高数强化NO3|函数性质及运算|函数极限的计算

函数性质及运算

$$\{\begin{array}{rl}& \text{概念：定义域，值域，对应规则 （例1）}\\ & \text{性质：单调性，有界性，周期性，奇偶性 （例3）}\\ & \text{运算：四则，复合，反 （例2）}\end{array}$$

$$\begin{array}{rl}& \text{解：}f(\phi (x))=e^{\phi (x)}=1-x\\ & \\ & \phi (x{)}^2=\ln (1-x),\phi (x)=\sqrt{\ln (1-x)}\\ & \\ & \text{为满足真数位置大于}0，\text{需满足}1-x>0，\text{即}x<1\\ & \\ & \text{为满足根号位置大于等于}0，\text{需满足}\ln (1-x)\ge 0，1-x\ge 1，x\le 0\\ & \\ & \text{将}x<1\text{和}x\le 0\text{取交集，}x\le 0.\\ & \\ & \text{则}\phi (x)=\sqrt{\ln (1-x)},x\le 0\end{array}$$
$$\begin{array}{rl}& \text{小结：① 定义域是使得所有函数均有意义的范围，取交集；}\\ & ②\text{若没让写定义域，也必须要留，如果没写，默认自然定义域.}\end{array}$$

$$\begin{array}{rl}& \text{解：}f(f(x))=\{\begin{array}{ll}1& |f(x)|\le 1\\ 0& |f(x)|>1\end{array}=\{\begin{array}{ll}1& x\in \mathbb{R}\\ 0& x\in \varnothing \end{array}=1\\ & \\ & f(f(f(x)))=f(1)=1\end{array}$$
$$\begin{array}{rl}& \text{法2：特殊值法，}f(f(f(0)))=f(f(1))=f(1)=1,\text{排除A, D}\\ & \\ & f(f(f(2)))=f(f(1))=f(1)=1,\text{排除C}.\end{array}$$
$$\begin{array}{rl}& \text{【小结】分段函数求复合函数的步骤是：先整体代入，再求解不等式；}\\ & \text{其中求解不等式是难点，关键是将分段函数的每一段的表达式满足不等式的范围和}\\ & \text{自变量的范围取交集，之后将每个交集再取并集即可.当有多层复合关系时，先求内}\\ & \text{层复合关系的表达式，然后再求外层复合关系的表达式，外层复合关系的表达式往}\\ & \text{往就比较简单了.}\end{array}$$

$$\text{解：显然，}f(x)=f(-x)，\text{故}f(x)\text{为偶函数，选(D)}$$
$$\begin{array}{rl}& \text{【小结】常见的奇函数：}\\ & y=x^k（k\text{为奇数}）,y=x^{\frac{1}{k}}（k\text{为奇数}）,y=\sin x,y=\arcsin x,\\ & y=\tan x,y=\arctan x,y=\ln (x+\sqrt{1+x^2}),\frac{1-e^x}{1+e^x},f(x)-f(-x),\ln \frac{f(x)}{f(-x)}；\\ & \\ & \text{常见的偶函数：}\\ & y=x^k（k\text{为偶数}）,y=\cos x,y=\sec x,y=|x|,f(x)+f(-x)\\ & \text{与}f(x)f(-x)。\end{array}$$
$$\begin{array}{rl}f(-x)& =\ln \left(-x+\sqrt{1+(-x{)}^2}\right)=\ln \frac{1}{\sqrt{1+x^2}+x}\\ & =-\ln \left(\sqrt{1+x^2}+x\right)=-f(x)\end{array}$$

函数极限的计算

$$\begin{array}{rl}& \text{四种方法，七种题型}\\ & \{\begin{array}{l}\text{等替}\{\begin{array}{l}\text{替换技巧：凑1，例4}\\ \text{加法的替换：例5、6}\end{array}\\ \text{泰勒}\{\begin{array}{l}\text{上下同阶：例7}\\ \text{反求参数：例8}\end{array}\\ \text{洛必达}\{\begin{array}{l}\text{变限积分：例9、10}\\ \text{幂指函数作为一部分：例11}\end{array}\\ \text{单侧极限：例12}\end{array}\end{array}$$

