4.3 习题课讲解

4.3.1 无穷大量与无穷小量

解 提示：设$x\to a$，$u\in \mathcal{O}(f)$、$v\in o(f)$，利用定义

• $\mathrm{\exists }M>0$，$\mathrm{\exists }\stackrel{\circ }{U}(a,{\delta }_1)$使得$x\in \stackrel{\circ }{U}(a,{\delta }_1)⟹|u(x)|\le M|f(x)|$。
• $\mathrm{\forall }\epsilon >0$，$\mathrm{\exists }\stackrel{\circ }{U}(a,{\delta }_2)$使得$x\in \stackrel{\circ }{U}(a,{\delta }_2)⟹|v(x)|\le \epsilon |f(x)|$。

解 提示：(1)$$\begin{array}{}\text{(4)}& \begin{array}{rl}|f(x)|& =|Bg(x)+o(g(x))-o(f(x))|\le |B||g(x)|+\frac{1}{2}|g(x)|+\frac{1}{2}|f(x)|\\ ⟹|f(x)|& \le (2|B|+1)|g(x)|,x\to a\\ |f(x)-Bg(x)|& =|o(g(x))-o(f(x))|\le \epsilon |f(x)|+\epsilon |g(x)|\le \underset{{\epsilon }^{\prime }}{\underbrace{2\epsilon (|B|+1)}}|g(x)|\\ ⟹|f(x)-Bg(x)|& \le {\epsilon }^{\prime }|g(x)|,x\to a\end{array}\end{array}$$ (2) 若$f(x)=g(x)+o(g(x))$，则$f=f+0=f+o(f)$，由(1)知$g(x)=f(x)+o(f(x))$。 $◻$

解 提示：确定主项。由例4.3.2(1)可知$$\begin{array}{}\text{(6)}& y=Ax+B{x}^k+o(x^k)=Ax+o(x)⟹x=\frac{y}{A}+o(y)\end{array}$$ 确定第二项。设$x=\frac{y}{A}+f(y)$，其中$f(y)=o(y)$，则$$\begin{array}{}\text{(7)}& \begin{array}{rl}y& =A(\frac{y}{A}+f(y))+B{(\frac{y}{A}+f(y))}^k+o\left({(\frac{y}{A}+f(y))}^k\right)\\ 0& =Af(y)+\frac{B}{A^k}{y}^k(1+o\left(1\right))+o(y^k)=Af(y)+\frac{B}{A^k}{y}^k+o(y^k)\\ f(y)& =-\frac{B}{A^{k+1}}{y}^k+o(y^k)\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(8)}& x=\frac{y}{A}-\frac{B}{A^{k+1}}{y}^k+o(y^k),y\to 0\end{array}$$ $◻$

解 Newton当年得到的结果为：$$\begin{array}{}\text{(9)}& (1+x{)}^{\frac{m}{n}}=\sum _{i=0}^k(\genfrac{}{}{0ex}{}{\frac{m}{n}}{i})x^i+o(x^k),x\to 0\end{array}$$ 其中广义二项式系数的定义为：

• $(\genfrac{}{}{0ex}{}{\alpha }{0})=1$；
• $(\genfrac{}{}{0ex}{}{\alpha }{i})=\frac{\alpha (\alpha -1)\cdots (\alpha -i+1)}{i!}$。

