1计算.

(1)  \dps{ \lim_{x\to 0}\frac{\int_0^{x^2}\sin^\frac{3}{2}t\rd t}{\int_0^xt\sex{t-\sin t}\rd t} }<script type="math/tex">\dps{ \lim_{x\to 0}\frac{\int_0^{x^2}\sin^\frac{3}{2}t\rd t}{\int_0^xt\sex{t-\sin t}\rd t} }</script>.

解答: \bex \mbox{原式}&=&\lim_{x\to 0}\frac{2x\sin^\frac{3}{2}x^2}{x\sex{x-\sin x}}\\ &=&\lim_{x\to 0}\frac{2x^4}{x\cdot\frac{x^3}{6}}\\ &=&12.  \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\lim_{x\to 0}\frac{2x\sin^\frac{3}{2}x^2}{x\sex{x-\sin x}}\\ &=&\lim_{x\to 0}\frac{2x^4}{x\cdot\frac{x^3}{6}}\\ &=&12. \eex</script>

(2)  \dps{ \int \arctan \sqrt{x}\rd x.  }<script type="math/tex">\dps{ \int \arctan \sqrt{x}\rd x. }</script>.

解答: \bex \mbox{原式}&=&x\arctan\sqrt{x} -\int \frac{1}{1+x}\cdot \frac{1}{2}x^{-\frac{1}{2}}\cdot x\rd x\\ &=&x\arctan\sqrt{x} -\frac{1}{2}\int \frac{x^\frac{1}{2}}{1+x}\rd x\\ &=&x\arctan\sqrt{x} -\int \frac{y}{1+y^2}\cdot y \rd y\sex{y=\sqrt{x}}\\ &=&x\arctan\sqrt{x} -y+\arctan y+C\\ &=&(x+1)\arctan\sqrt{x}-\sqrt{x}+C. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&x\arctan\sqrt{x} -\int \frac{1}{1+x}\cdot \frac{1}{2}x^{-\frac{1}{2}}\cdot x\rd x\\ &=&x\arctan\sqrt{x} -\frac{1}{2}\int \frac{x^\frac{1}{2}}{1+x}\rd x\\ &=&x\arctan\sqrt{x} -\int \frac{y}{1+y^2}\cdot y \rd y\sex{y=\sqrt{x}}\\ &=&x\arctan\sqrt{x} -y+\arctan y+C\\ &=&(x+1)\arctan\sqrt{x}-\sqrt{x}+C. \eex</script>

(3)  \dps{ \int_1^2 \rd x\int_{\sqrt{x}}^x \frac{1}{y}e^{-x}\rd y +\int_2^4 \rd x\int_{\sqrt{x}}^2 \frac{1}{y}e^{-x}\rd y.  }<script type="math/tex">\dps{ \int_1^2 \rd x\int_{\sqrt{x}}^x \frac{1}{y}e^{-x}\rd y +\int_2^4 \rd x\int_{\sqrt{x}}^2 \frac{1}{y}e^{-x}\rd y. }</script>

解答: \bex \mbox{原式}&=&\int_1^2\rd y \int_y^{y^2}\frac{1}{y}e^{-x}\rd x\\ &=&\int_1^2\sex{y-1}e^{-y}\rd y\\ &=&\left. -ye^{-y}\right|^2_1\\ &=&-2e^{-2}+e^{-1}. \eex<script type="math/tex; mode=display">\bex \mbox{原式}&=&\int_1^2\rd y \int_y^{y^2}\frac{1}{y}e^{-x}\rd x\\ &=&\int_1^2\sex{y-1}e^{-y}\rd y\\ &=&\left. -ye^{-y}\right|^2_1\\ &=&-2e^{-2}+e^{-1}. \eex</script>

(4)  求抛物线 \dps{ y^2=4x }<script type="math/tex">\dps{ y^2=4x }</script> 与它在 (1,2)<script type="math/tex">(1,2)</script> 处的法线所围成的有限区域的面积.