$$\begin{array}{rl}\text{解：}& \underset{x\to 0}{\lim }\frac{e\left(1-e^{\cos x-1}\right)}{\frac{1}{3}{x}^2}\\ & =\underset{x\to 0}{\lim }\frac{e\cdot \left(-(\cos x-1)\right)}{\frac{1}{3}{x}^2}\\ & =\underset{x\to 0}{\lim }\frac{e\cdot \frac{1}{2}{x}^2}{\frac{1}{3}{x}^2}\\ & =\frac{3}{2}e\end{array}$$
$$\begin{array}{rl}& \text{【小结】常用的替换技巧：}\\ & （1）\text{若}\alpha (x)\to 1，\text{则}\ln \alpha (x)=\ln \left[1+\alpha (x)-1\right]\sim \alpha (x)-1；\\ & \\ & （2）\text{若}\alpha (x),\beta (x)\to a，\text{则}{e}^{\alpha (x)}-e^{\beta (x)}=e^{\beta (x)}\left[e^{\alpha (x)-\beta (x)}-1\right]\sim e^a\left[\alpha (x)-\beta (x)\right]；\\ & \\ & （3）\text{若}\alpha (x)\to 1，\text{则}\sqrt[n]{\alpha (x)}-1={\left[1+\alpha (x)-1\right]}^{\frac{1}{n}}-1\sim \frac{1}{n}\left(\alpha (x)-1\right)。\end{array}$$
$$\begin{array}{rl}& \text{改：}\underset{x\to 0}{\lim }\frac{(1+x^2{)}^{\frac{1}{2}}-(1+x{)}^{\frac{1}{2}}}{\ln (1+x)}\\ & =\underset{x\to 0}{\lim }\frac{1+x^2-(1+x)}{\ln (1+x)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\\ & =-\frac{1}{2}\end{array}$$
$$\begin{array}{rl}& \text{改：}\underset{x\to 0}{\lim }\frac{(1+x^2{)}^{\frac{1}{2}}-(1+x{)}^{\frac{1}{3}}}{\ln (1+x)}\\ & =\underset{x\to 0}{\lim }\frac{(1+x{)}^{\frac{1}{3}}\left(\frac{(1+x^2{)}^{\frac{1}{2}}}{(1+x{)}^{\frac{1}{3}}}-1\right)}{\ln (1+x)}\\ & =\underset{x\to 0}{\lim }\frac{e^{\ln \frac{(1+x^2{)}^{\frac{1}{2}}}{(1+x{)}^{\frac{1}{3}}}-1}}{x}=\underset{x\to 0}{\lim }\frac{\ln \frac{(1+x^2{)}^{\frac{1}{2}}}{(1+x{)}^{\frac{1}{3}}}}{x}\\ & =\underset{x\to 0}{\lim }\frac{\frac{1}{2}\ln (1+x^2)-\frac{1}{3}\ln (1+x)}{x}=-\frac{1}{3}\end{array}$$
只要两项相减，两项的数相等，就可以把后一项提出去，剩下的部分用对数恒等变形，

$$\begin{array}{rl}& \text{解：① 若}a=0,c\ne 0,\underset{x\to 0}{\lim }\frac{b\cdot \frac{x^2}{2}}{c\ln (1-2x)+d(1-e^{-x^2})}=\underset{x\to 0}{\lim }\frac{\frac{b}{2}}{\frac{c\ln (1+x)}{x^2}+\frac{d(1-e^{-x^2})}{x^2}}=0\\ & \ne 2,\text{不符合}\\ & \text{② 若}a\ne 0,c=0,\text{极限为}\infty \\ & \text{故只能}a\ne 0,c\ne 0,\underset{x\to 0}{\lim }\frac{a\tan x+b(1-\cos x)}{c\ln (1-2x)+d(1-e^{-x^2})}=\underset{x\to 0}{\lim }\frac{a\tan ax}{c\ln (1+2x)}=\frac{a}{-2c}=2,\text{故：}\\ & a=-4c\\ & \underset{x\to 0}{\lim }\frac{a\tan x+b(1-\cos x)}{a\tan x}=1,\underset{x\to \infty }{\lim }\frac{c\ln (1-2x)+d(1-e^{-x^2})}{c\ln (1-2x)}=1\end{array}$$
$$\begin{array}{rl}& \text{【小结】加法的常用替换法则：}\\ & （1）\text{如果}\alpha (x)=o(\beta (x))，\text{则}\alpha (x)+\beta (x)\sim \beta (x)(\text{低阶+高阶等价于低阶})；\\ & \\ & （2）\text{如果}\alpha (x)\sim k{x}^m,\beta (x)\sim l{x}^m，\text{且}k+l\ne 0，\text{则}\alpha (x)+\beta (x)\sim (k+l)x^m(\text{同阶相加，}\\ & \text{则当系数和不为零时，可以替换})。\end{array}$$
$$\begin{array}{rl}& \text{低阶吸高阶与抓大头的区别与联系}\\ & \{\begin{array}{l}\text{区别：低阶吸高阶（无穷小）、抓大头（无穷大）}\\ \\ \text{联系：本质都是留大舍小.}\text{（只不过}x\to \infty \text{时，次数高的大，}x\to 0\text{时，次数低的大）}\end{array}\end{array}$$
$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}\text{不能等价替换}\\ & \underset{x\to 0}{\lim }\frac{1-\cos x+\arcsin x(e^x-1)}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}+\underset{x\to 0}{\lim }\frac{\arcsin x(e^x-1)}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\frac{x^2}{2}}{x^2}+\underset{x\to 0}{\lim }\frac{\cdot x\cdot x}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\frac{x^2}{2}+x^2}{x^2}=\frac{3}{2}\end{array}$$

