[尊重原有作者劳动成果]
一.
（1）证明：由${ {x}_{1}}=\frac{1}{2},{ {x}_{2}}=\frac{3}{8},{ {x}_{3}}=\frac{55}{128},\cdots $，猜测$\{ { {x}_{2n+1}}\}$单调递减，$\{ { {x}_{2n}}\}$单调递增
下用数归法先证$\sqrt{2}-1\le { {x}_{2n+1}}\le \frac{1}{2},\frac{3}{8}\le { {x}_{2n+1}}\le \sqrt{2}-1$
（1）当$n=1$时，$\sqrt{2}-1{ {x}_{1}}=\frac{1}{2}$恒成立
（2）设$n=2k-1$时，$\sqrt{2}-1\le { {x}_{2k-1}}\le \frac{1}{2}$，则
\[{ {x}_{2k+1}}=\frac{1}{2}(1-x_{2k}^{2})=\frac{1}{2}[1-\frac{1}{4}{ {(1-x_{2k-1}^{2})}^{2}}]\in [\sqrt{2}-1,\frac{1}{2})\]
即当$n=2k+1$时也恒成立
由（1）（2）可知：$\sqrt{2}-1\le { {x}_{2n+1}}\le \frac{1}{2}$
同理可证$\frac{3}{8}\le { {x}_{2n+1}}\le \sqrt{2}-1$
再证$\{ { {x}_{2n+1}}\}$单调递减，$\{ { {x}_{2n}}\}$单调递增，同样采用数学归纳法
（1）当$n=1$时，${ {x}_{1}}=\frac{1}{2}\frac{55}{128}={ {x}_{3}}$恒成立
（2）设$n=2k-1$时，${ {x}_{2k-1}}\ge { {x}_{2k+1}}$，则
${ {x}_{2k+3}}-{ {x}_{2k+1}}=\frac{1}{8}[{ {(1-x_{2k-1}^{2})}^{2}}-{ {(1-x_{2k+1}^{2})}^{2}}]=\frac{1}{8}(x_{2k+1}^{2}-x_{2k-1}^{2})(2-x_{2k+1}^{2}-x_{2k-1}^{2})\le 0$
即当$n=2k+1$时也恒成立
由（1）（2）可知：$\{ { {x}_{2n+1}}\}$单调递减
同理可证：$\{ { {x}_{2n}}\}$单调递增
于是$\{ { {x}_{2n+1}}\}$单调递减，$\sqrt{2}-1\le { {x}_{2n+1}}\le \frac{1}{2}$，由单调有界原理可知：$\{ { {x}_{2n+1}}\}$收敛
于是不妨设$\underset{n\to +\infty }{\mathop{\lim }}\,{ {x}_{2n+1}}=l\in [\sqrt{2}-1,\frac{1}{2}]$，由${ {x}_{2k+3}}=\frac{1}{2}[1-\frac{1}{4}{ {(1-x_{2k+1}^{2})}^{2}}]$，两边对$n\to +\infty $
则$l=\frac{1}{2}[1-\frac{1}{4}{ {(1-{ {l}^{2}})}^{2}}]$，求得$l=\sqrt{2}-1$
即$\underset{x\to \infty }{\mathop{\lim }}\,{ {x}_{2n+1}}=\sqrt{2}-1=A$
同理可证$\underset{x\to \infty }{\mathop{\lim }}\,{ {x}_{2n}}=\sqrt{2}-1=A$
于是$\underset{n\to +\infty }{\mathop{\lim }}\,{ {x}_{n}}=\underset{n\to +\infty }{\mathop{\lim }}\,{ {x}_{2n}}=\underset{n\to +\infty }{\mathop{\lim }}\,{ {x}_{2n+1}}=A$
（3）证明：不妨设${ {a}_{n}}={ {x}_{n}}-A={ {x}_{n}}-(\sqrt{2}-1)$
于是
$\underset{n\to +\infty }{\mathop{\lim }}\,\left| \frac{ { {a}_{n+1}}}{ { {a}_{n}}} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\left| \frac{ { {x}_{n+1}}-A}{ { {x}_{n}}-A} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\left| \frac{\frac{1}{2}(1-x_{n}^{2})-A}{ { {x}_{n}}-A} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{2}\left| \frac{1-x_{n}^{2}-(2\sqrt{2}-2)}{ { {x}_{n}}-(\sqrt{2}-1)} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{2}\left| \frac{x_{n}^{2}-{ {(\sqrt{2}-1)}^{2}}}{ { {x}_{n}}-(\sqrt{2}-1)} \right|=A1 $
于是$\sum\limits_{n=1}^{+\infty }{({ {x}_{n}}}-A)$绝对收敛
二、
（1）成立，理由如下：