解答: 在 (1,2)<script type="math/tex">(1,2)</script> 处, \frac{\rd x}{\rd y}=\frac{1}{2}\cdot 2=1, <script type="math/tex; mode=display"> \frac{\rd x}{\rd y}=\frac{1}{2}\cdot 2=1, </script> 而法线方程为 y-2=-\sex{x-1},\ x=3-y, <script type="math/tex; mode=display"> y-2=-\sex{x-1},\ x=3-y, </script> 其与抛物线交于 \sex{1,2},\ \sex{9,-6}. <script type="math/tex; mode=display"> \sex{1,2},\ \sex{9,-6}. </script> 于是所求面积 \bex S&=&\int_{-6}^2\rd y\int_{\frac{y^2}{4}}^{3-y}\rd x\\ &=&\int_{-6}^2\sex{3-y-\frac{y^2}{4}}\rd y\\ &=&\left. \sex{2y-\frac{y^2}{2}-\frac{y^3}{12}}\right|^2_{-6}\\ &=&3\cdot 8-\frac{1}{2}\cdot\sex{-32}-\frac{1}{12}\cdot224\\ &=&24+16-\frac{56}{3}\\ &=&\frac{64}{3}. \eex<script type="math/tex; mode=display">\bex S&=&\int_{-6}^2\rd y\int_{\frac{y^2}{4}}^{3-y}\rd x\\ &=&\int_{-6}^2\sex{3-y-\frac{y^2}{4}}\rd y\\ &=&\left. \sex{2y-\frac{y^2}{2}-\frac{y^3}{12}}\right|^2_{-6}\\ &=&3\cdot 8-\frac{1}{2}\cdot\sex{-32}-\frac{1}{12}\cdot224\\ &=&24+16-\frac{56}{3}\\ &=&\frac{64}{3}. \eex</script>

(5)  求幂级数 \dps{ \sum_{n=1}^\infty (-1)^{n-1}\frac{x^{2n-1}}{n} }<script type="math/tex">\dps{ \sum_{n=1}^\infty (-1)^{n-1}\frac{x^{2n-1}}{n} }</script> 的收敛域与和函数.

解答: (1)回忆对于级数 \bee\label{lz09sf_1_5_a_n} \sum_{n=1}^\infty a_n, \eee<script type="math/tex; mode=display">\bee\label{lz09sf_1_5_a_n} \sum_{n=1}^\infty a_n, \eee</script> 若 \limsup_{n\to\infty}\sev{\frac{a_{n+1}}{a_n}}<1, <script type="math/tex; mode=display"> \limsup_{n\to\infty}\sev{\frac{a_{n+1}}{a_n}}<1, </script> 则 \eqref{lz09sf_1_5_a_n}<script type="math/tex">\eqref{lz09sf_1_5_a_n}</script> 收敛. 实际上, \bex & &1>\limsup_{n\to\infty}\sev{\frac{a_{n+1}}{a_n}} =\inf_k\sup_{n\geq k}\sev{\frac{a_{n+1}}{a_n}}\\ &\ra&\mbox{对 }1>r>0,\ \exists\ N_r>0,\ s. t. \ k\geq N_r\rightsquigarrow \sev{\frac{a_{n+1}}{a_n}}\leq r\\ &\ra&\sev{a_n}\leq r\sev{a_{n-1}}\leq\cdots \leq r^{n-N_r}\sev{a_{N_r}}\\ &\ra&\sum_{n=1}^\infty a_n\mbox{ 有优级数 }a_1+\cdots+a_{N_r-1}+\sev{a_{N_r}}\sum_{n=N_r}^\infty r^{n-N_r}\\ &\ra&\sum_{n=1}^\infty a_n\mbox{ 收敛}. \eex<script type="math/tex; mode=display">\bex & &1>\limsup_{n\to\infty}\sev{\frac{a_{n+1}}{a_n}} =\inf_k\sup_{n\geq k}\sev{\frac{a_{n+1}}{a_n}}\\ &\ra&\mbox{对 }1>r>0,\ \exists\ N_r>0,\ s. t. \ k\geq N_r\rightsquigarrow \sev{\frac{a_{n+1}}{a_n}}\leq r\\ &\ra&\sev{a_n}\leq r\sev{a_{n-1}}\leq\cdots \leq r^{n-N_r}\sev{a_{N_r}}\\ &\ra&\sum_{n=1}^\infty a_n\mbox{ 有优级数 }a_1+\cdots+a_{N_r-1}+\sev{a_{N_r}}\sum_{n=N_r}^\infty r^{n-N_r}\\ &\ra&\sum_{n=1}^\infty a_n\mbox{ 收敛}. \eex</script>

