来源 [尊重原有作者劳动成果]
一、
(1)证明:由于${{x}_{1}}\in (0,\frac{\pi }{2}),{{x}_{n+1}}=\sin {{x}_{n}}$,则${{x}_{n}}\in (0,\frac{\pi }{2}),n=1,2,\cdots $
且${{x}_{n+1}}=\sin {{x}_{n}}\le {{x}_{n}}$
于是$\{{{x}_{n}}\}$单调递减且${{x}_{n}}\in (0,\frac{\pi }{2})$
由单调有界原理可知:$\{{{x}_{n}}\}$收敛
不妨设$\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{n}}=l\in [0,\frac{\pi }{2}]$,由${{x}_{n+1}}=\sin {{x}_{n}}$,两边取极限,得$l=\sin l$
解得$l=0$,即$\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{n}}=0$
(2)证明:由(1)知:$\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{n}}=0$,则$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{x_{n}^{2}}=+\infty $
由$Stolz$公式可知:
$\underset{n\to +\infty }{\mathop{\lim }}\,n{{\sin }^{2}}{{x}_{n}}=\underset{n\to +\infty }{\mathop{\lim }}\,nx_{n+1}^{2}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{n}{\frac{1}{x_{n+1}^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{n-(n-1)}{\frac{1}{x_{n+1}^{2}}-\frac{1}{x_{n}^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{\frac{1}{{{\sin }^{2}}{{x}_{n}}}-\frac{1}{x_{n}^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{x_{n}^{2}{{\sin }^{2}}{{x}_{n}}}{x_{n}^{2}-{{\sin }^{2}}{{x}_{n}}}$
$=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{x_{n}^{4}+o(x_{n}^{4})}{x_{n}^{2}-{{[{{x}_{n}}-\frac{x_{n}^{3}}{3!}+o(x_{n}^{3})]}^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{x_{n}^{4}+o(x_{n}^{4})}{\frac{x_{n}^{4}}{3}+o(x_{n}^{4})}=3$
(3)证明:由(2)知:
$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{x_{n+1}^{2}}{\frac{1}{n}}=3$
于是
$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{x_{n+1}^{p}}{{{(\frac{1}{n})}^{\frac{p}{2}}}}=\underset{n\to +\infty }{\mathop{\lim }}\,{{(\frac{x_{n+1}^{2}}{\frac{1}{n}})}^{\frac{p}{2}}}={{3}^{\frac{p}{2}}}$
则$\sum\limits_{n=1}^{+\infty }{x_{n+1}^{p}}$与$\sum\limits_{n=1}^{+\infty }{(\frac{1}{n}}{{)}^{\frac{p}{2}}}$收敛性相同
而$p2$时,$\sum\limits_{n=1}^{+\infty }{(\frac{1}{n}}{{)}^{\frac{p}{2}}}$收敛;$p\le 2$时,$\sum\limits_{n=1}^{+\infty }{(\frac{1}{n}}{{)}^{\frac{p}{2}}}$发散
于是$p2$时,\[\sum\limits_{n=1}^{+\infty }{x_{n+1}^{p}}\]收敛;
$p\le 2$时,\[\sum\limits_{n=1}^{+\infty }{x_{n+1}^{p}}\]发散
二、
(1)证明:由于$\{{{b}_{n}}\}$有界,则存在$M0$,对一切$n\in {{N}^{\text{*}}}$,有$\left| {{b}_{n}} \right|M$
由$\alpha {{a}_{n+1}}+\beta {{a}_{n}}={{b}_{n}}$
1:若$\alpha =0\ne \beta $,此时${{b}_{n}}=\beta {{a}_{n}}$,则$\underset{n\to +\infty }{\mathop{\lim }}\,{{b}_{n}}$存在$\Leftrightarrow \underset{n\to +\infty }{\mathop{\lim }}\,{{a}_{n}}$存在