我们证明$k\le 2$的情况。$k=0$显然成立，设$(1+x{)}^{\frac{m}{n}}=1+f(x)$，其中$f(x)=o(1)$，则$$\begin{array}{}\text{(10)}& \begin{array}{rl}(1+x{)}^m& =(1+f(x){)}^n\\ 1+mx+o(x)& =1+nf(x)+o(f(x))\\ ⟹f(x)& =\frac{mx}{n}+o(x)\end{array}\end{array}$$ 再设$(1+x{)}^{\frac{m}{n}}=1+\frac{m}{n}x+g(x)$，其中$g(x)=o(x)$，则$$\begin{array}{}\text{(11)}& \begin{array}{rl}(1+x{)}^m& ={(1+\frac{m}{n}x+g(x))}^n\\ 1+mx+\frac{m(m-1)}{2}{x}^2+o(x^2)& =1+n(\frac{m}{n}x+g(x))+\frac{n(n-1)}{2}{(\frac{m}{n}x+g(x))}^2+o(x^2)\\ & =1+mx+ng(x)+\frac{m^2(n-1)}{2n}{x}^2\\ ⟹g(x)& =\frac{\frac{m}{n}(\frac{m}{n}-1)}{2}{x}^2+o(x^2)\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(12)}& (1+x{)}^{\frac{m}{n}}=1+\frac{m}{n}x+\frac{\frac{m}{n}(\frac{m}{n}-1)}{2}{x}^2+o(x^2),x\to 0\end{array}$$ $◻$

解 设$$\begin{array}{}\text{(14)}& u(x)=\frac{(1+x{)}^{\alpha }-1-\alpha x}{x}=o(1),x\to 0\end{array}$$ 则$$\begin{array}{}\text{(15)}& \begin{array}{rl}(1+2x{)}^{\alpha }& =[(1+x{)}^2-x^2{]}^{\alpha }=[(1+x{)}^{\alpha }{]}^2{[1-\frac{x^2}{(1+x{)}^2}]}^{\alpha }\\ 1+2\alpha x+2xu(2x)& =[1+\alpha x+xu(x){]}^2[1-\frac{\alpha x^2}{(1+x{)}^2}+o(x^2)]\\ & =[1+2\alpha x+2xu(x)+{\alpha }^2{x}^2][1-\alpha x^2+o(x^2)]\\ & =1+2\alpha x+2xu(x)+\alpha (\alpha -1)x^2+o(x^2)\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(16)}& u(2x)-u(x)-\frac{\alpha (\alpha -1)}{2}x=o(x),x\to 0\end{array}$$ 猜测$u(x)=A{x}^{\beta }+o(x^{\beta })$，其中$\beta \ge 1$，代入上式得$$\begin{array}{}\text{(17)}& \begin{array}{rl}A(2x{)}^{\beta }+o((2x{)}^{\beta })-A{x}^{\beta }+o(x^{\beta })-\frac{\alpha (\alpha -1)}{2}x& =o(x)\\ (2^{\beta }-1)A{x}^{\beta }+o(x^{\beta })& =\frac{\alpha (\alpha -1)}{2}x+o(x)\end{array}\end{array}$$ 对比等式两侧可得$$\begin{array}{}\text{(18)}& \beta =1,A=\frac{\alpha (\alpha -1)}{2(2^{\beta }-1)}=\frac{\alpha (\alpha -1)}{2}\end{array}$$ 因此$$\begin{array}{}\text{(19)}& u(x)=\frac{\alpha (\alpha -1)}{2}x+o(x),x\to 0\end{array}$$ $◻$