要证明${ {\sin }^{3}}\left| f(x) \right|$在$I$上一致连续，只需证明$\left| f(x) \right|$在$I$上一致连续即可
由于$f(x)$在$I$上一致连续，则对任意的$\varepsilon 0$，任意的${ {x}_{1}},{ {x}_{2}}\in I$，存在$\delta 0$，当
$\left| { {x}_{1}}-{ {x}_{2}} \right|\delta $时，有$\left| f({ {x}_{1}})-f({ {x}_{2}}) \right|\varepsilon $
于是对任意的$\varepsilon 0$，任意的${ {x}_{1}},{ {x}_{2}}\in I$，存在$\delta 0$，当
$\left| { {x}_{1}}-{ {x}_{2}} \right|\delta $时，有$\left| \left| f({ {x}_{1}}) \right|-\left| f({ {x}_{2}}) \right| \right|\le \left| f({ {x}_{1}})-f({ {x}_{2}}) \right|\varepsilon $
即$\left| f(x) \right|$在$I$上一致连续即可
对任意的$\varepsilon 0$，任意的${ {x}_{1}},{ {x}_{2}}\in I$，存在$\delta 0$，当
$\left| { {x}_{1}}-{ {x}_{2}} \right|\delta $时，有
$\left| { {\sin }^{3}}\left| f({ {x}_{1}}) \right|-{ {\sin }^{3}}\left| f({ {x}_{1}}) \right| \right|=\left| (\sin \left| f({ {x}_{1}}) \right|-\sin \left| f({ {x}_{1}}) \right|)({ {\sin }^{2}}\left| f({ {x}_{1}}) \right|+\sin \left| f({ {x}_{1}}) \right|\sin \left| f({ {x}_{2}}) \right|+{ {\sin }^{2}}\left| f({ {x}_{2}}) \right| \right|$$\le 3\left| \sin \left| f({ {x}_{1}}) \right|-\sin \left| f({ {x}_{1}}) \right| \right|\le 3\left| \sin f({ {x}_{1}})-\sin f({ {x}_{2}}) \right|\le 3\left| f({ {x}_{1}})-f({ {x}_{2}}) \right|$
于是由复合函数的一致收敛性可知：${ {\sin }^{3}}\left| f(x) \right|$在$I$上一致连续
（4）成立，理由如下
要证明${ {\sin }^{3}}f(x)$在$I$上一致连续，只需证明$f(x)$在$I$上一致连续即可
由于$\left| f(x) \right|$在$I$上一致连续，则对任意的$\varepsilon 0$，任意的${ {x}_{1}},{ {x}_{2}}\in I$，存在$\delta 0$，当
$\left| { {x}_{1}}-{ {x}_{2}} \right|\delta $时，由$\left| \left| f({ {x}_{1}}) \right|-\left| f({ {x}_{2}}) \right| \right|\frac{\varepsilon }{2}$
1：若$f({ {x}_{1}}),f({ {x}_{2}})$同号，则有$\left| f({ {x}_{1}})-f({ {x}_{2}}) \right|\frac{\varepsilon }{2}\varepsilon $
2：若$f({ {x}_{1}}),f({ {x}_{2}})$同号，由$f(x)$连续，则存在$y$在${ {x}_{1}},{ {x}_{2}}$之间使得$f(y)=0$
于是$\left| f({ {x}_{1}})-f({ {x}_{2}}) \right|\le \left| f({ {x}_{1}}) \right|+\left| f({ {x}_{2}}) \right|=\left| \left| f({ {x}_{1}}) \right|-\left| f(y) \right| \right|+\left| \left| f({ {x}_{2}}) \right|-\left| f(y) \right| \right|\varepsilon $
由1，2可知，$f(x)$在$I$上一致连续
再利用复合函数的一致连续性可知，${ {\sin }^{3}}f(x)$在$I$上一致连续
三、证明：
充分性：反证法：假设对任意的$x\in (a,b)$都有$f(x)\le \frac{f(b)-f(a)}{b-a}$
令$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$
则当$x\in (a,b)$时有$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}\le 0$
于是$g(x)$在$[a,b]$上单调递减
而$g(a)=g(b)=f(a)$
从而当$x\in [a,b]$时，$g(x)=f(a)$
于是$f(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$矛盾