2:若$\alpha \ne 0$,则${{a}_{n+1}}=\frac{1}{\alpha }{{b}_{n}}-\frac{\beta }{\alpha }{{a}_{n}}$
令$r=\frac{\beta }{\alpha }\ne 1$,则
${{a}_{n}}=\frac{1}{\alpha }{{b}_{n-1}}-r{{a}_{n-1}}=\frac{1}{\alpha }{{b}_{n-1}}-r(\frac{1}{\alpha }{{b}_{n-2}}-r{{a}_{n-2}})=\frac{1}{\alpha }{{b}_{n-1}}-\frac{r}{\alpha }{{b}_{n-2}}+{{r}^{2}}{{a}_{n-2}}$
$=\cdots =\frac{1}{\alpha }[{{b}_{n-1}}-r{{b}_{n-2}}+\cdots +{{(-r)}^{n-2}}]+{{a}_{1}}\cdot {{(-r)}^{n-1}}$
若$r=-1$,此时${{a}_{n}}=\frac{1}{\alpha }({{b}_{n-1}}+{{b}_{n-2}}+\cdots +{{b}_{1}})+{{a}_{1}}$不一定恒有界(如${{b}_{n}}=\frac{1}{n}$)
若$\left| r \right|1$,由$\underset{n\to +\infty }{\mathop{\lim }}\,{{\left| r \right|}^{n-2}}=+\infty $,则$\{{{a}_{n}}\}$无界
从而$\left| r \right|=\left| \frac{\beta }{\alpha } \right|1$,才使得$\{{{a}_{n}}\}$恒有界
于是$\left| \alpha +\beta \right|\ge \left| \alpha \right|-\left| \beta \right|0\Rightarrow \alpha +\beta \ne 0$
必要性:先给出法一,是种非常复杂的方法,但是比较实用
法一:
若$\{{{b}_{n}}\}$收敛,不妨设$\underset{n\to +\infty }{\mathop{\lim }}\,{{b}_{n}}=b$,进一步设${{b}_{n}}=b+{{\varepsilon }_{n}}$,其中$\underset{n\to +\infty }{\mathop{\lim }}\,{{\varepsilon }_{n}}=0$
于是
${{a}_{n+1}}=\frac{1}{\alpha }{{b}_{n}}-r{{a}_{n}}=\frac{b+{{\varepsilon }_{n}}}{\alpha }-r{{a}_{n}}$
${{a}_{n+2}}=\frac{1}{\alpha }{{b}_{n+1}}-r{{a}_{n+1}}=\frac{b+{{\varepsilon }_{n+1}}}{\alpha }-r{{a}_{n+1}}=\frac{b+{{\varepsilon }_{n+1}}}{\alpha }-r(\frac{b+{{\varepsilon }_{n}}}{\alpha }-r{{a}_{n}})=\frac{b(1-r)}{\alpha }+\frac{{{\varepsilon }_{n+1}}-r{{\varepsilon }_{n}}}{\alpha }+{{r}^{2}}{{a}_{n}}$
$\vdots $
${{a}_{n+k}}=\frac{b}{\alpha }[1-r+{{r}^{2}}+\cdots +{{(-r)}^{k-1}}]+\frac{{{\varepsilon }_{n+k-1}}-r{{\varepsilon }_{n+k-2}}+\cdots +{{(-1)}^{k-1}}{{\varepsilon }_{n}}}{\alpha }+{{(-1)}^{k}}{{a}_{n}}$
由于
$\frac{b}{\alpha }[1-r+{{r}^{2}}+\cdots +{{(-r)}^{k-1}}]=\frac{b}{\alpha }\cdot \frac{1-{{(-r)}^{k}}}{1+r}=\frac{b}{\alpha +\beta }[1-{{(-r)}^{k}}]$
另一方面,由$\underset{n\to +\infty }{\mathop{\lim }}\,{{\varepsilon }_{n}}=0$,则对任意的$\varepsilon 0$,存在${{N}_{1}}0$,当$n{{N}_{1}}$时,$\left| {{\varepsilon }_{n}} \right|\frac{(1-\left| r \right|)\varepsilon }{2}$
于是$\left| {{a}_{n+k}}-\frac{b}{\alpha +\beta } \right|\le \left| \frac{\left| b \right|}{\alpha +\beta }+\left| {{a}_{n}} \right| \right|\cdot {{\left| r \right|}^{k}}+\frac{(1-\left| r \right|)\varepsilon }{2}\cdot \frac{1-{{\left| r \right|}^{k}}}{1-\left| r \right|}$
