北大本科06数分期中试题(李伟固命题)

前些年在百度文库(http://wenku.baidu.com/link?url=Y_HDeeYMcyEGuUJWt3fNqC7N08AEqMVfleNVGcv7hC2t9EVO0-MFFHWuqnLYyiDJ4H7ATgg1fOQGN1Lta2RW4Z4pOrm6aZ468WkrTqNqZzG)找到一份李伟固命制的06年北大期中考试题，当时感觉难度稍大，正好结识了Veer大神，便把这份题发给他。事实证明，V神秒得还是很顺利！

1.给定实数$\lambda_i(1\leq i\leq n)$,满足$\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$.令$f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}}$.证明: $f^{(k)}(0)>0,k=1,2,3,\cdots$.

证明：令$$g\left( x \right) = \ln f\left( x \right) = - \ln \left( {1 - {\lambda _1}x} \right)-\ln \left( {1 - {\lambda _2}x} \right) -\cdots -\ln \left( {1 - {\lambda _n}x} \right)$$$\left( {x \in U\left( {0;\delta } \right)\text{使得对}\forall {\lambda _i},\text{有}1 - {\lambda _i}x > 0} \right)$,则

\[g\left( x \right) = \left( {\sum\limits_{i = 1}^n {{\lambda _i}} } \right)x + \left( {\frac{1}{2}\sum\limits_{i = 1}^n {\lambda _i^2} } \right){x^2} + \cdots + \left( {\frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} } \right){x^k} + \cdots \]由函数幂级数展开的唯一性可知${g^{\left( k \right)}}\left( 0 \right) = \frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {x \in U\left( {0;\delta } \right)} \right)$.

另一方面$f\left( x \right) = {e^{g\left( x \right)}}\left( {x \in U\left( {0;\delta } \right)} \right)$.首先注意到对任意可导函数$F(x)$,有${\left( {{e^{F\left( x \right)}}} \right)^\prime } = F'\left( x \right){e^{F\left( x \right)}}$.其次注意到对可导函数组$F_1,F_2,\cdots,F_s$,有${\left( {{F_1}{F_2}{F_3} \cdots {F_s}} \right)^\prime } = {{F'}_1}{F_2}{F_3} \cdots {F_s} + {F_1}{F_2}^\prime {F_3} \cdots {F_s} + \cdots + {F_1}{F_2}{F_3} \cdots {F_s}^\prime$,从而归纳可证

\[{f^{\left( k \right)}}\left( x \right) = {\left( {{e^{g\left( x \right)}}} \right)^{\left( k \right)}} = \left( {\sum\limits_{j \in {N_ + },{k_i} \in {N_ + }} {{g^{\left( {{k_1}} \right)}}\left( x \right){g^{\left( {{k_2}} \right)}}\left( x \right) \cdots {g^{\left( {{k_j}} \right)}}\left( x \right)} } \right){e^{g\left( x \right)}}.\]由${g^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$且$g(0)=0$,所以${f^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$.

补充：可知\[f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}} = \left( {1 + {\lambda _1}x + \lambda _1^2{x^2} + \cdots } \right)\left( {1 + {\lambda _2}x + \lambda _2^2{x^2} + \cdots } \right) \cdots \left( {1 + {\lambda _n}x + \lambda _n^2{x^2} + \cdots } \right).\]由幂级数的乘积公式归纳可得

\[{f^{\left( k \right)}}\left( 0 \right) = \sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}} \left( \text{其中}{{k_i} \in N} \right).\]若能通过$\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$得出$\sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}}>0$即可得到证明.

2.令$D = \left\{ {u = \left( {x,y} \right) \in {\mathbb{R}^2}\left| {\left\| u \right\| = \sqrt {{x^2} + {y^2}} \le \frac{1}{2}} \right.} \right\}$. $f(u)=f(x,y)$是全平面上的连续可微函数满足$\left\| {\nabla f\left( {0,0} \right)} \right\| = 1,\left\| {\nabla f\left( u \right) - \nabla f\left( v \right)} \right\| \le \left\| {u - v} \right\|$.那么对于任意的$u,v\in D$,证明函数$f|_D$在$D$中唯一点处达到其最大值.

证明：对$u\in D$,有$\left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| {u - \left( {0,0} \right)} \right\|$,即

\[1 - \left\| {\nabla f\left( u \right)} \right\| = \left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( u \right)} \right\| \le \left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| u \right\|,\]亦即$1 - \left\| u \right\| \le \left\| {\nabla f\left( u \right)} \right\|$,则$\nabla f\left( u \right) \ne \left( {0,0} \right)$,所以$f$的最大值只可能在边界上取得.