从而必存在$\xi \in (a,b)$，使得$f(\xi )\frac{f(b)-f(a)}{b-a}$
必要性：
令$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$
则$g(a)=g(b)=f(a)$
反证法：
1：若$f(x)$为常函数，则$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}=-\frac{f(b)-f(a)}{b-a},x\in [a,b]$
这与存在$\xi \in (a,b) $，使得$f(\xi )\frac{f(b)-f(a)}{b-a}$矛盾
2：若$f(x)$为线性函数，可知$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}$为常数
又由于存在$\xi \in (a,b) $，使得$f(\xi )\frac{f(b)-f(a)}{b-a}$可知，$g(x)0$
于是$g(x)$在$[a,b]$上严格单调递增
而$g(a)=g(b)=f(a)$矛盾
从而由1,2可知，$f(x)$不为常函数或线性函数
四、
（1）证明：设$F\left( x,y \right)={ {x}^{2}}+y-\cos \left( xy \right) $，
显然，有$F\left( 0,1 \right)=0 $，
${ {F}_{y}}\left( x,y \right)=1+x\sin \left( xy \right) $，
${ {F}_{y}}\left( 0,1 \right)=1\ne 0 $，由隐函数存在定理，
存在$\delta 0 $，存在$\left[ -\delta ,\delta \right] $上的连续可微的函数$y=y\left( x \right)$，$y\left( 0 \right)=1 $，
满足$F\left( x,y\left( x \right) \right)\equiv 0 $，$x\in U(0) $
（3）证明：由（1）可知：
${ {F}_{x}}\left( x,y \right)=2x+y\sin \left( xy \right) $，
${y}\left( x \right)=-\frac{ { {F}_{x}}\left( x,y \right)}{ { {F}_{y}}\left( x,y \right)}=-\frac{2x+y\sin \left( xy \right)}{1+x\sin \left( xy \right)} $，
当$0x\delta $，（$\delta 0 $充分小）时，有${y}\left( x \right)0 $，$y\left( x \right) $在$\left[ 0,\delta \right] $上严格单调递减；
当$-\delta x0 $时，有${y}\left( x \right)0 $，$y\left( x \right) $在$\left[ -\delta ,0 \right] $上严格单调递增，
五、
（1）解：（1）由偏导数的定义：
${ {f}_{x}}(0,0)=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\Delta x\cos \frac{1}{\left| \Delta x \right|}=0 $
${ {f}_{y}}(0,0)=\underset{\Delta y\to 0}{\mathop{\lim }}\,\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=\underset{\Delta y\to 0}{\mathop{\lim }}\,\Delta y\cos \frac{1}{\left| \Delta y \right|}=0\ $
（2）当$(x,y)\ne (0,0)$时，
${ {f}_{x}}(x,y)=2x\cos \frac{1}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}}+\frac{x}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}}\sin \frac{1}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}} $
${ {f}_{y}}(x,y)=2y\cos \frac{1}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}}+\frac{y}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}}\sin \frac{1}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}} $
于是${f_x}(x,y)=\left\{\begin{array}{ll}2x\cos \frac{1}{ {\sqrt { {x^2} + {y^2}} }} + \frac{x}{ {\sqrt { {x^2} + {y^2}} }}\sin \frac{1}{ {\sqrt { {x^2} + {y^2}} }}, \hbox{${x^2} + {y^2} \ne 0$} \\0, \hbox{${x^2} + {y^2} = 0$.}\end{array}\right.$
