_1} + frac{{{4^2}}}{{2!}}{A_2} + frac{{{4^3}}}{{3!}}{A_3} + frac{{{4^4}}}{{4!}}{A_4} = 0, end{array} ight.$ $Rightarrowleft{egin{array}{ll} {A_1} + frac{1}{{2!}}{A_2} + frac{1}{{3!}}{A_3} + frac{1}{{4!}}{A_4} = 0, \ {A_1} + frac{2}{{2!}}{A_2} + frac{{{2^2}}}{{3!}}{A_3} + frac{{{2^3}}}{{4!}}{A_4} = 0, \ {A_1} + frac{3}{{2!}}{A_2} + frac{{{3^2}}}{{3!}}{A_3} + frac{{{3^3}}}{{4!}}{A_4} = 0, \ {A_1} + frac{4}{{2!}}{A_2} + frac{{{4^2}}}{{3!}}{A_3} + frac{{{4^3}}}{{4!}}{A_4} = 0, end{array} ight.$
方程组的系数行列式为
$ left|egin{array}{cccc} 1 {frac{1}{{2!}}} {frac{1}{{3!}}} {frac{1}{{4!}}} \ 1 {frac{2}{{2!}}} {frac{{{2^2}}}{{3!}}} {frac{{{2^3}}}{{4!}}} \ 1 {frac{3}{{2!}}} {frac{{{3^2}}}{{3!}}} {frac{{{3^3}}}{{4!}}} \ 1 {frac{4}{{2!}}} {frac{{{4^2}}}{{3!}}} {frac{{{4^3}}}{{4!}}} end{array} ight| $$ =frac{1}{{1!2!3!4!}}left|egin{array}{cccc} 1 1 1 1 \ 1 2 {{2^2}} {{2^3}} \ 1 3 {{3^2}} {{3^3}} \ 1 4 {{4^2}} {{4^3}} end{array} ight| $
=$frac{1}{{1!2!3!4!}}(2 - 1)(3 - 1)(3 - 2)(4 - 1)(4 - 2)(4 - 3) = frac{1}{{4!}} e 0$
即$underset{x o +infty }{mathop{lim }}\,{{f}^{(k)}}(x)=0,k=1,2,3,4$
（或利用第二数学归纳法证明）
四、证明：证明：由题设知$f(x)$在$ [0,+infty ) $上必有界,设$left| f(x) ight|le M$
对$forall varepsilon 0$,有
$left| frac{1}{x}int_{0}^{x}{f(t)dt}-A ight|$
$=left| int_{0}^{1}{(f(yx)-A)dy} ight|$
$le int_{0}^{frac{varepsilon }{2(M+A)}}{left| f(yx)-A ight|}dy+int_{frac{varepsilon }{2(M+A)}}^{1}{left| f(yx)-A ight|}dy$
由$underset{x o +infty }{mathop{lim }}\,f(x)=A$知
对上述$varepsilon 0,exists {{X}_{1}}0, $使得当$x{{X}_{1}}$时有$left| f(x)-A ight|frac{varepsilon }{2}$
令$ ext{X}=frac{2(M+A)}{varepsilon }{{X}_{1}}$,则当$xX$时,有$int_{frac{varepsilon }{2(M+A)}}^{1}{left| f(yx)-A ight|}dyfrac{varepsilon }{2}$
于是
$left| frac{1}{x}int_{0}^{x}{f(t)dt}-A ight|frac{varepsilon }{2}+frac{varepsilon }{2}=varepsilon $
因此
$underset{x o +infty }{mathop{lim }}\,frac{1}{x}int_{0}^{x}{f(t)dt}=A$.