易知$f$在其边界上的函数可设为\[g\left( t \right) = f\left( {\frac{1}{2}\cos \theta ,\frac{1}{2}\sin \theta } \right),g'\left( \theta \right) = \frac{1}{2}\left( { - \sin \theta {f_x} + \cos \theta {f_y}} \right),\theta \in \left[ {0,2\pi } \right).\]现假设$f$在其边界上有两点取得最大值,不妨设为$\theta=\theta_1$和$\theta=\theta_2$,记${u_1} = \left( {\frac{1}{2}\cos {\theta _1},\frac{1}{2}\sin {\theta _1}} \right),{u_2} = \left( {\frac{1}{2}\cos {\theta _2},\frac{1}{2}\sin {\theta _2}} \right)$,则由$g'\left( {{\theta _1}} \right) = g'\left( {{\theta _2}} \right) = 0$可得$- \sin {\theta _1}{f_x}\left( {{u_1}} \right) + \cos {\theta _1}{f_y}\left( {{u_1}} \right) = 0$,即$\nabla f\left( {{u_1}} \right)$与$\left( { - \sin {\theta _1},\cos {\theta _1}} \right)$垂直即可.设$\nabla f\left( {{u_1}} \right) = {a_1}\left( {\cos {\theta _1},\sin {\theta _1}} \right)$,由于$f$在$u_1$处取得最大值,则$f$在$u_1$点沿方向$(\cos\theta_1,\sin \theta_1)$的方向导数需大于等于$0$,所以$a_1\geq 0$.

另一方面由$\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( {{u_1}} \right)} \right\| \le \left\| {\nabla f\left( {0,0} \right) - \nabla f\left( {{u_1}} \right)} \right\|$当且仅当${\nabla f\left( {{0,0}} \right)}$与${\nabla f\left( {{u_1}} \right)}$异向取等可知

\[\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {{u_1}} \right\| \le \left\| {\nabla f\left( {{u_1}} \right)} \right\| = \left| {{a_1}} \right|,\]

即$a_1\geq \frac12$.同理$\nabla f\left( {{u_2}} \right) = {a_2}\left( {\cos {\theta _2},\sin {\theta _2}} \right),a_2\geq \frac12$.由于不等式取等需要与${\nabla f\left( {0,0} \right)}$异向,故$a_1\geq\frac12,a_2\geq\frac12$中有一个是严格的.不妨设为$a_1>\frac12$,再设$\overrightarrow {{r_1}} = \left( {\cos {\theta _1},\sin {\theta _1}} \right),\overrightarrow {{r_2}} = \left( {\cos {\theta _2},\sin {\theta _2}} \right)$,则

\begin{align*}&{\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2}\\= &{\left\| {{a_1}\overrightarrow {{r_1}} - {a_2}\overrightarrow {{r_2}} } \right\|^2} - {\left\| {\frac{1}{2}\overrightarrow {{r_1}} - \frac{1}{2}\overrightarrow {{r_2}} } \right\|^2} = a_1^2 + a_2^2 - 2{a_1}{a_2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} - \left( {\frac{1}{4} - \frac{1}{2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} + \frac{1}{4}} \right)\\= &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right)\overrightarrow {{r_1}} \overrightarrow {{r_2}} \left( \text{由于}{\overrightarrow {{r_1}}\text{与} \overrightarrow {{r_2}} \text{不同向},\text{所以}\overrightarrow {{r_1}} \overrightarrow {{r_2}} < 1,\text{且}2{a_1}{a_2} - \frac{1}{2} > 0} \right)\\> &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right) = {\left( {{a_1} - {a_2}} \right)^2} \ge 0,\end{align*}

因此${\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2} > 0$,即$\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\| > \left\| {{u_1} - {u_2}} \right\|$,矛盾,从而$f$在边界上只有一点取得最大值,即函数$f|_D$在$D$中唯一点处达到其最大值.

3.讨论级数$\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}},\alpha\in \mathbb{R}$的收敛性.

解：(1)当$\alpha\leq0$时,我们可知$\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}$发散,从而级数$\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}}$发散.

(2)当$\alpha>0$时,由于$\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}} = 0$,我们讨论其前$2n$项和数列的收敛性即可,也就是级数$\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$的收敛性.