解 $\mathrm{\forall }\epsilon >0$，$\mathrm{\exists }\delta >0$，使得$\mathrm{\forall }x\in \mathbb{R}$$$\begin{array}{}\text{(33)}& 0<|x|<\delta ⟹|f(\lambda x)-Af(x)-B{x}^{\alpha }|<\epsilon |x{|}^{\alpha }\end{array}$$ $\mathrm{\forall }k\in {\mathbb{N}}^{\ast }$，$0<|{\lambda }^{-(k-1)}x|\le |x|<\delta$，从而$$\begin{array}{}\text{(34)}& |A^{k-1}f\left({\lambda }^{-(k-1)}x\right)-A^{k}f\left({\lambda }^{-k}x\right)-B{A}^{k-1}{\left({\lambda }^{-k}x\right)}^{\alpha }|<\epsilon |A{|}^{k-1}{\left|{\lambda }^{-k}x\right|}^{\alpha }\end{array}$$ 因此$\mathrm{\forall }n\in {\mathbb{N}}^{\ast }$$$\begin{array}{}\text{(35)}& |f(x)-A^{n}f\left({\lambda }^{-n}x\right)-\frac{B}{{\lambda }^{\alpha }}{x}^{\alpha }\sum _{k=0}^{n-1}{\left(\frac{A}{{\lambda }^{\alpha }}\right)}^k|\le \epsilon |x{|}^{\alpha }\frac{1}{{\lambda }^{\alpha }}\sum _{k=0}^{n-1}{\left|\frac{A}{{\lambda }^{\alpha }}\right|}^k\end{array}$$ 令$n\to +\mathrm{\infty }$可得$$\begin{array}{}\text{(36)}& |f(x)-\frac{B}{{\lambda }^{\alpha }}\frac{1}{1-\frac{A}{{\lambda }^{\alpha }}}|\le \frac{1}{{\lambda }^{\alpha }}\frac{\epsilon |x{|}^{\alpha }}{1-\frac{|A|}{{\lambda }^{\alpha }}}=\frac{\epsilon |x{|}^{\alpha }}{{\lambda }^{\alpha }-|A|}\stackrel{?}{\le }{\epsilon }^{\prime }\frac{|B||x{|}^{\alpha }}{{\lambda }^{\alpha }-A}\end{array}$$ 取$\epsilon$满足$$\begin{array}{}\text{(37)}& \frac{\epsilon }{{\epsilon }^{\prime }}\le |B|\frac{{\lambda }^{\alpha }-|A|}{{\lambda }^{\alpha }-A}\in (0,|B|]\end{array}$$ 即可，因此$$\begin{array}{}\text{(38)}& f(x)=\frac{B}{{\lambda }^{\alpha }-A}{x}^{\alpha }+o(x^{\alpha }),x\to 0\end{array}$$

在进行严谨证明时，不能猜测$f(x)=C{x}^{\beta }+o(x^{\beta })$，因为这样仅仅证明了蕴涵式“若$f$具有这个形式则结论成立”为真，与“结论成立”为真无关。 $◻$

错解 (1) 错误使用等价无穷小替换。$$\begin{array}{}\text{(39)}& \underset{x\to 0}{lim}\frac{\tan x-x}{x^3}=\underset{x\to 0}{lim}\frac{x-x+o(x)}{x^3}=\underset{x\to 0}{lim}\frac{o(x)}{x^3}=?\end{array}$$ (2) 预先假设极限存在。$$\begin{array}{}\text{(40)}& \begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{\tan x-x}{x^3}=\underset{x\to 0}{lim}\frac{\tan 2x-2x}{(2x{)}^3}=\frac{1}{4}\underset{x\to 0}{lim}\frac{\frac{1}{2}\tan 2x-x}{x^3}\\ 4L-L& =\underset{x\to 0}{lim}\frac{\frac{1}{2}\tan 2x-\tan x}{x^3}=\underset{x\to 0}{lim}\frac{{\tan }^{3}x}{x^3(1-{\tan }^{2}x)}=1\\ ⟹L& =\frac{1}{3}\end{array}\end{array}$$ 类似(2)的思路，我们可以证明很多离谱的“极限”，如$$\begin{array}{}\text{(41)}& L=\underset{x\to 0}{lim}\frac{1}{x}⟹\frac{L}{2}=\underset{x\to 0}{lim}\frac{1}{2x}=L⟹L=0\end{array}$$