而$\underset{k\to +\infty }{\mathop{\lim }}\,{{\left| r \right|}^{k}}=0$,则对上述$\varepsilon 0$,存在${{N}_{2}}0$,当$n{{N}_{2}}$时,有
$\left| \frac{\left| b \right|}{\alpha +\beta }+\left| {{a}_{n}} \right| \right|\cdot {{\left| r \right|}^{k}}\frac{\varepsilon }{2}$,$\frac{(1-\left| r \right|)\varepsilon }{2}\cdot \frac{1-{{\left| r \right|}^{k}}}{1-\left| r \right|}$
令$N=\max \{{{N}_{1}},{{N}_{2}}\}$,当$nN$时,有$\left| {{a}_{n}}-\frac{b}{\alpha +\beta } \right|\varepsilon $
即$\underset{n\to +\infty }{\mathop{\lim }}\,{{a}_{n}}=\frac{b}{\alpha +\beta }$
法二:
首先以下三个命题成立:
$\underset{n\to +\infty }{\mathop{\lim }}\,({{u}_{n}}+{{v}_{n}})=\underset{n\to +\infty }{\mathop{\lim }}\,{{u}_{n}}+\underset{n\to +\infty }{\mathop{\lim }}\,{{v}_{n}},\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,({{u}_{n}}+{{v}_{n}})=\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{u}_{n}}+\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{v}_{n}},\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{u}_{n}}=-\overline{\underset{n\to +\infty }{\mathop{\lim }}\,}{{u}_{n}}$
不妨设\[\overline{\underset{n\to \infty }{\mathop{\lim }}\,}{{a}_{n}}={{a}_{1}},\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{a}_{n}}={{a}_{2}},\underset{n\to +\infty }{\mathop{\lim }}\,{{b}_{n}}=b\]
由于$\{{{a}_{n}}\},\{{{b}_{n}}\}$都是有界数列
则${{a}_{1}},{{a}_{2}},b$都是有限值,且${{b}_{n}}=\alpha {{a}_{n+1}}+\beta {{a}_{n}}$可知:
取上极限得:$\alpha {{a}_{1}}=b+\beta {{a}_{2}}$
取下极限得:$\beta {{a}_{1}}=b+\alpha {{a}_{2}}$
由$\alpha \ne \beta $可知:${{a}_{1}}={{a}_{2}}$,即$\{{{a}_{n}}\}$存在
充分性:
若$\{{{a}_{n}}\}$收敛,则$\{{{a}_{n+1}}\}$也收敛,于是$\{{{b}_{n}}\}$收敛
法二:首先以下三个命题成立:
$\underset{n\to +\infty }{\mathop{\lim }}\,({{u}_{n}}+{{v}_{n}})=\underset{n\to +\infty }{\mathop{\lim }}\,{{u}_{n}}+\underset{n\to +\infty }{\mathop{\lim }}\,{{v}_{n}},\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,({{u}_{n}}+{{v}_{n}})=\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{u}_{n}}+\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{v}_{n}},\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{u}_{n}}=-\overline{\underset{n\to +\infty }{\mathop{\lim }}\,}{{u}_{n}}$
不妨设\[\overline{\underset{n\to \infty }{\mathop{\lim }}\,}{{a}_{n}}={{a}_{1}},\underset{n\to +\infty }{\mathop{\underline{\lim }}}\,{{a}_{n}}={{a}_{2}},\underset{n\to +\infty }{\mathop{\lim }}\,{{b}_{n}}=b\]
由于$\{{{a}_{n}}\},\{{{b}_{n}}\}$都是有界数列
则${{a}_{1}},{{a}_{2}},b$都是有限值,且${{b}_{n}}=\alpha {{a}_{n+1}}+\beta {{a}_{n}}$可知:
取上极限得:$\alpha {{a}_{1}}=b+\beta {{a}_{2}}$
取下极限得:$\beta {{a}_{1}}=b+\alpha {{a}_{2}}$
由$\alpha \ne \beta $可知:${{a}_{1}}={{a}_{2}}$,即$\{{{a}_{n}}\}$存在
(2)若$\alpha =\beta $,则必要性不成立,充分性成立
只需取${{a}_{n}}={{(-\frac{\beta }{\alpha })}^{n}}={{(-1)}^{n}}$,此时$\{{{a}_{n}}\}$不收敛,但
${{b}_{n}}=\alpha {{a}_{n+1}}+\beta {{a}_{n}}=0$收敛
三、
(1)解:不存在,理由如下:
反证法:不妨设存在满足条件的$f(x)$
由$f(x)0$且$f(x)=f(f(x))0$,则$f(x)$严格单调递增
且$f(x)=f(f(x))f(0)$
令$a1$,则
$\int_{-a}^{0}{f(x)dx=f(0)-f(-a)f(0)[0-(-a)]=af(0)}$
于是$f(-a)f(0)-af(0)=(1-a)f(0)0$与$f(x)0$矛盾
于是不存在满足条件的$f(x)$
(2)只有零解,同样利用反证法
若$f(x)$有零点但不恒为$0$(没有零点则回归到(1))
记${{x}_{0}}=\sup \{x\in R|f(x)=0\}$
由$f(x)$有零点(集合非空)但不恒为$0$且单调不减(集合有上界)
则$f(x)=0,\forall x\in (-\infty ,{{x}_{0}})$,由上确界的定义以及$f(x)$的连续性可知:
$f({{x}_{0}})=0,f(x)0,x\in ({{x}_{0}},+\infty )$
于是$f(x)=0,x\in (-\infty ,{{x}_{0}})$,下证${{x}_{0}}=0$
若不然,不妨设${{x}_{0}}0=f({{x}_{0}})$,由局部有界性可知:存在${{\delta }_{0}}0$,
$\forall x\in ({{x}_{0}},{{x}_{0}}+{{\delta }_{0}}),f(x){{x}_{0}},f(x)=f(f(x))=0$
于是$f(x)=0,\forall x\in (-\infty ,{{x}_{0}}+{{\delta }_{0}})$矛盾
同理可证${{x}_{0}}0$的情况也不成立
于是${{x}_{0}}=0$
于是$f(x)0,x\in (0,+\infty )$,则$f(x)=f(f(x))0,x\in (0,+\infty )$
由微分中值定理可知,
$f(x)=f(f(x))=f(f(x))-f(0)=f(\eta )(f(x)-0)f(x)f(x),\eta \in (0,x)$
于是$f(x)1,x\in (0,+\infty )$,且$f(x)$在$x=0$右连续
于是$f(0)\ge 1$与$f(0)=0$矛盾
于是满足条件的$f(x)\equiv 0,x\in R$
四、解:
(1)不满足,考虑到不存在区域$D:\left| x \right|\le ,\left| y \right|\le b$,使得${{F}_{x}}$在$D$上连续且${{F}_{y}}(0,0)=0$
于是$F(x,y)$在$(0,0)$附近不满足$F(x,y)=0$的隐函数存在定理条件
从而$F(x,y)$在$(0,0)$附近不满足$F(x,y)=0$
(2)存在,由于$y$严格单调升,$\sin (\left| x \right|y)$在$(0,0)$附近关于$y$严格单调升,故在$(0,0)$附近关于$y$严格单调升。
而$F(x,y)$在$(0,0)$附近区域$D:\left| x \right|\le a,\left| y \right|\le b, $上连续。
固定$x=0$因$F(0,y) $在$ [-y,y] $上是的严格单调增函数,且$F(0,0)=0$
由函数的连续性知有$F(0,-b)0,F(0,b)0. $
考虑一元连续函数$F(x,-b) $
由于$F(0,-b)0, $所以存在${{\eta }_{1}}0$,使得$F(x,-b)0,x\in U(0,{{\eta }_{1}})$
明显也存在${{\eta }_{2}}0, $使得$F(x,b)0,x\in U(0,{{\eta }_{2}})$
令$\eta =\min ({{\eta }_{1}},{{\eta }_{2}}),$则对$\forall x\in U(0,\eta ) $有$F(x,-b)0,F(x,b)0$
设$\bar{x}$为$U(0,\eta ) $中任一点,则 $F(\bar{x},-b)\cdot F(\bar{x},b)0$
又由于作为$y$的函数$F(\bar{x},y) $严格增加
故必存在唯一的点$\bar{y}\in (-b,b) $使得$F(\bar{x},\bar{y})=0$。
由$\bar{x}$的任意性就确定了唯一的隐函数$y=f(x) $,显然$0=f(0) $。
再设${{x}_{1}}$是$U(0,\eta ) $的任意一点,记 ${{y}_{1}}=f({{x}_{1}}),$
对$\forall \varepsilon 0$作两根平行线$y={{y}_{1}}+\varepsilon ,y={{y}_{1}}-\varepsilon $