(2)现求收敛域. 由于 \lim_{n\to\infty}\frac{\sev{x}^{2n+1}}{n+1}\cdot\frac{n}{\sev{x}^{2n-1}} =\sev{x}^2, <script type="math/tex; mode=display">\lim_{n\to\infty}\frac{\sev{x}^{2n+1}}{n+1}\cdot\frac{n}{\sev{x}^{2n-1}} =\sev{x}^2, </script> 而

(a)当 \sev{x}<1<script type="math/tex">\sev{x}<1</script> 时, 原幂级数绝对收敛;

(b)当 x=-1<script type="math/tex">x=-1</script> 时, 原幂级数 \dps{=\sum_{n=1}^\infty \frac{\sex{-1}^n}{n}}<script type="math/tex">\dps{=\sum_{n=1}^\infty \frac{\sex{-1}^n}{n}}</script>, 为条件收敛;

(c)当 x=1<script type="math/tex">x=1</script> 时, 原幂级数 \dps{=\sum_{n=1}^\infty \frac{\sex{-1}^{n-1}}{n}}<script type="math/tex">\dps{=\sum_{n=1}^\infty \frac{\sex{-1}^{n-1}}{n}}</script>, 亦为条件收敛;

(d)当 \sev{x}>1<script type="math/tex">\sev{x}>1</script> 时, 原幂级数发散.

(3)再求和函数. \bex \mbox{原幂级数} &=&\frac{2}{x}\sum_{n=1}^\infty \sex{-1}^n\frac{x^{2n}}{2n}\\ &=&\frac{-2}{x}\int_0^x \sez{\sum_{n=1}^\infty \sex{-1}^nt^{2n-1}}\rd t\\ &=&\frac{-2}{x}\int_0^x \frac{1}{t} \sum_{n=1}^\infty \sex{-t^2}^n\rd t\\ &=&\frac{-2}{x}s\int_0^x\frac{1}{t}\cdot \frac{t^2}{1+t^2}\rd t\\ &=&\frac{-1}{x}\int_0^x \frac{2t}{1+t^2}\rd t\\ &=&-\frac{1}{x}\ln\sex{1+t^2}|^x_0\\ &=&-\frac{\ln\sex{1+x^2}}{x},\ x\in (-1,1). \eex<script type="math/tex; mode=display">\bex \mbox{原幂级数} &=&\frac{2}{x}\sum_{n=1}^\infty \sex{-1}^n\frac{x^{2n}}{2n}\\ &=&\frac{-2}{x}\int_0^x \sez{\sum_{n=1}^\infty \sex{-1}^nt^{2n-1}}\rd t\\ &=&\frac{-2}{x}\int_0^x \frac{1}{t} \sum_{n=1}^\infty \sex{-t^2}^n\rd t\\ &=&\frac{-2}{x}s\int_0^x\frac{1}{t}\cdot \frac{t^2}{1+t^2}\rd t\\ &=&\frac{-1}{x}\int_0^x \frac{2t}{1+t^2}\rd t\\ &=&-\frac{1}{x}\ln\sex{1+t^2}|^x_0\\ &=&-\frac{\ln\sex{1+x^2}}{x},\ x\in (-1,1). \eex</script>

注记: 当 x=0<script type="math/tex">x=0</script> 时, \lim_{x\to 0}\frac{\ln\sex{1+x^2}}{x}=\lim_{x\to 0}x=0.  <script type="math/tex; mode=display"> \lim_{x\to 0}\frac{\ln\sex{1+x^2}}{x}=\lim_{x\to 0}x=0. </script>

(6)  计算曲面积分 \dps{ \int_L \sex{e^x\sin y-b\sex{x-y}}\rd x+\sex{e^x\cos y-ax}\rd y, }<script type="math/tex">\dps{ \int_L \sex{e^x\sin y-b\sex{x-y}}\rd x+\sex{e^x\cos y-ax}\rd y, }</script> 其中 L<script type="math/tex">L</script> 是从 (2a,0)<script type="math/tex">(2a,0)</script> 沿曲线 y=\sqrt{2ax-x^2}<script type="math/tex">y=\sqrt{2ax-x^2}</script> 到点 (0,0)<script type="math/tex">(0,0)</script> 的一段.