解 $$\begin{array}{}\text{(42)}& \begin{array}{rl}\frac{\tan x-x}{x^3}& =\frac{\sin x-x\cos x}{x^3\cos x}\\ & =\frac{x-\frac{x^3}{6}+o(x^3)-x(1-\frac{x^2}{2}+o(x^2))}{x^3(1+o(1))}\\ & =\frac{\frac{x^3}{3}+o(x^3)}{x^3(1+o(1))}=\frac{1}{3}+o(1)\to \frac{1}{3},x\to 0\end{array}\end{array}$$ 也可类似前题的思路：设$\tan x=x+u(x)$，则$$\begin{array}{}\text{(43)}& \begin{array}{rl}\sin x& =[x+u(x)]\cos x\\ x+o(x)& =[x+u(x)][1+o(1)]=x+u(x)+o(x)\end{array}\end{array}$$ 故可设$u(x)=xv(x)$，其中$v(x)=o(1)$，则$$\begin{array}{}\text{(44)}& \begin{array}{rl}\tan 2x& =\frac{2\tan x}{1-{\tan }^{2}x}\\ 2x+2xv(2x)& =\frac{2(x+xv(x))}{1-(x+xv(x){)}^2}\\ 2x+2xv(x)& =2x[1+v(2x)][1-(x+xv(x){)}^2]=2x[1+v(2x)][1-x^2+o(x^2)]\\ & =2x[1+v(2x)-x^2+o(x^2)]=2x+2xv(2x)-2x^3+o(x^3)\\ ⟹v(2x)-v(x)-x^2& =o(x),x\to 0\\ ⟹v(x)& =\frac{x^2}{3}+o(x^2),x\to 0\end{array}\end{array}$$ $◻$

解 记$h=1-x$，$t=\frac{\pi }{2}-\arcsin \frac{2x}{1+x^2}\in [0,\pi ]$，则$h\to 0^{\pm }$当且仅当$x\to 1^{\mp }$，$$\begin{array}{}\text{(46)}& \cos t=\sin (\frac{\pi }{2}-t)=\frac{2x}{1+x^2}=\frac{2(1-h)}{1+(1-h{)}^2}=\frac{1}{1+\frac{h^2}{2(1-h)}}\to 1^{-}\end{array}$$ 于是$t=0^{+}$，上式两边展开可得$$\begin{array}{}\text{(47)}& 1-\frac{t^2}{2}+o(t^2)=1-\frac{h^2}{2(1-h)}+o(h^2)=1-\frac{h^2}{2}+o(h^2)⟹t^2=h^2+o(h)\end{array}$$ 从而$$\begin{array}{}\text{(48)}& t=\sqrt{h^2+o(h^2)}=|h|\sqrt{1+o(1)}=|h|(1+o(1))\end{array}$$ 因此$$\begin{array}{}\text{(49)}& \begin{array}{rl}\underset{x\to 1^{-}}{lim}\frac{\arcsin \frac{2x}{1+x^2}-\frac{\pi }{2}}{x-1}& =\underset{h\to 0^{+}}{lim}\frac{-t}{-h}=\underset{h\to 0^{+}}{lim}\frac{h(1+o(1))}{h}=1\\ \underset{x\to 1^{+}}{lim}\frac{\arcsin \frac{2x}{1+x^2}-\frac{\pi }{2}}{x-1}& =\underset{h\to 0^{-}}{lim}\frac{-t}{-h}=\underset{h\to 0^{-}}{lim}\frac{-h(1+o(1))}{h}=-1\end{array}\end{array}$$ $◻$

4.3.2 极限的综合练习

解 $$\begin{array}{}\text{(51)}& \begin{array}{rl}\sqrt{x^2+2x}-\sqrt[3]{x^3-x^2}& =x{(1+\frac{2}{x})}^{\frac{1}{2}}-x{(1-\frac{1}{x})}^{\frac{1}{3}}\\ & =x(1+\frac{1}{x}+o\left(\frac{1}{x}\right))-x(1-\frac{1}{3x}+o\left(\frac{1}{x}\right))\\ & =\frac{4}{3}+o(1)\to \frac{4}{3},x\to +\mathrm{\infty }\end{array}\end{array}$$ $◻$

解 $$\begin{array}{}\text{(53)}& \frac{\sin (\tan x)}{\tan (\sin x)}=\frac{\tan x+o(\tan x)}{\sin x+o(\sin x)}=\frac{x+o(x)}{x+o(x)}\to 1,x\to 0\end{array}$$ $◻$

解 记$t=\frac{1}{x}$，则$x\to +\mathrm{\infty }$当且仅当$t\to 0^{+}$，因此$$\begin{array}{}\text{(55)}& x^2\ln (\cos \frac{1}{x})=\frac{\ln (\cos t)}{t^2}=\frac{\ln (1-\frac{t^2}{2}+o(t^2))}{t^2}=\frac{-\frac{t^2}{2}+o(t^2)}{t^2}\to -\frac{1}{2}\end{array}$$ $◻$