类似上面的证明知 存在$U({{x}_{1}},\delta ) $使得对$\forall x\in ({{x}_{1}},\delta ) $有$F(x,{{y}_{1}}+\varepsilon )0,F(x,{{y}_{1}}-\varepsilon )0. $由于作为$y$的连续函数$F(x,y) $的严格增加性,在$ ({{y}_{1}}-\varepsilon ,{{y}_{1}}+\varepsilon ) $上存在唯一的一个实点$y_{{}}^{{}}$使得$F(x,y)=0$,这说明对于$\forall x\in ({{x}_{1}},\delta ) $有$\left| f(x)-f({{x}_{1}}) \right|\varepsilon . $故$y=f(x) $在$U(0,\eta ) $内连续。
显然,$F(x,0)=0$于是由$F(x,y)=0$在$(0,0)$附近确定隐函数$y=f(x) $的唯一性知$y=f(x)\equiv 0$ ,其导函数${y}\equiv 0$
五、
(1)解:由于$\left| \frac{{{x}^{3}}-{{y}^{3}}}{{{x}^{2}}+{{y}^{2}}} \right|\le \left| \frac{{{x}^{3}}}{{{x}^{2}}+{{y}^{2}}} \right|+\left| \frac{{{y}^{3}}}{{{x}^{2}}+{{y}^{2}}} \right|\le \frac{1}{2}({{x}^{2}}+{{y}^{2}})$
而$\underset{(x,y)\to (0,0)}{\mathop{\lim }}\,\frac{1}{2}({{x}^{2}}+{{y}^{2}})=0$可知,$\underset{(x,y)\to (0,0)}{\mathop{\lim }}\,f(x,y)=\underset{(x,y)\to (0,0)}{\mathop{\lim }}\,\frac{{{x}^{3}}-{{y}^{3}}}{{{x}^{2}}+{{y}^{2}}}=0=f(0,0) $
从而$f(x,y)$在$(0,0)$处连续
(2)解:当$(x,y)\ne (0,0)$时,由于
${{f}_{x}}(0,0)=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=1,{{f}_{y}}(0,0)=\underset{\Delta y\to 0}{\mathop{\lim }}\,\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=-1$
可知$f(x,y)$的偏导数${{f}_{x}}(x,y),{{f}_{y}}(x,y)$在$(0,0)$存在
(3)解:考虑到
$\underset{\rho \to 0}{\mathop{\lim }}\,\frac{f(\Delta x,\Delta y)-f(0,0)-{{f}_{x}}(0,0)\Delta x-{{f}_{y}}(0,0)\Delta y}{\rho }=\underset{\rho \to 0}{\mathop{\lim }}\,\frac{\frac{{{(\Delta x)}^{3}}-{{(\Delta y)}^{3}}}{{{(\Delta x)}^{2}}+{{(\Delta y)}^{2}}}-\Delta x+\Delta y}{\sqrt{{{(\Delta x)}^{2}}+{{(\Delta y)}^{2}}}}$
令$\Delta y=k\Delta x$,于是
$\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{\frac{{{(\Delta x)}^{3}}-{{k}^{3}}{{(\Delta x)}^{3}}}{(1+{{k}^{2}}){{(\Delta x)}^{2}}}-\Delta x+k\Delta x}{\sqrt{1+{{k}^{2}}}\Delta x}=\frac{(1-{{k}^{3}})+(k-1)(1+{{k}^{2}})}{{{(1+{{k}^{2}})}^{\frac{3}{2}}}}=\frac{k-{{k}^{2}}}{{{(1+{{k}^{2}})}^{\frac{3}{2}}}}$与$k$有关
从而$f(x,y)$在$(0,0)$不可微
六、
(1)证明:当${{x}_{n}}=1+\frac{1}{n}$时,此时考虑$\sum\limits_{n=1}^{+\infty }{\frac{\ln (n+2)}{{{n}^{1+\frac{1}{n}}}}}$
由$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\frac{\ln (n+2)}{{{n}^{1+\frac{1}{n}}}}}{\frac{1}{n}}=+\infty $且$\sum\limits_{n=1}^{+\infty }{\frac{1}{n}}$发散
从而$\sum\limits_{n=1}^{+\infty }{\frac{\ln (1+nx\text{)}}{{{n}^{x}}}}$在$(1,+\infty )$上不一致收敛
(2)证明:由于$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\frac{\ln [1+(n+1)x]}{{{(n+1)}^{x}}}}{\frac{\ln (1+nx\text{)}}{{{n}^{x}}}}=\underset{n\to +\infty }{\mathop{\lim }}\,$
七、
(1)证明:由积分第二中值定理可知:
$\left| \int_{0}^{a}{f(x)\sin nxdx} \right|=\left| f(0)\int_{0}^{b}{\sin nxdx+f(a)\int_{b}^{a}{\sin nxdx}} \right|=\left| f(0)\frac{1-cosnb}{n}+f(a)\frac{\cos nb-\cos na}{n} \right|$$\le \frac{2\left| f(0) \right|+\left| f(a) \right|}{n}$
而$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{2\left| f(0) \right|+\left| f(a) \right|}{n}=0$可知$\underset{n\to +\infty }{\mathop{\lim }}\,\int_{0}^{a}{f(x)\sin nxdx}=0$
(2)不妨设$f(x)$在$[0,+\infty )$上单调递增,由
$\int_{0}^{+\infty }{f(x)\sin nxdx=}\int_{0}^{2\pi }{f(x)\sin nxdx+\int_{2\pi }^{3\pi }{f(x)\sin nxdx+\int_{3\pi }^{4\pi }{f(x)\sin nxdx+}\cdots }}$
$\ge \int_{0}^{2\pi }{f(x)\sin nxdx+}f(2\pi )\int_{2\pi }^{3\pi }{\sin nxdx}+f(3\pi )\int_{3\pi }^{4\pi }{\sin nxdx+\cdots }$
$=\int_{0}^{2\pi }{f(x)\sin nxdx}$
$\int_{0}^{+\infty }{f(x)\sin nxdx=}\int_{0}^{\pi }{f(x)\sin nxdx+\int_{\pi }^{2\pi }{f(x)\sin nxdx+\int_{2\pi }^{3\pi }{f(x)\sin nxdx+}\cdots }}$
$\le \int_{0}^{\pi }{f(x)\sin nxdx+}f(2\pi )\int_{\pi }^{2\pi }{\sin nxdx}+f(3\pi )\int_{2\pi }^{3\pi }{\sin nxdx+\cdots }$
$=\int_{0}^{\pi }{f(x)\sin nxdx}$
于是$\int_{0}^{2\pi }{f(x)\sin nxdx\le }\int_{0}^{+\infty }{f(x)\sin nxdx\le }\int_{0}^{\pi }{f(x)\sin nxdx}$
两边取极限,由夹逼原理及(1)知$\underset{n\to +\infty }{\mathop{\lim }}\,\int_{0}^{+\infty }{f(x)\sin nxdx=0}$
八、证明:$u\in {{C}^{2}}(\overline{\Omega })$及$u$为非常值函数知$\iint\limits_{\Omega }{\left[ {{\left( \frac{\partial u}{\partial x} \right)}^{2}}+{{\left( \frac{\partial u}{\partial y} \right)}^{2}} \right]}dxdy0$,令
$P=-u\frac{\partial u}{\partial y},Q=u\frac{\partial u}{\partial x},$则因$u\in {{C}^{2}}(\overline{\Omega }),$利用格林公式
$\iint\limits_{\Omega }{\left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dxdy=\iint\limits_{\Omega }{\left[ {{\left( \frac{\partial u}{\partial x} \right)}^{2}}+{{\left( \frac{\partial u}{\partial y} \right)}^{2}}+u\left( \frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}} \right) \right]}dxdy=\oint\limits_{\partial \Omega }{\left( -u\frac{\partial u}{\partial y} \right)dx+u\frac{\partial u}{\partial x}dy}}$
因$u\left| _{\partial \Omega }=0,\therefore \oint\limits_{\partial \Omega }{\left( -u\frac{\partial u}{\partial y} \right)dx+u\frac{\partial u}{\partial x}dy=0} \right. $
于是$\iint\limits_{\Omega }{u\left( \frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}} \right)dxdy=-\iint\limits_{\Omega }{\left[ {{\left( \frac{\partial u}{\partial x} \right)}^{2}}+{{\left( \frac{\partial u}{\partial y} \right)}^{2}} \right]}dxdy}0$