解答: (1)曲线 y=\sqrt{2ax-x^2}<script type="math/tex">y=\sqrt{2ax-x^2}</script>, 即 x^2-2ax+y^2=0,\ y\geq 0, <script type="math/tex; mode=display"> x^2-2ax+y^2=0,\ y\geq 0, </script> \sex{x-a}^2+y^2=a^2,\ y\geq 0.  <script type="math/tex; mode=display"> \sex{x-a}^2+y^2=a^2,\ y\geq 0. </script>

(2)记 P=e^x\sin y-b\sex{x-y},\ Q=e^x\cos y-ax, <script type="math/tex; mode=display"> P=e^x\sin y-b\sex{x-y},\ Q=e^x\cos y-ax, </script> 则 \frac{\p P}{\p y}=e^x\cos y-b,\ \frac{\p Q}{\p x}=e^x\cos y-a, <script type="math/tex; mode=display"> \frac{\p P}{\p y}=e^x\cos y-b,\ \frac{\p Q}{\p x}=e^x\cos y-a, </script> 而 \frac{\p Q}{\p x}-\frac{\p P}{\p y}=b-a.  <script type="math/tex; mode=display"> \frac{\p Q}{\p x}-\frac{\p P}{\p y}=b-a. </script>

(3)由 Green<script type="math/tex">Green</script> 公式, \bex \mbox{原曲线积分}&=&\iint\limits_{\sex{x-a}^2+y^2\leq a^2\atop y\geq 0} \sex{b-a}\rd x\rd y-\int_0^{2a}\sex{-bx}\rd x\\ &=&\sex{b-a}\cdot\frac{1}{2}\cdot \pi a^2 +b\cdot \frac{1}{2}\cdot \sex{2a}^2\\ &=&\frac{\sex{\pi+4}a^2b-\pi a^3}{2}. \eex<script type="math/tex; mode=display">\bex \mbox{原曲线积分}&=&\iint\limits_{\sex{x-a}^2+y^2\leq a^2\atop y\geq 0} \sex{b-a}\rd x\rd y-\int_0^{2a}\sex{-bx}\rd x\\ &=&\sex{b-a}\cdot\frac{1}{2}\cdot \pi a^2 +b\cdot \frac{1}{2}\cdot \sex{2a}^2\\ &=&\frac{\sex{\pi+4}a^2b-\pi a^3}{2}. \eex</script>

2  证明: \dps{\lim_{n\to\infty}\sin n}<script type="math/tex">\dps{\lim_{n\to\infty}\sin n}</script> 不存在.

证明:

(1)由于区间 \sez{2k\pi+\frac{\pi}{4},\ 2k\pi+\frac{3\pi}{4}},\ k=0,1,2,\cdots <script type="math/tex; mode=display"> \sez{2k\pi+\frac{\pi}{4},\ 2k\pi+\frac{3\pi}{4}},\ k=0,1,2,\cdots </script> 长度为 \dps{\frac{\pi}{2}>1}<script type="math/tex">\dps{\frac{\pi}{2}>1}</script>, 而存在整数 n_k\in \sez{2k\pi+\frac{\pi}{4},\ 2k\pi+\frac{3\pi}{4}}; <script type="math/tex; mode=display"> n_k\in \sez{2k\pi+\frac{\pi}{4},\ 2k\pi+\frac{3\pi}{4}}; </script>

(2)同理, 存在整数 m_k\in \sez{2k\pi+\frac{5\pi}{4},\ 2k\pi+\frac{7\pi}{4}}; <script type="math/tex; mode=display"> m_k\in \sez{2k\pi+\frac{5\pi}{4},\ 2k\pi+\frac{7\pi}{4}}; </script>

(3)于是若 \dps{\lim_{n\to\infty}\sin n=a}<script type="math/tex">\dps{\lim_{n\to\infty}\sin n=a}</script> 存在, 则 \sez{\frac{\sqrt{2}}{2},\ 1} \ni \lim_{k\to\infty}\sin n_k =a =\lim_{k\to\infty}\sin m_k \in \sez{-1,\ -\frac{\sqrt{2}}{2}}, <script type="math/tex; mode=display"> \sez{\frac{\sqrt{2}}{2},\ 1} \ni \lim_{k\to\infty}\sin n_k =a =\lim_{k\to\infty}\sin m_k \in \sez{-1,\ -\frac{\sqrt{2}}{2}}, </script> 矛盾!