解 $$\begin{array}{}\text{(57)}& \begin{array}{rl}{(2\sin x+\cos x)}^{\frac{1}{x}}& =\exp \frac{\ln (2\sin x+\cos x)}{x}=\exp \frac{\ln (1+2x+o(x))}{x}\\ & =\exp \frac{2x+o(x)}{x}=\exp (2+o(1))\to {\mathrm{e}}^2,x\to 0\end{array}\end{array}$$ $◻$

解 $$\begin{array}{}\text{(59)}& \begin{array}{rl}{\mathrm{e}}^{-n}{(1+\frac{1}{n})}^{n^2}& =\exp [n^2\ln (1+\frac{1}{n})-n]\\ & =\exp [n^2(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right))-n]\\ & =\exp (-\frac{1}{2}+o(1))\to \frac{1}{\sqrt{\mathrm{e}}},n\to +\mathrm{\infty }\end{array}\end{array}$$ $◻$

解 记$t=\frac{1}{x}$，则$x\to +\mathrm{\infty }$当且仅当$t\to 0^{+}$，因此$$\begin{array}{}\text{(61)}& \begin{array}{rl}& x^p(a^{\frac{1}{x}}-a^{\frac{1}{x+1}})=\frac{1}{t^p}({\mathrm{e}}^{t\ln a}-\exp \frac{t\ln a}{1+t})\\ =& \frac{1}{t^p}(1+t\ln a+\frac{t^2{\ln }^{2}a}{2}+o(t^2)-1-\frac{t\ln a}{1+t}-\frac{t^2{\ln }^{2}a}{2(1+t{)}^2}+o(t^2))\\ =& \frac{1}{t^p}(t\ln a+\frac{t^2{\ln }^{2}a}{2}-t(1-t)\ln a-\frac{t^2{\ln }^{2}a}{2}+o(t^2))\\ =& \frac{\ln a+o(1)}{t^{p-2}}\to \ln a\ne 0,t\to 0^{+}\end{array}\end{array}$$ 所以$p=2$，并且上述极限为$\ln a$。

或者利用$$\begin{array}{}\text{(62)}& \begin{array}{rl}{x}^p{a}^{\frac{1}{x+1}}(a^{\frac{1}{x}-\frac{1}{x+1}}-1)& =x^p(1+o(1))[\ln a\cdot (\frac{1}{x}-\frac{1}{x+1})+o(\frac{1}{x}-\frac{1}{x+1})]\\ & =x^{p-2}\ln a+o(x^{p-2})=\{\begin{array}{ll}\ln a,& p=2\\ 0,& p<2\\ \mathrm{\infty },& p>2\end{array}\end{array}\end{array}$$ $◻$

解 $$\begin{array}{}\text{(64)}& \begin{array}{rl}{(1+\frac{1}{n})}^{n+\alpha }& =\exp ((n+\alpha )\ln (1+\frac{1}{n}))\\ & =\exp ((n+\alpha )(\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}+o\left(\frac{1}{n^3}\right)))\\ & =\exp (1+\frac{\alpha -\frac{1}{2}}{n}+\frac{-\frac{\alpha }{2}+\frac{1}{3}}{n^2}+o\left(\frac{1}{n^2}\right))\\ & =\mathrm{e}(1+\frac{\alpha -\frac{1}{2}}{n}+\frac{\frac{1}{2}{(\alpha -\frac{1}{2})}^2+(\frac{1}{3}-\frac{\alpha }{2})}{n^2}+o\left(\frac{1}{n^2}\right)),n\to +\mathrm{\infty }\end{array}\end{array}$$ 故当$\alpha \ne \frac{1}{2}$时，${(1+\frac{1}{n})}^{n+\alpha }$与$\frac{1}{n}$同阶；当$\alpha =\frac{1}{2}$时，${(1+\frac{1}{n})}^{n+\alpha }$与$\frac{1}{n^2}$同阶；即在这一数列族中，${(1+\frac{1}{n})}^{n+\frac{1}{2}}$收敛最快，但它的收敛速度远不如$\sum _{k=0}^n\frac{1}{k!}$，因为$$\begin{array}{}\text{(65)}& \sum _{k=0}^n\frac{1}{k!}=\mathrm{e}(1+o\left(\frac{2}{(n+1)!}\right)),n\to +\mathrm{\infty }\end{array}$$ $◻$