而

\begin{align*}&\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\= &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)\left( {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right) + \frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right) - \sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)} \right|\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\frac{{2\cos \left[ {\frac{{\ln \left( {2k - 1} \right) + \ln \left( {2k} \right)}}{2}} \right]\sin \left[ {\frac{{\ln \left( {2k - 1} \right) - \ln \left( {2k} \right)}}{2}} \right]}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha}}}.\end{align*}

已知$\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}} \sim \frac{1}{{{{\left( {2k} \right)}^{\alpha + 1}}}}$,从而$\sum\limits_{k = 1}^{ + \infty } {\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}}}$收敛.又$\sum\limits_{k = 1}^{ + \infty } {\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$是收敛的,故$\sum\limits_{n = 1}^{ + \infty } {\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|}$收敛,因此$\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$亦收敛.

综上所述, 当$\alpha\leq 0$时,级数发散;当$\alpha>0$时级数收敛.

3.求\[\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\}.\]

解：首先, $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^1 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \cos x = 0,\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \left( {\cos x + \frac{{\cos 2x}}{2}} \right) = - \frac{1}{2}$.

现记${f_n}\left( x \right) = \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}}$,则$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = 0 \ge - \frac{1}{2},\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = - \frac{1}{2} \ge - \frac{1}{2}$.

现假设$n=k-1(k\geq2,k\in\mathbb{N}_+)$时,有$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_{k - 1}}\left( x \right) \ge - \frac{1}{2}$成立;

当$n=k$时,注意到\[{f_n}^\prime \left( x \right) = \sum\limits_{k = 1}^n {\left( { - \sin kx} \right)} = \sum\limits_{i = 1}^k {\left( { - \sin ix} \right)} = \sum\limits_{i = 1}^k {\frac{{\cos \left( {i + \frac{1}{2}} \right)x - \cos \left( {i - \frac{1}{2}} \right)x}}{{2\sin \frac{x}{2}}}} ,\]

则\[{f_k}^\prime \left( x \right) = \frac{{\cos \left( {k + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = - \frac{{\sin \frac{{k + 1}}{2}x\sin \frac{k}{2}x}}{{\sin \frac{x}{2}}},x \in \left[ {0,\frac{\pi }{2}} \right].\]

由$f_k(x)$的连续性可知$\min f_k(x)$的点只可能在端点或稳定点,从而令$f'_k(x)=0$,则$x = \frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k},j \in \mathbb{Z}$且$\frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k} \in \left( {0,\frac{\pi }{2}} \right)$,而

\[{f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{{k + 1}}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2kj\pi }}{{k + 1}}}}{k} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k}.\]

由于$\frac{{2j\pi }}{{k + 1}} \in \left[ {0,\frac{\pi }{2}} \right]$,所以$\frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} > 0$,则\[{f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} \ge - \frac{1}{2}.\]

而${f_k}\left( {\frac{{2j\pi }}{k}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{k}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{k}} \right) + \frac{1}{k} \ge - \frac{1}{2}$,又

${f_k}\left( 0 \right) = \sum\limits_{i = 1}^k {\frac{1}{i}} > - \frac{1}{2},{f_k}\left( {\frac{\pi }{2}} \right) = \sum\limits_{i = 1}^k {\frac{{\cos i\frac{\pi }{2}}}{i}} = \left\{ \begin{array}{l}\frac{1}{2}\left[ { - 1 + \left({\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s + 1\\\frac{1}{2}\left[ { - 1 + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s\end{array} \right.$,从而${f_k}\left( {\frac{\pi }{2}} \right) \ge - \frac{1}{2}$.

综上归纳可知$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} \ge - \frac{1}{2}$,又$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = - \frac{1}{2}$,因此

\[\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\} = - \frac{1}{2}.\]

4.函数$f(x)$在$[0,1]$上二次可导, $f(0)=2,f'(0)=-2,f(1)=1$.证明存在$c\in (0,1)$,使得$f(c)f'(c)+f''(c)=0$.

证明：令$F\left( x \right) = \frac{1}{2}{f^2}\left( x \right) + f'\left( x \right)$,则$F(0)=\frac12\times 2^2-2=0$.现假设不存在$\xi\in (0,1]$使得$F(\xi)=0$,则$F$在$(0,1]$上不变号.倘若$\forall x\in (0,1]$都有$F(x)<0$,即$\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) < 0$,则$f'\left( x \right) < - \frac{1}{2}{f^2}\left( x \right) \le 0$,则$f$在$[0,1]$上单调递减,由$f(0)=2,f(1)=1$可知$1\leq f(x)\leq 2,f(x)\neq 0$,从而$\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} < - \frac{1}{2}$.由$f,f'$的连续性可知$\int_0^1 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx} < \int_0^1 {\left( { - \frac{1}{2}} \right)dx}$,即$\left. { - \frac{1}{{f\left( x \right)}}} \right|_0^1 < - \frac{1}{2}$,得到$-\frac12<-frac12$,矛盾.