3  设函数 f:[a,b]\to [a,b]<script type="math/tex">f:[a,b]\to [a,b]</script> 满足 \dps{\sev{f(x)-f(y)}\leq L\sev{x-y}^\alpha}<script type="math/tex">\dps{\sev{f(x)-f(y)}\leq L\sev{x-y}^\alpha}</script>, 其中 L<script type="math/tex">L</script>, \alpha<script type="math/tex">\alpha</script> 为正常数. 证明:

(1)当 \alpha>1<script type="math/tex">\alpha>1</script> 时, f(x)<script type="math/tex">f(x)</script> 恒为常数;

(2)当 L<1<script type="math/tex">L<1</script>, \alpha=1<script type="math/tex">\alpha=1</script> 时, 存在唯一的 \xi\in [a,b]<script type="math/tex">\xi\in [a,b]</script>, 使得 f\sex{\xi}=\xi<script type="math/tex">f\sex{\xi}=\xi</script>.

证明:

(1)当 \alpha>1<script type="math/tex">\alpha>1</script> 时, 由 0\leq \sev{\frac{f(y)-f(x)}{y-x}} \leq L\sev{y-x}^{\alpha-1} \to 0,\ \sex{y\to x}, <script type="math/tex; mode=display"> 0\leq \sev{\frac{f(y)-f(x)}{y-x}} \leq L\sev{y-x}^{\alpha-1} \to 0,\ \sex{y\to x}, </script> 知 f‘(x)=0,\ \forall\ x\in [a,b] <script type="math/tex; mode=display"> f‘(x)=0,\ \forall\ x\in [a,b] </script> (<script type="math/tex">(</script>这里左端点左导数, 右端点右导数)<script type="math/tex">)</script>, 于是 f<script type="math/tex">f</script> 恒为常数.

(2)当 L<1<script type="math/tex">L<1</script>, \alpha=1<script type="math/tex">\alpha=1</script> 时, 任取 \xi_0\in [a,b]<script type="math/tex">\xi_0\in [a,b]</script>, 做数列 \bee\label{lz09sf_3_xi_n} \xi_n=f\sex{\xi_{n-1}},\ n\geq 1.  \eee<script type="math/tex; mode=display">\bee\label{lz09sf_3_xi_n} \xi_n=f\sex{\xi_{n-1}},\ n\geq 1. \eee</script> 由 \sev{\xi_n-\xi_{n-1}}\leq L\sev{\xi_{n-1}-\xi_{n-2}}, <script type="math/tex; mode=display"> \sev{\xi_n-\xi_{n-1}}\leq L\sev{\xi_{n-1}-\xi_{n-2}}, </script> 知 \sed{\xi_n}_{n=1}^\infty<script type="math/tex">\sed{\xi_n}_{n=1}^\infty</script> 是压缩数列, 而极限存在, 设为 \xi<script type="math/tex">\xi</script>. 于 \eqref{lz09sf_3_xi_n}<script type="math/tex">\eqref{lz09sf_3_xi_n}</script> 中令 n\to\infty<script type="math/tex">n\to\infty</script>, 利用 f<script type="math/tex">f</script> 之连续性, 有 \xi=f\sex(\xi). <script type="math/tex; mode=display"> \xi=f\sex(\xi). </script> 下证唯一性. 设 \eta<script type="math/tex">\eta</script> 也满足 \eta=f(\eta)<script type="math/tex">\eta=f(\eta)</script>, 则 \sev{\xi-\eta} =\sev{f\sex{\xi}-f\sex{\eta}} <L\sev{\xi-\eta}, <script type="math/tex; mode=display"> \sev{\xi-\eta} =\sev{f\sex{\xi}-f\sex{\eta}} \xi=\eta<script type="math/tex">\xi=\eta</script>.