${f_y}(x,y)=\left\{\begin{array}{ll}2y\cos \frac{1}{ {\sqrt { {x^2} + {y^2}} }} + \frac{y}{ {\sqrt { {x^2} + {y^2}} }}\sin \frac{1}{ {\sqrt { {x^2} + {y^2}} }},, \hbox{${x^2} + {y^2} \ne 0$} \\ 0, \hbox{${x^2} + {y^2} = 0$.}\end{array}\right.$（2）设$y=kx$，于是$\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sqrt{ { {x}^{2}}+{ {y}^{2}}}}=\frac{1}{\sqrt{1+{ {k}^{2}}}}$与$k$有关
于是$\underset{x\to 0}{\mathop{\lim }}\,{ {f}_{x}}(x,0) $不存在，故${ {f}_{x}}(x,y) $在（0，0）不连续。
同理${ {f}_{y}}(x,y) $ 在（0，0）也不连续。
（3）设$u=f(x,y) $，则在（0，0）点有
$\Delta u-du=[f(\Delta x,\Delta y)-f(0,0)]-[{ {f}_{x}}(0,0)\Delta x+{ {f}_{y}}(0,0)\Delta y]=(\Delta { {x}^{2}}+\Delta { {y}^{2}})\cos \frac{1}{\sqrt{\Delta { {x}^{2}}+\Delta { {y}^{2}}}} $
因$\underset{\Delta x\to 0,\Delta y\to 0}{\mathop{\lim }}\,\frac{\Delta u-du}{\sqrt{\Delta { {x}^{2}}+\Delta { {y}^{2}}}}=\underset{\Delta x\to 0,\Delta y\to 0}{\mathop{\lim }}\,\sqrt{\Delta { {x}^{2}}+\Delta { {y}^{2}}}\cos \frac{1}{\sqrt{\Delta { {x}^{2}}+\Delta { {y}^{2}}}}=0$
故$f(x,y) $在（0，0）可微。
六、解：由于$y=x+y,y(0)=1$
利用常数变易法求得$y=-1-x+2{ {e}^{x}}$
记${ {a}_{n}}=y(\frac{1}{n})-1-\frac{1}{n}=2({ {e}^{\frac{1}{n}}}-1-\frac{1}{n})$
由$\underset{n\to +\infty }{\mathop{\lim }}\,\sqrt[n]{ { {a}_{n}}}={ {e}^{\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\ln { {a}_{n}}}{n}}}={ {e}^{\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\ln 2({ {e}^{\frac{1}{n}}}-1-\frac{1}{n})}{n}}}\overset{n=\frac{1}{x}}{\mathop{=}}\,{ {e}^{\underset{x\to +{ {0}^{+}}}{\mathop{\lim }}\,\frac{\ln 2({ {e}^{x}}-1-x)}{\frac{1}{x}}}}={ {e}^{-\underset{x\to +{ {0}^{+}}}{\mathop{\lim }}\,\frac{ { {x}^{2}}({ {e}^{x}}-1)}{ { {e}^{x}}-1-x}}}={ {e}^{-\underset{x\to +{ {0}^{+}}}{\mathop{\lim }}\,\frac{ { {x}^{2}}[x+o(x)]}{\frac{ { {x}^{2}}}{2}+o({ {x}^{2}})}}}=1 $
于是幂级数的收敛半径$R=1$
又由于$x=1$时，$y(\frac{1}{n})-1-\frac{1}{n}=2({ {e}^{\frac{1}{n}}}-1-\frac{1}{n})=\sum\limits_{n=1}^{+\infty }{\frac{2}{ { {n}^{2}}}+}\sum\limits_{n=1}^{+\infty }{o(\frac{2}{ { {n}^{2}}}})$收敛
$x=-1$时，由莱布利兹判别法可知级数收敛
于是该幂级数的收敛域为$[-1,1]$
七、
（1）证明：令$u={ {t}^{2}}\Rightarrow t=\sqrt{u},dt=\frac{1}{2\sqrt{u}}du$，于是$\int_{x}^{x+c}{\sin { {t}^{2}}}dt=\frac{1}{2}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\frac{\sin u}{\sqrt{u}}}du $
而函数$\frac{1}{\sqrt{u}}$在$[{ {x}^{2}},{ {(x+c)}^{2}}]$上递减，且$\frac{1}{\sqrt{u}}\ge 0$，由积分第二中值定理可知：
存在$\xi \in [{ {x}^{2}},{ {(x+c)}^{2}}]$，使得
$\int_{x}^{x+c}{\sin { {t}^{2}}}dt=\frac{1}{2}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\frac{\sin u}{\sqrt{u}}}du=\frac{1}{2x}\int_{ { {x}^{2}}}^{\xi }{\sin udu=\frac{1}{2x}(\cos { {x}^{2}}-\cos \xi )} $