倘若$\forall x\in (0,1]$都有$F(x)>0$,即$\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) > 0$,由于$f(1)=1\neq 0$,则$\forall x\in U^-(1,\delta)$,有$\int_x^1 {\frac{{f'\left( t \right)}}{{{f^2}\left( t \right)1}}dt} > \int_x^1 {\left( { - \frac{1}{2}} \right)dt}$,即$\left. { - \frac{1}{{f\left( t \right)}}} \right|_x^1 > \frac{{x - 1}}{2}$,从而$\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2}$.

当$x\in U^-(1,\delta)$时,有$\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2} > 0\left( {\delta \le 1} \right)$,故$f(x)$在$[0,1]$上不为$0$,则$\frac{1}{{f\left( 0 \right)}} > \frac{{0 + 1}}{2}$,则$\frac12>\frac12$,矛盾.

因此存在$\xi\in (0,1]$使得$F(\xi)=0$,从而$F(0)=F(\xi)=0$,由罗尔定理可知$\exists c\in (0,1)$,使得

\[F'\left( c \right) = f\left( c \right)f'\left( c \right) + f''\left( c \right) = 0.\]

5.$A$和$B$是自然数$\mathbb{N}$的两个无穷子集,满足$A\cap B=\text{空集},A\cup B=\mathbb{N}$,对于任意的自然数$c>0$,是否存在两个递增的数列$\{a_n\},\{b_n\},\{a_n\}\in A,\{b_n\}\in B$,使得$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$.

证明：对$\forall l\geq s\in \mathbb{N}_+$.记$s\sim l=\{s,s+1,\cdots,l\}$,则若对$\forall N\in \mathbb{N}_+$, $\exists n_2\geq n_1\geq N\geq c^2$有$x_{n_1}\sim x_{n_2}\subset B$.令$x_{n_1}=v_1,x_{n_2}=u_1$,则$v_1\sim u_1\subset B$.若$cv_1\sim cu_1+c[\sqrt{u_1}+1]\not\subset B$,则$\exists a\in A,b\in B$,使得$|a-cb|\leq c$或$|a-cb|\leq \sqrt{b}+1$.

若$cv_1\sim cu_1+c[\sqrt{u_1}+1]\subset B$,令$u_2=u_1+[\sqrt{u_1}+1]$,则$cv_1\sim cu_2\subset B$.以此类推,若始终成立$c^kv_1 \sim c^k u_{k+1}\subset B$,易知$u_{k+1}=u_k+[\sqrt{u_k}+1]\geq u_k+\sqrt{u_1}$,即$u_{k+1}-u_k\geq \sqrt{u_1}$,所以$u_k$有趋于无穷大的趋势,所以存在某个$k_0\in \mathbb{N}_+$,使得$\frac{u_{k_0}}{v_1}>c$.而$c^{k_0}v_1\sim c^{k_0} u_{k_0+1}\subset B$.令$x_{n_{k_0}}\in B$满足$c^{k_0}v_1\sim c^{k_0}u_{k_0+1}\subset c^{k_0}v_1\sim x_{n_{k_0}}\subset B,x_{n_{k_0}}+1\in A$,令$a_1=x_{n_{k_0}}+1$,则$c^{k_0}v_1<\frac{a_1}{c}\leq x_{n_{k_0}+1}$,则$\exists b_1\in B$,使得$\left| {\frac{{{a_1}}}{c} - {b_1}} \right| \le 1$,即$\left| {{a_1} - c{b_1}} \right| \le c$.

若不始终成立$c^kv_1\sim c^ku_{k+1}$,则$\exists b_1\in B,a_1\in A$有$|a_1-cb_1|\leq c$或$|a_1-cb_1|\leq \sqrt{b_1}+1$,即$|a_1-cb_1|\leq \sqrt{a_1}+1$.接着再取$n_1\geq n_2\geq N$,有$x_{n_2}\geq x_{n_1}>a_1,b_1$且$x_{n_1}\sim x_{n_2}\subset B$,则$\exists a_2\in A,b_2\in B$有$|a_2-cb_2|\leq \sqrt{b_2}+1,\cdots$,故存在递增数列$\{a_n\}\subset A,\{b_n\}\subset B$,有$|a_n-cb_n|\leq \sqrt{b_n}+1$,则$\left| {\frac{{{a_n}}}{{{b_n}}} - c} \right| \le \frac{1}{{\sqrt {{b_n}} }} + \frac{1}{{{b_n}}}$,所以$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$.