4  证明: 有界函数 f(x)<script type="math/tex">f(x)</script> 在区间 I<script type="math/tex">I</script> 上一致连续的充分必要条件是对任给的 \ve>0<script type="math/tex">\ve>0</script> 和 x,y\in I<script type="math/tex">x,y\in I</script>, 总存在正数 M<script type="math/tex">M</script>, 使得当 \dps{\sev{\frac{f(x)-f(y)}{x-y}}>M}<script type="math/tex">\dps{\sev{\frac{f(x)-f(y)}{x-y}}>M}</script> 时就有 \sev{f(x)-f(y)}<\ve<script type="math/tex">\sev{f(x)-f(y)}<\ve</script>.

证明:

(1)这一题目大概是说 f<script type="math/tex">f</script> 一致连续等价于 f<script type="math/tex">f</script> 在大范围内不能抖动得太厉害 (<script type="math/tex">(</script>否则太伤身体了)<script type="math/tex">)</script>.

(2)原题目没有对 f<script type="math/tex">f</script> 加{\bf 有界}条件. 一个简单的例子 f(x)=x,\ x\in \bbR<script type="math/tex">f(x)=x,\ x\in \bbR</script> 说明有界是必须的.

(3)必要性. 用反证法. 若结论不成立, 则 \exists\ \ve_0>0,\ \forall\ n\in\bbN, \ \exists\ \sed{x_n},\ \sed{y_n},\ s. t. <script type="math/tex">\exists\ \ve_0>0,\ \forall\ n\in\bbN, \ \exists\ \sed{x_n},\ \sed{y_n},\ s. t. </script> \sev{\frac{f(x_n)-f(y_n)}{x_n-y_n}}>n,\ \sev{f(x_n)-f(y_n)}>\ve_0.  <script type="math/tex; mode=display"> \sev{\frac{f(x_n)-f(y_n)}{x_n-y_n}}>n,\ \sev{f(x_n)-f(y_n)}>\ve_0. </script> 即有 \sev{x_n-y_n}<\frac{2K}{n},\ \sex{K=\sup_{x\in I}\sev{f(x)}}, <script type="math/tex; mode=display"> \sev{x_n-y_n}<\frac{2K}{n},\ \sex{K=\sup_{x\in I}\sev{f(x)}}, </script> 这与 f<script type="math/tex">f</script> 一致连续矛盾.

(4)充分性. 同样是用反证法. 若结论不成立, 则 \exists\ \ve_0>0,\ \sed{x_n},\ \sed{y_n},\ s. t. <script type="math/tex">\exists\ \ve_0>0,\ \sed{x_n},\ \sed{y_n},\ s. t. </script> \sev{x_n-y_n}<\frac{1}{n},\ \sev{f(x_n)-f(y_n)}>\ve_0, <script type="math/tex; mode=display"> \sev{x_n-y_n}<\frac{1}{n},\ \sev{f(x_n)-f(y_n)}>\ve_0, </script> 即有 \sev{\frac{f(x_n)-f(y_n)}{x_n-y_n}}>n\ve_0.  <script type="math/tex; mode=display"> \sev{\frac{f(x_n)-f(y_n)}{x_n-y_n}}>n\ve_0. </script> 由假设, 对 \dps{\frac{\ve_0}{2}>0}<script type="math/tex">\dps{\frac{\ve_0}{2}>0}</script>, 只需 n<script type="math/tex">n</script> 充分大, 就有 \sev{f(x_n)-f(y_n)}<\frac{\ve_0}{2}, <script type="math/tex; mode=display"> \sev{f(x_n)-f(y_n)}<\frac{\ve_0}{2}, </script> 矛盾!

5  设 f:\bbR^2\to \bbR^2<script type="math/tex">f:\bbR^2\to \bbR^2</script> 是连续映射, 若对 \bbR^2<script type="math/tex">\bbR^2</script> 中任何有界闭集 K<script type="math/tex">K</script>, f^{-1}\sex{K}<script type="math/tex">f^{-1}\sex{K}</script> 均是有界的, 证明 f\sex{\bbR^2}<script type="math/tex">f\sex{\bbR^2}</script> 是闭集.

证明:

(1)古怪得很.