解 利用AM-GM不等式可得$$\begin{array}{}\text{(67)}& 1<n^{1/k}=\sqrt[n]{\underset{k-2}{\underbrace{1\cdots 1}}\cdot \sqrt{n}\cdot \sqrt{n}}<\frac{k-2+2\sqrt{n}}{k}<1+\frac{2\sqrt{n}}{k},k\ge 2\end{array}$$ 因此$$\begin{array}{}\text{(68)}& \begin{array}{rl}\mathrm{Ans}& >\frac{n+1+1+\cdots +1}{n}=\frac{2n-1}{n}=2-\frac{1}{n}\\ \mathrm{Ans}& <1+\frac{1}{n}\sum _{k=2}^n(1+\frac{2\sqrt{n}}{k})=2-\frac{1}{n}+\frac{2}{\sqrt{n}}\sum _{k=2}^n\frac{1}{k}<2-\frac{1}{n}+\frac{2\ln n}{\sqrt{n}}\end{array}\end{array}$$ 由夹挤定理可得极限存在且等于$2$。 $◻$

解 首先考虑分母。易见$$\begin{array}{}\text{(70)}& n^1+n^2+\cdots +n^n=n^n(1+o(1))\end{array}$$

然后考虑分子。注意到$$\begin{array}{}\text{(71)}& \sum _{k=1}^n\frac{k^n}{n^n}=\sum _{k=0}^{n-1}{(1-\frac{k}{n})}^n\end{array}$$ 可以证明$$\begin{array}{}\text{(72)}& 0\le {\mathrm{e}}^{-k}-{(1-\frac{k}{n})}^n\le \frac{k^2{\mathrm{e}}^{-k}}{n}\end{array}$$ 因此$$\begin{array}{}\text{(73)}& \left|\sum _{k=0}^{n-1}[{(1-\frac{k}{n})}^n-{\mathrm{e}}^{-k}]\right|\le \sum _{k=0}^{n-1}\frac{k^2{\mathrm{e}}^{-k}}{n}\le \sum _{k<9}\frac{k^2{\mathrm{e}}^{-k}}{n}+\sum _{k\ge 9}\frac{k^2\cdot k^{-4}}{n}\le \frac{A}{n}\end{array}$$ 从而$$\begin{array}{}\text{(74)}& \begin{array}{rl}|\sum _{k=0}^{n-1}{(1-\frac{k}{n})}^n-\frac{\mathrm{e}}{\mathrm{e}-1}|& \le |\sum _{k=0}^{n-1}{(1-\frac{k}{n})}^n-{\mathrm{e}}^{-k}|+|\sum _{k=0}^{n-1}{\mathrm{e}}^{-k}-\frac{1}{1-{\mathrm{e}}^{-1}}|\\ & \le \frac{A}{n}+\sum _{k=n}^{+\mathrm{\infty }}{\mathrm{e}}^{-k}=\frac{A}{n}+\frac{{\mathrm{e}}^{-n}}{1-{\mathrm{e}}^{-1}}\to 0,n\to +\mathrm{\infty }\end{array}\end{array}$$ 亦即$$\begin{array}{}\text{(75)}& \underset{n\to +\mathrm{\infty }}{lim}\frac{1^n+2^n+\cdots +n^n}{n^1+n^2+\cdots +n^n}=\underset{n\to +\mathrm{\infty }}{lim}\frac{n^n}{n^n(1+o(1))}\underset{n\to +\mathrm{\infty }}{lim}\sum _{k=0}^{n-1}{(1-\frac{k}{n})}^n=\frac{\mathrm{e}}{\mathrm{e}-1}\end{array}$$ $◻$