（或直接利用定义）
五、证明：
充分性：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛，令$0S=underset{n o +infty }{mathop{lim }}\,{{S}_{n}}=sumlimits_{n=1}^{+infty }{{{a}_{n}}}+infty $
由于
$underset{n o +infty }{mathop{lim }}\,frac{frac{{{a}_{n}}}{{{S}_{n}}}}{{{a}_{n}}}=underset{n o +infty }{mathop{lim }}\,frac{1}{{{S}_{n}}}=frac{1}{S}+infty $
由比较判别法可知：$sumlimits_{n=1}^{+infty }{frac{{{a}_{n}}}{{{S}_{n}}}}$收敛
必要性：反证法：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$发散
考虑
$sumlimits_{k=n+1}^{n+p}{frac{{{a}_{k}}}{{{S}_{k}}}}=frac{{{a}_{n+1}}}{{{S}_{n+1}}}+frac{{{a}_{n+2}}}{{{S}_{n+2}}}+cdots +frac{{{a}_{n+p}}}{{{S}_{n+p}}}frac{sumlimits_{k=n+1}^{n+p}{{{a}_{k}}}}{{{S}_{n+p}}}=frac{{{S}_{n+p}}-{{S}_{n}}}{{{S}_{n+p}}}=1-frac{{{S}_{n}}}{{{S}_{n+p}}}$
由于$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$发散，则对任意的$nin {{N}^{*}}$，存在$pin {{N}^{*}}$，使得
$frac{{{S}_{n}}}{{{S}_{n+p}}}frac{1}{2}$
从而$sumlimits_{k=n+1}^{n+p}{frac{{{a}_{k}}}{{{S}_{k}}}}frac{1}{2}$
由柯西收敛准则知：$sumlimits_{n=1}^{+infty }{frac{{{a}_{n}}}{{{S}_{n}}}}$发散，矛盾
从而$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛
六、
（1）证明：由于
$2sin frac{x}{2}(frac{1}{2}+sumlimits_{k=1}^{n}{cos kx)=sin }frac{x}{2}+(sin frac{3x}{2}-sin frac{x}{2})+cdots +[sin (n+frac{1}{2})x-sin (n-frac{1}{2})x]=sin (n+frac{1}{2})x$
当$xin [alpha ,2pi -alpha ]$时，$sin frac{x}{2} e 0$
于是
$frac{1}{2}+sumlimits_{k=1}^{n}{cos kx}=frac{sin (n+frac{1}{2})x}{2sin frac{x}{2}}$
于是$sumlimits_{n=1}^{+infty }{cos nx}$部分和在$xin [alpha ,2pi -alpha ]$上一致有界
令${{u}_{n}}(x)=cos nx,{{v}_{n}}(x)={{a}_{n}}$
由狄利克雷判别法知：$sumlimits_{n=1}^{+infty }{{{a}_{n}}}cos nx$在$xin [alpha ,2pi -alpha ]$上一致收敛
（2）证明：
充分性：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛，由于$left| {{a}_{n}}cos nx ight|le left| {{a}_{n}} ight|={{a}_{n}}$
由$M$判别法可知：$sumlimits_{n=1}^{+infty }{{{a}_{n}}}cos nx$在$[0,2pi ]$上一致收敛
必要性：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}cos nx$在$[0,2pi ]$上一致收敛，令$x=0$
即$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛收敛
七、证明：由于
${{u}_{x}}=z+x{{z}_{x}}+yf(z){{z}_{x}}+g(z){{z}_{x}}=z$
${{u}_{xx}}={{z}_{x}}$
${{u}_{xy}}={{z}_{y}}$
${{u}_{y}}=x{{z}_{y}}+f(z)+yf(z){{z}_{y}}+g(z){{z}_{y}}=f(z)$
${{u}_{yy}}=f(z){{z}_{y}}$
${{u}_{yx}}=f(z){{z}_{x}}$
于是
$frac{{{partial }^{2}}u}{partial {{x}^{2}}}cdot frac{{{partial }^{2}}u}{partial {{y}^{2}}}-{{left( frac{{{partial }^{2}}u}{partial xpartial y} ight)}^{2}}=0$
八、证明：
（1）由于$frac{partial f}{partial n}=frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)+frac{partial f}{partial z}cos (n,z)$
由第一型和第二型曲面积分的关系和$Guass$公式可知：
$iintlimits_{S}{frac{partial f}{partial n}}dS=iintlimits_{S}{frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)+frac{partial f}{partial z}cos (n,z)}dS$
$=iintlimits_{S}{frac{partial f}{partial x}dydz+frac{partial f}{partial y}dzdx+frac{partial f}{partial z}dxdy}$
$=iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}$
即$iintlimits_{S}{frac{partial f}{partial n}}dS=iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}$
（2）充分性：若对任意的$S$，$iintlimits_{S}{frac{partial f}{partial n}}dS$，总有$iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}=0$
不妨设$g(x,y,z)=frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}$，由积分中值定理可知：存在$(xi ,eta ,zeta )in V$
[iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}})dxdydz=g(xi ,eta ,zeta )V=0]
于是由$V$的任意性可知：$g(xi ,eta ,zeta )=0$，再由$(xi ,eta ,zeta )$的任意性知：
$frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}=0$
必要性：若$frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}=0$，则$iintlimits_{S}{frac{partial f}{partial n}}dS=iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}=0$