$$\begin{array}{rl}& x-\sin x\sim \frac{1}{6}{x}^3,\underset{x\to 0}{\lim }\frac{(x-\sin x)\left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)}{x^2}=0,(x-\sin x)\left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)=o(x^2)\\ & \text{解：原式}=\underset{x\to 0}{\lim }\frac{\cos x-e^{x^2}}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{(\cos x-1)-(e^{x^2}-1)}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\left(-\frac{x^2}{2}\right)-\left(x^2\right)}{x^2}\\ & =-\frac{3}{2}\\ & \end{array}$$

$$\begin{array}{rl}& \text{解：法1：原式=}{e}^{{\lim }_{x\to 0}\frac{\cos 2x+2x\sin x-1}{x^4}}\\ & \text{其中}\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+o(x^4),\sin x=x-\frac{x^3}{6}+o(x^3)\\ & =e^{{\lim }_{x\to 0}\frac{1-\frac{(2x{)}^2}{2}+\frac{(2x{)}^4}{24}+o(x^4)+2x\left(x-\frac{x^3}{6}+o(x^3)\right)-1}{x^4}}(\text{展开})\\ & \text{合并：}=e^{{\lim }_{x\to 0}\frac{\frac{2}{3}{x}^4-\frac{1}{3}{x}^4+o(x^4)}{x^4}}\\ & \text{替换}=e^{{\lim }_{x\to 0}\frac{\frac{1}{3}{x}^4}{x^4}}=e^{\frac{1}{3}}\end{array}$$
$$\begin{array}{rl}& \text{法2：口算}\\ & e^{{\lim }_{x\to 0}\frac{\cos 2x+2x\sin x-1}{x^4}}\\ & =e^{{\lim }_{x\to 0}\frac{-2{\sin }^{2}x+2x\sin x}{x^4}}\\ & =e^{{\lim }_{x\to 0}\frac{2\sin x(x-\sin x)}{x^4}}=e^{{\lim }_{x\to 0}\frac{2x\cdot \frac{x^3}{6}}{x^4}}=e^{\frac{1}{3}}\end{array}$$
$$\begin{array}{rl}& \text{【小结】确定展开阶数的基本原则：上下同阶（分子分母有一项阶数已知），}\\ & \text{多退少补（展开到没有完全抵消的第一项）。}\end{array}$$

$$\begin{array}{rl}& \text{正解：}\underset{x\to 0}{\lim }\frac{e^x(1+Bx+C{x}^2)-1-Ax}{x^3}=\underset{x\to 0}{\lim }\frac{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(1+Bx+C{x}^2\right)-1-Ax}{x^3}\\ & =\underset{x\to 0}{\lim }\frac{(B+1-A)x+\left(B+C+\frac{1}{2}\right)x^2+\left(C+\frac{B}{2}+\frac{1}{6}\right)x^3+o(x^3)}{x^3}=0\\ & \text{由极限为0，分子各阶系数需为0：}\\ & \{\begin{array}{l}B+1-A=0\\ B+C+\frac{1}{2}=0\\ C+\frac{B}{2}+\frac{1}{6}=0\end{array}⟹\{\begin{array}{l}A=\frac{1}{3}\\ B=-\frac{2}{3}\\ C=\frac{1}{6}\end{array}\\ & \\ & \text{展开过程验证：}\\ & 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)+Bx+B{x}^2+\frac{B{x}^3}{2}+\frac{B{x}^4}{6}+Bx\cdot o(x^3)+C{x}^2+C{x}^3+\frac{C{x}^4}{2}+\frac{C{x}^5}{6}+C{x}^2\cdot o(x^3)\\ & -1-Ax+o(x^3)\\ & =1+x+\frac{x^2}{2}+\frac{x^3}{6}+Bx+B{x}^2+\frac{B{x}^3}{2}+C{x}^2+C{x}^3-1-Ax+o(x^3)\\ & =(B+1-A)x+\left(\frac{1}{2}+B+C\right)x^2+\left(\frac{1}{6}+\frac{B}{2}+C\right)x^3+o(x^3)\end{array}$$