故$\left| \int_{x}^{x+c}{\sin { {t}^{2}}}dt \right|\le \frac{1}{x}$
（2）能，理由如下
证明：
令$f(x)=\int_{x}^{x+c}{\sin { {t}^{2}}}dt=\frac{1}{2}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\frac{\sin u}{\sqrt{u}}}du=-\frac{1}{2}\frac{\cos u}{\sqrt{u}}|_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}-\frac{1}{4}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\frac{\sin u}{ { {u}^{\frac{3}{2}}}}}du$
$=\frac{\cos { {x}^{2}}}{2x}-\frac{\cos { {(x+c)}^{2}}}{2(x+c)}-\frac{1}{4}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\frac{\sin u}{ { {u}^{\frac{3}{2}}}}}du$
存在${ {u}_{0}}\in [{ {x}^{2}},{ {(x+c)}^{2}}]$，使得$\left| \frac{\cos { {u}_{0}}}{u_{0}^{\frac{3}{2}}} \right|\frac{1}{ { {u}^{\frac{3}{2}}}}$，因此当$x0$时，有
$\left| f(x) \right|\le \left| \frac{\cos { {x}^{2}}}{2x} \right|+\left| \frac{\cos { {(x+c)}^{2}}}{2(x+c)} \right|+\frac{1}{4}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\left| \frac{\cos u}{ { {u}^{\frac{3}{2}}}} \right|}du\frac{1}{2x}+\frac{1}{2(x+c)}+\frac{1}{4}\int_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}{\frac{1}{ { {u}^{\frac{3}{2}}}}}du $
$=\frac{1}{2x}+\frac{1}{2(x+c)}+\frac{1}{4}(-2{ {u}^{-\frac{1}{2}}})|_{ { {x}^{2}}}^{ { {(x+c)}^{2}}}=\frac{1}{x}$
八、证明：对${ {R}^{2}}$上任意一点$({ {x}_{0}},{ {y}_{0}})$，令${ {L}_{1}}=\{x|{ {(x-{ {x}_{0}})}^{2}}={ {r}^{2}}\}$，方向取逆时针
由格林公式可知：
$\int_{L}{Pdx+Qdy=}\int_{L+{ {L}_{1}}}{Pdx+Qdy=\iint_{D}{[\frac{\partial Q}{\partial x}}}-\frac{\partial P}{\partial y}]dxdy=[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}]{ {|}_{M}}\pi { {r}^{2}}=0$
其中$D$是由$L+{ {L}_{1}}$包围的图形，$M\in D$
另一方面由积分中值定理可知：
$\int_{L}{Pdx+Qdy=}-\int_{ { {L}_{1}}}{P(x,{ {y}_{0}}})dx=-P({ {x}_{1}},{ {y}_{0}})\cdot 2r$
其中$({ {x}_{1}},{ {y}_{0}})\in { {L}_{1}}$
比较这两个式子知：
$[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}]{ {|}_{M}}\pi { {r}^{2}}=-P({ {x}_{1}},{ {y}_{0}})\cdot 2r\Rightarrow [\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}]{ {|}_{M}}\frac{\pi r}{2}=-P({ {x}_{1}},{ {y}_{0}})$
令$r\to 0$可知：$P({ {x}_{0}},{ {y}_{0}})=0$，由$({ {x}_{0}},{ {y}_{0}})$的任意性可知，$P(x,y)=0$
从而有 $[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}]{ {|}_{M}}$，令$r\to 0$可知$[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}]{ {|}_{({ {x}_{0}},{ {y}_{0}})}}=0$，由$({ {x}_{0}},{ {y}_{0}})$的任意性可知，$\frac{\partial Q}{\partial x}=0$