(2)用反证法. 若 f\sex{\bbR^2}<script type="math/tex">f\sex{\bbR^2}</script> 不是闭集, 则 \exists\ \sed{x_n}\subset\bbR^2,\ s. t. \ f(x_n)\to y\notin f\sex{\bbR^2}. <script type="math/tex; mode=display"> \exists\ \sed{x_n}\subset\bbR^2,\ s. t. \ f(x_n)\to y\notin f\sex{\bbR^2}. </script> 而集合 A=\sed{f(x_n)}_{n=1}^\infty\cup \sed{y} <script type="math/tex; mode=display"> A=\sed{f(x_n)}_{n=1}^\infty\cup \sed{y} </script> 作为 \bbR^2<script type="math/tex">\bbR^2</script> 中的有界闭集 (<script type="math/tex">(</script>有界是因为极限存在, 而闭性是由于极限唯一)<script type="math/tex">)</script>, 其原像 f^{-1}\sex{A}<script type="math/tex">f^{-1}\sex{A}</script> 是有界的. 现因 x_n\in f^{-1}(A), <script type="math/tex; mode=display"> x_n\in f^{-1}(A), </script> 及 Weierstrass<script type="math/tex">Weierstrass</script> 聚点定理, \exists\ \sed{n_k}\subset \sed{n},\ s. t. \ x_{n_k}\to x.  <script type="math/tex; mode=display"> \exists\ \sed{n_k}\subset \sed{n},\ s. t. \ x_{n_k}\to x. </script> 由 f<script type="math/tex">f</script> 之连续性, \lim_{k\to\infty}f\sex{x_{n_k}}=f(x)=y\in f\sex{\bbR^2}, <script type="math/tex; mode=display"> \lim_{k\to\infty}f\sex{x_{n_k}}=f(x)=y\in f\sex{\bbR^2}, </script> 矛盾!

6  证明二元函数 f(x,y)=\sqrt{xy}<script type="math/tex">f(x,y)=\sqrt{xy}</script> 在点 (0,0)<script type="math/tex">(0,0)</script> 处连续, f_x(0,0)<script type="math/tex">f_x(0,0)</script>, f_y(0,0)<script type="math/tex">f_y(0,0)</script> 存在, 但 f(x,y)<script type="math/tex">f(x,y)</script> 在点 (0,0)<script type="math/tex">(0,0)</script> 处不可微.

证明:

(1)f(x,y)=\sqrt{xy}<script type="math/tex">f(x,y)=\sqrt{xy}</script> 在点 (0,0)<script type="math/tex">(0,0)</script> 处连续. 事实上, \forall\ \ve>0,\ \exists\ \delta=\ve>0,\ s. t. <script type="math/tex">\forall\ \ve>0,\ \exists\ \delta=\ve>0,\ s. t. </script> \left.  \ba{ll} 0<x<\delta\\ 0<y<\delta \ea\right\}\ra \sev{f(x,y)-f(0,0)}=\sqrt{xy}<\ve.  <script type="math/tex; mode=display"> \left. \ba{ll} 0

(2)f_x(0,0)<script type="math/tex">f_x(0,0)</script>, f_y(0,0)<script type="math/tex">f_y(0,0)</script> 存在. 由 x>0\ra \sev{\frac{f(x,0)-f(0,0)}{x}}=0 =\sev{\frac{f(0,y)-f(0,0)}{y}}, <script type="math/tex; mode=display"> x>0\ra \sev{\frac{f(x,0)-f(0,0)}{x}}=0 =\sev{\frac{f(0,y)-f(0,0)}{y}}, </script> 知 f_x(0,0)=0=f_y(0,0). <script type="math/tex; mode=display"> f_x(0,0)=0=f_y(0,0). </script>

(3)f(x,y)<script type="math/tex">f(x,y)</script> 在点 (0,0)<script type="math/tex">(0,0)</script> 处不可微. 这是因为 \bex & &\frac{1}{n}\to 0\\ &\ra&\sev{\frac{\dps{f\sex{\frac{1}{n},\frac{1}{n}}-f(0,0)}} {\sqrt{\sex{\frac{1}{n}}^2+\sex{\frac{1}{n}}^2}}} =\frac{\dps{\frac{1}{n}}}{\sqrt{2}\dps{\frac{1}{n}}} =\frac{1}{\sqrt{2}}\nrightarrow 0.  \eex<script type="math/tex; mode=display">\bex & &\frac{1}{n}\to 0\\ &\ra&\sev{\frac{\dps{f\sex{\frac{1}{n},\frac{1}{n}}-f(0,0)}} {\sqrt{\sex{\frac{1}{n}}^2+\sex{\frac{1}{n}}^2}}} =\frac{\dps{\frac{1}{n}}}{\sqrt{2}\dps{\frac{1}{n}}} =\frac{1}{\sqrt{2}}\nrightarrow 0. \eex</script>