$$\begin{array}{rl}& \text{解：①：}{\left({\int }_0^x\sqrt{x-t}{e}^{t}dt\right)}^{\prime }\stackrel{u=x-t}{=}{\left({\int }_x^0\sqrt{u}{e}^{x-u}d(x-u)\right)}^{\prime }\\ & ={\left({\int }_0^x\sqrt{u}{e}^{x-u}du\right)}^{\prime }={\left(e^x{\int }_0^x\sqrt{u}{e}^{-u}du\right)}^{\prime }\\ & =e^x{\int }_0^x\sqrt{u}{e}^{-u}du+e^x\cdot \sqrt{x}{e}^{-x}=e^x{\int }_0^x\sqrt{u}{e}^{-u}du+\sqrt{x}\\ & \\ & \text{原式}\stackrel{\text{洛必达}}{=}\underset{x\to 0}{\lim }\frac{{\left({\int }_0^x\sqrt{x-t}{e}^{t}dt\right)}^{\prime }}{(\sqrt{x^3}{)}^{\prime }}=\underset{x\to 0}{\lim }\frac{e^x{\int }_0^x\sqrt{u}{e}^{-u}du+\sqrt{x}}{\frac{3}{2}{x}^{\frac{1}{2}}}\\ & =\underset{x\to 0}{\lim }\frac{e^x{\int }_0^x\sqrt{u}{e}^{-u}du}{\frac{3}{2}{x}^{\frac{1}{2}}}+\frac{2}{3}\\ & =\underset{x\to 0}{\lim }\frac{{\int }_0^x\sqrt{u}{e}^{-u}du}{\frac{3}{2}{x}^{\frac{1}{2}}}+\frac{2}{3}\\ & \stackrel{\text{洛必达}}{=}\underset{x\to 0}{\lim }\frac{\sqrt{x}{e}^{-x}}{\frac{3}{2}\cdot \frac{1}{2}{x}^{-\frac{1}{2}}}+\frac{2}{3}=\frac{2}{3}\end{array}$$
$$\begin{array}{rl}& \text{洛必达前一定要先化简：}\\ & ①\text{非零因子直接代入}\\ & ②\text{零因子直接替换}\end{array}$$
$$\begin{array}{rl}& \text{法2：}{\int }_0^x\sqrt{x-t}{e}^{t}dt\stackrel{u=x-t}{=}{e}^x{\int }_0^x\sqrt{u}{e}^{-u}du\\ & \underset{x\to 0}{\lim }\frac{e^x{\int }_0^x\sqrt{u}{e}^{-u}du}{x^{\frac{3}{2}}}=\underset{x\to 0}{\lim }\frac{{\int }_0^x\sqrt{u}{e}^{-u}du}{x^{\frac{3}{2}}}\\ & \stackrel{\text{洛必达}}{=}\underset{x\to 0}{\lim }\frac{\sqrt{x}{e}^{-x}}{\frac{3}{2}{x}^{\frac{1}{2}}}=\frac{2}{3}\end{array}$$
$$\begin{array}{rl}& \text{法3：}\underset{x\to 0}{\lim }\frac{{\int }_0^x\sqrt{x-t}{e}^{t}dt}{x^{\frac{3}{2}}}=\underset{x\to 0}{\lim }\frac{{\int }_0^x\sqrt{x-t}dt}{x^{\frac{3}{2}}}\\ & =\underset{x\to 0}{\lim }\frac{-\frac{2}{3}(x-t{)}^{\frac{3}{2}}{|}_0^x}{x^{\frac{3}{2}}}=\underset{x\to 0}{\lim }\frac{\frac{2}{3}{x}^{\frac{3}{2}}}{x^{\frac{3}{2}}}=\frac{2}{3}\end{array}$$
$$\begin{array}{rl}& \text{【小结】极限式中如果出现变限积分，基本的思路是运用洛必达法则进行计算，计算时}\\ & \text{注意积分号下的}x\text{的处理方式：}\\ & 1.\text{如果积分变限函数积分号下出现关于}x\text{的函数}u(x),\text{类似于本题,则可以将}u(x)\text{移至积}\\ & \text{分号外,再进行求导,即}{\int }_a^{x}u(x)f(t)dt=u(x){\int }_a^{x}f(t)dt，\text{故}\\ & {\left({\int }_a^{x}u(x)f(t)dt\right)}^{\prime }=u^{\prime }(x){\int }_a^{x}f(t)dt+u(x)f(x).\\ & \\ & 2.\text{如果}x\text{出现在积分号下被积函数的自变量中,则可以先进行换元, 再进行求导, 即形如}\\ & {\int }_a^{x}f(u(x,t))dt\text{的积分变限函数, 令}u=u(x,t)\text{进行换元.}\end{array}$$

