高数强化NO5|极限的概念与性质

$$\begin{array}{rl}& \text{解1：当}n\text{充分大时，}{a}_n\text{与}\alpha \text{无限接近，故排除}B.\\ & \text{令}{a}_n=\alpha +\frac{2}{n}\text{，故排除}D；a_n=\alpha -\frac{2}{n}\text{，故排除}C\text{，故选}A.\end{array}$$
$$\begin{array}{rl}& \text{解2：}\underset{n\to \infty }{\lim }{a}_n=\alpha ,\underset{n\to \infty }{\lim }|a_n|=|\alpha |>\frac{|\alpha |}{2},\text{由极限的保号性可知：当}n\text{充分大时，}|a_n|>\frac{|\alpha |}{2},\text{故选}A.\\ & \\ & \text{解3：}\exists N,\text{当}n>N\text{时，}\forall \epsilon >0,\text{有}||a_n|-|\alpha ||<\epsilon ,\text{即}|\alpha |-\epsilon <|a_n|<|\alpha |+\epsilon \\ & \text{令}\epsilon =\frac{|\alpha |}{2},\text{故}|a_n|>\frac{|\alpha |}{2},\text{故选}A\end{array}$$
$$\begin{array}{rl}& \text{改题}\\ & \text{设}\underset{n\to \infty }{\lim }{a}_n=a\text{且}a\ne 0,\text{则当}n\text{充分大时，有（}\boxed{B}\text{）}\\ & (A)|a_n|>\frac{3}{2}|a|(B)|a_n|<\frac{3}{2}|a|\\ & (C)a_n>a-\sin \frac{1}{n}(D)a_n<a+\sin \frac{1}{n}\end{array}$$

$$\begin{array}{rl}& \text{解：}(A)(B)\text{：当}n\text{充分大时，}{a}_n<b_n,b_n<c_n\\ & \text{也即}\exists N>0,\text{当}n>N\text{时，有}{a}_n<b_n,b_n<c_n.\\ & \text{但不是任意的}n,\text{故排除}AB.\\ & \\ & (C)0\cdot \infty \text{是未定式，极限不一定存在，有可能存在。例如}{a}_n=\frac{1}{n},c_n=n,\underset{n\to \infty }{\lim }\frac{1}{n}\cdot n=1.\\ & \\ & (D)\underset{n\to \infty }{\lim }{b}_n{c}_n=\underset{n\to \infty }{\lim }{b}_n\cdot \underset{n\to \infty }{\lim }{c}_n=1\cdot \underset{n\to \infty }{\lim }{c}_n=\infty \end{array}$$
$$\begin{array}{rl}& \text{【小结】1、由极限的四则运算法则我们可以得到如下结论：}\\ & \text{收敛+收敛=收敛，收敛+发散=发散，发散+发散=？}\\ & \text{收敛×收敛=收敛，收敛×发散=}\{\begin{array}{l}\text{发散，收敛}\ne 0\\ \text{？，收敛}=0\end{array}\text{，发散×发散=？}\\ & \\ & 2、\text{该结论可以类比推广到所有由极限定义的概念}\\ & （1）\text{连续性}\\ & \text{连续+连续=连续，连续+间断=间断，间断+间断=？}\\ & \text{连续×连续=连续，连续×间断=}\{\begin{array}{l}\text{间断，连续}\ne 0\\ \text{？，连续}=0\end{array}\text{，间断×间断=？}\\ & \\ & （2）\text{可导性}\\ & \text{可导+可导=可导，可导+不可导=不可导，不可导+不可导=？}\\ & \text{可导×可导=可导，可导×不可导=}\{\begin{array}{l}\text{不可导，可导}\ne 0\\ \text{？，可导}=0\end{array}\text{，不可导×不可导=？}\\ & \\ & （3）\text{可积性}\\ & \text{可积+可积=可积，可积+不可积=不可积，不可积+不可积=？}\\ & \text{可积×可积=可积，可积×不可积=}\{\begin{array}{l}\text{不可积，可积}\ne 0\\ \text{？，可积}=0\end{array}\text{，不可积×不可积=？}\\ & \\ & （4）\text{级数的敛散性}\\ & \text{收敛+收敛=收敛，收敛+发散=发散，发散+发散=？}\\ & \\ & （5）\text{有界性}\\ & \text{有界+有界=有界，有界+无界=无界，无界+无界=？}\\ & \text{有界×有界=有界，收敛×无界=}\{\begin{array}{l}\text{无界，收敛}\ne 0\\ \text{？，收敛}=0\end{array}\text{，无界×无界=？}\end{array}$$