7  设 \dps{f(x)=\sum_{n=1}^\infty \frac{1}{2^n+x}}<script type="math/tex">\dps{f(x)=\sum_{n=1}^\infty \frac{1}{2^n+x}}</script>, 证明:

(1)f(x)<script type="math/tex">f(x)</script> 在 [0,\infty)<script type="math/tex">[0,\infty)</script> 上可导, 且一致连续;

(2)反常积分 \dps{\int_0^\infty f(x)\rd x}<script type="math/tex">\dps{\int_0^\infty f(x)\rd x}</script> 发散.

证明:

(1)f(x)<script type="math/tex">f(x)</script> 在 [0,\infty)<script type="math/tex">[0,\infty)</script> 上可导, 且一致连续. 记 u_n=\frac{1}{2^n+x}, <script type="math/tex; mode=display"> u_n=\frac{1}{2^n+x}, </script> 则 u_n‘=-\frac{1}{\sex{2^n+x}^2}, <script type="math/tex; mode=display"> u_n‘=-\frac{1}{\sex{2^n+x}^2}, </script> 而(<script type="math/tex">(</script>表递进)<script type="math/tex">)</script>

(a)\dps{\sum_{n=1}^\infty u_n}<script type="math/tex">\dps{\sum_{n=1}^\infty u_n}</script> 收敛,

(b)u_n‘<script type="math/tex">u_n‘</script> 连续,

(c)\dps{\sum_{n=1}^\infty u_n‘}<script type="math/tex">\dps{\sum_{n=1}^\infty u_n‘}</script> 一致收敛, 由逐项求导, f‘(x)=-\sum_{n=1}^\infty \frac{1}{\sex{2^n+x}^2},\ \forall\ x\geq 0.  <script type="math/tex; mode=display"> f‘(x)=-\sum_{n=1}^\infty \frac{1}{\sex{2^n+x}^2},\ \forall\ x\geq 0. </script> 又 \sev{f‘(x)} \leq \sum_{n=1}^\infty \frac{1}{2^{2n}}=\frac{1}{4}\cdot\frac{1}{\dps{1-\frac{1}{4}}}=\frac{1}{3}, <script type="math/tex; mode=display"> \sev{f‘(x)} \leq \sum_{n=1}^\infty \frac{1}{2^{2n}}=\frac{1}{4}\cdot\frac{1}{\dps{1-\frac{1}{4}}}=\frac{1}{3}, </script> 知 f<script type="math/tex">f</script> 一致连续.

(2)反常积分\dps{\int_0^\infty f(x)\rd x}<script type="math/tex">\dps{\int_0^\infty f(x)\rd x}</script> 发散. 事实上, 由 \bex \sev{\int_{2^{k-1}}^{2^k}\sum_{k=1}^\infty \frac{1}{2^n+x}\rd x} &\geq& \int_{2^{k-1}}^{2^k}\frac{1}{2^k+x}\rd x\\ &\geq& \frac{2 \cdot 2^{k-1}}{2\cdot 2^k}\\ &=&\frac{1}{2}>0,\ \forall\ k\geq 1, \eex<script type="math/tex; mode=display">\bex \sev{\int_{2^{k-1}}^{2^k}\sum_{k=1}^\infty \frac{1}{2^n+x}\rd x} &\geq& \int_{2^{k-1}}^{2^k}\frac{1}{2^k+x}\rd x\\ &\geq& \frac{2 \cdot 2^{k-1}}{2\cdot 2^k}\\ &=&\frac{1}{2}>0,\ \forall\ k\geq 1, \eex</script> 及 Cauchy<script type="math/tex">Cauchy</script> 收敛准则, 知所讨论之反常积分确实反常, 的确发散.