$$\begin{array}{rl}& \text{解：原式}=\underset{x\to +\infty }{\lim }\frac{{\int }_1^x\left(t^2\left(e^{\frac{1}{t}}-1\right)-t\right)dt}{x}\\ & \stackrel{\text{洛必达}}{=}\underset{x\to +\infty }{\lim }\frac{x^2\left(e^{\frac{1}{x}}-1\right)-x}{1}\\ & \text{法1：令}t=\frac{1}{x},\underset{t\to 0^{+}}{\lim }\frac{e^t-1-t}{t^2}=\frac{1}{2}\\ & 法2:=\underset{x\to +\infty }{\lim }{x}^2\left(1+\frac{1}{x}+\frac{1}{2}\cdot \frac{1}{x^2}+o\left(\frac{1}{x^2}\right)-1\right)-x=\frac{1}{2}\end{array}$$
$$\begin{array}{rl}& \text{【小结】（1）对于}\frac{\infty }{\infty }\text{型未定式，运用洛必达法则时，只需要验证分母的极限为无穷，就}\\ & \text{可以进行计算；}\\ & \\ & （2）\text{对于}\infty -\infty \text{型未定式，如果没有分母，则往往可以通过做变量代换制造出分母再通}\\ & \text{分进行计算，最常用的变量代换方法是令}x=\frac{1}{t}。\end{array}$$

$$\begin{array}{rl}& \text{解：}\underset{x\to +\infty }{\lim }x\left(\frac{1}{(x+1{)}^x}-e^{-1}\right)\\ & =\underset{x\to +\infty }{\lim }x\left(e^{-x\ln (1+\frac{1}{x})}-e^{-1}\right)\\ & =e^{-1}\underset{x\to +\infty }{\lim }x\left(e^{1-x\ln (1+\frac{1}{x})}-1\right)\\ & =e^{-1}\underset{x\to +\infty }{\lim }x\cdot \left(1-x\ln (1+\frac{1}{x})\right)\\ & \stackrel{t=\frac{1}{x}}{=}{e}^{-1}\underset{t\to 0^{+}}{\lim }\frac{1-\frac{\ln (1+t)}{t}}{t}=e^{-1}\underset{t\to 0^{+}}{\lim }\frac{t-\ln (1+t)}{t^2}\\ & =e^{-1}\cdot \frac{1}{2}=\frac{1}{2e}\end{array}$$
$$\begin{array}{rl}& \text{【小结】如果极限式的一部分为幂指函数，则无论幂指函数是什么类型的，都需要用对数}\\ & \text{恒等式变形。}\end{array}$$

$$\begin{array}{rl}& \underset{x\to 0^{+}}{\lim }a\cdot \arctan \frac{1}{x}+(1+x{)}^{\frac{1}{x}}=\frac{\pi }{2}a+e\\ & \underset{x\to 0^{-}}{\lim }a\cdot \arctan \frac{1}{x}+(1-x{)}^{\frac{1}{x}}=-\frac{\pi }{2}a+e^{-1}\\ & \text{由极限存在，左右极限相等：}\\ & \frac{\pi }{2}a+e=-\frac{\pi }{2}a+e^{-1}⟹a=\frac{1}{\pi }(e^{-1}-e)\end{array}$$
$$\begin{array}{rl}& \text{【小结】需要分左右求极限的情况：（总结）}\\ & 一\text{是函数本身是分段的（包括隐式的分段函数，如绝对值、最大值和最小值、取整等）；}\\ & 二\text{是出现}{e}^{\infty }\text{和}\arctan \infty \text{（这里的无穷均要求不能确定正负）。}\end{array}$$
$$\begin{array}{rl}& \text{当}a>1\text{时：}{a}^{+\infty }=+\infty ,a^{-\infty }=0\\ & \text{当}0<a<1\text{时：}{a}^{+\infty }=0,a^{-\infty }=+\infty \end{array}$$