$$\begin{array}{rl}& \frac{1}{f(x)}\cdot \phi (x)\\ & \text{（连续(不等于0)*间断）=间断}\\ & 选D\end{array}$$
$$\begin{array}{rl}& (A)f(x)\text{的值域未必在}\phi (x)\text{的取间断点的点.}\\ & \text{eg:}\phi (x)=\{\begin{array}{ll}1& x\ge 0\\ -1& x<0\end{array},f(x)=e^x,\phi (f(x))=1\text{排除}A\\ & \\ & (B)\phi (x)=\{\begin{array}{ll}1& x\ge 0\\ -1& x<0\end{array},{\phi }^2(x)=1.\\ & (C)f(\phi (x))\text{不一定有间断点,}\\ & \text{例1：}f(x)=e^x,\phi (x)=\{\begin{array}{ll}1& x\ge 0\\ -1& x<0\end{array}\\ & f(\phi (x))=\{\begin{array}{ll}e& x\ge 0\\ e^{-1}& x<0\end{array}\text{有间断点；}\\ & \text{例2：令}f(x)=e^{|x|},\phi (x)=\{\begin{array}{ll}1& x\ge 0\\ -1& x<0\end{array}\\ & f(\phi (x))=e^1=e\text{无间断点.}\end{array}$$
$$\begin{array}{rl}& \text{问：}(f(x){)}^2\text{必连续？}\boxed{\surd }\\ & f(x)\phi (x)\text{必有间断点？}\boxed{\surd }\\ & \phi (x)\cdot \sin f(x)\text{必有间断点？}\boxed{\times }\text{因为}\sin f(x)\text{可能为}0\end{array}$$

解1：图像法，二阶导大于0，凹函数若u1>u2① u1,u2均在左边(收敛)，可能在右边（发散）② u1在左边，u2在右边，{un}发散(A)(B)均错误.若u1<u2① u1在左边，u2在右边，{un}发散② u1在右边，u2在右边，{un}发散故选(D) \begin{aligned} &\text{解1：图像法，二阶导大于0，凹函数} \\ &\text{若}u_1 > u_2 \\ &①\ u_1,u_2\text{均在左边(收敛)，可能在右边（发散）} \\ &②\ u_1\text{在左边，}u_2\text{在右边，}\{u_{n}\}\text{发散} \\ &(A)(B)\text{均错误.} \\ & \\ &\text{若}u_1 < u_2 \\ &①\ u_1\text{在左边，}u_2\text{在右边，}\{u_{n}\}\text{发散} \\ &②\ u_1\text{在右边，}u_2\text{在右边，}\{u_{n}\}\text{发散} \\ & \\ &\text{故选}(D) \end{aligned} ​解1：图像法，二阶导大于0，凹函数若u1​>u2​① u1​,u2​均在左边(收敛)，可能在右边（发散）② u1​在左边，u2​在右边，{un​}发散(A)(B)均错误.若u1​<u2​① u1​在左边，u2​在右边，{un​}发散② u1​在右边，u2​在右边，{un​}发散故选(D)​
解2：un=f(n),由拉格朗日可知 un+1−un=f(n+1)−f(n)=(n+1−n)f′(ξn)=f′(ξn)ξn∈(n,n+1)若f′′(x)>0,则f′(x)单调递增.f′(ξ1)<f′(ξ2)<f′(ξ3)<⋯<f′(ξn),故u2−u1<u3−u2<u4−u3<⋯<un+1−un若u1>u2,则u2−u1<0, un+1−un符号不确定；若u1<u2,则u2−u1>0, un+1−un>0, {un}单调递增.un+1=u1+∑i=1n(ui+1−ui)=u1+∑i=1nf′(ξi)>u1+∑i=1nf′(ξ1)=u1+nf′(ξ1)=u1+n(u2−u1)lim⁡n→∞[u1+n(u2−u1)]=+∞, 故lim⁡n→∞un+1=+∞, {un}发散. \begin{aligned} &\text{解2：}u_n = f(n),\text{由拉格朗日可知 }u_{n+1}-u_n = f(n+1)-f(n) = (n+1-n)f'(\xi_n) = f'(\xi_n) \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\xi_n \in (n,n+1) \\ &\text{若}f''(x)>0,\text{则}f'(x)\text{单调递增}. \\ &f'(\xi_1) < f'(\xi_2) < f'(\xi_3) < \dots < f'(\xi_n),\text{故}u_2-u_1 < u_3-u_2 < u_4-u_3 < \dots < u_{n+1}-u_n \\ & \\ &\text{若}u_1 > u_2,\text{则}u_2-u_1 < 0,\ u_{n+1}-u_n\text{符号不确定}； \\ &\text{若}u_1 < u_2,\text{则}u_2-u_1 > 0,\ u_{n+1}-u_n > 0,\ \{u_n\}\text{单调递增}. \\ & \\ &u_{n+1} = u_1 + \sum_{i=1}^n (u_{i+1}-u_i) = u_1 + \sum_{i=1}^n f'(\xi_i) > u_1 + \sum_{i=1}^n f'(\xi_1) = u_1 + nf'(\xi_1) = u_1 + n(u_2-u_1) \\ &\lim_{n \to \infty} \left[ u_1 + n(u_2-u_1) \right] = +\infty,\ \text{故}\lim_{n \to \infty} u_{n+1} = +\infty,\ \{u_n\}\text{发散}. \end{aligned} ​解2：un​=f(n),由拉格朗日可知 un+1​−un​=f(n+1)−f(n)=(n+1−n)f′(ξn​)=f′(ξn​)ξn​∈(n,n+1)若f′′(x)>0,则f′(x)单调递增.f′(ξ1​)<f′(ξ2​)<f′(ξ3​)<⋯<f′(ξn​),故u2​−u1​<u3​−u2​<u4​−u3​<⋯<un+1​−un​若u1​>u2​,则u2​−u1​<0, un+1​−un​符号不确定；若u1​<u2​,则u2​−u1​>0, un+1​−un​>0, {un​}单调递增.un+1​=u1​+i=1∑n​(ui+1​−ui​)=u1​+i=1∑n​f′(ξi​)>u1​+i=1∑n​f′(ξ1​)=u1​+nf′(ξ1​)=u1​+n(u2​−u1​)n→∞lim​[u1​+n(u2​−u1​)]=+∞, 故n→∞lim​un+1​=+∞, {un​}发散.​
改题：二阶导小于0，解：un=f(n),由拉格朗日可知 un+1−un=f(n+1)−f(n)=(n+1−n)f′(ξn)=f′(ξn)ξn∈(n,n+1)若f′′(x)<0,则f′(x)单调递减f′(ξ1)>f′(ξ2)>f′(ξ3)>⋯>f′(ξn),故u2−u1>u3−u2>u4−u3>⋯>un+1−un若u1<u2,则u2−u1>0, un+1−un符号不确定；若u1>u2,则u2−u1<0, un+1−un<0, {un}单调递减.un+1=u1+∑i=1n(ui+1−ui)=u1+∑i=1nf′(ξi)<u1+∑i=1nf′(ξ1)=u1+nf′(ξ1)=u1+n(u2−u1)lim⁡n→∞[u1+n(u2−u1)]=−∞, 故lim⁡n→∞un+1=−∞, {un}发散，选(B) \begin{aligned} &\text{改题：二阶导小于0，解：}\\&u_n = f(n),\text{由拉格朗日可知 }u_{n+1}-u_n = f(n+1)-f(n) = (n+1-n)f'(\xi_n) = f'(\xi_n) \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\xi_n \in (n,n+1) \\ &\text{若}f''(x)<0,\text{则}f'(x)\text{单调递减} \\ &f'(\xi_1) > f'(\xi_2) > f'(\xi_3) > \dots > f'(\xi_n),\text{故}u_2-u_1 > u_3-u_2 > u_4-u_3 > \dots > u_{n+1}-u_n \\ & \\ &\text{若}u_1 < u_2,\text{则}u_2-u_1 > 0,\ u_{n+1}-u_n\text{符号不确定}； \\ &\text{若}u_1 > u_2,\text{则}u_2-u_1 < 0,\ u_{n+1}-u_n < 0,\ \{u_n\}\text{单调递减}. \\ & \\ &u_{n+1} = u_1 + \sum_{i=1}^n (u_{i+1}-u_i) = u_1 + \sum_{i=1}^n f'(\xi_i) < u_1 + \sum_{i=1}^n f'(\xi_1) = u_1 + nf'(\xi_1) = u_1 + n(u_2-u_1) \\ &\lim_{n \to \infty} \left[ u_1 + n(u_2-u_1) \right] = -\infty,\ \text{故}\lim_{n \to \infty} u_{n+1} = -\infty,\ \{u_n\}\text{发散，选}(B) \end{aligned} ​改题：二阶导小于0，解：un​=f(n),由拉格朗日可知 un+1​−un​=f(n+1)−f(n)=(n+1−n)f′(ξn​)=f′(ξn​)ξn​∈(n,n+1)若f′′(x)<0,则f′(x)单调递减f′(ξ1​)>f′(ξ2​)>f′(ξ3​)>⋯>f′(ξn​),故u2​−u1​>u3​−u2​>u4​−u3​>⋯>un+1​−un​若u1​<u2​,则u2​−u1​>0, un+1​−un​符号不确定；若u1​>u2​,则u2​−u1​<0, un+1​−un​<0, {un​}单调递减.un+1​=u1​+i=1∑n​(ui+1​−ui​)=u1​+i=1∑n​f′(ξi​)<u1​+i=1∑n​f′(ξ1​)=u1​+nf′(ξ1​)=u1​+n(u2​−u1​)n→∞lim​[u1​+n(u2​−u1​)]=−∞, 故n→∞lim​un+1​=−∞, {un​}发散，选(B)​