【数学】高昆轮高数上强化

函数极限

1.方法：等价代换，洛必达法则，泰勒公式，导数定义，拉格朗日中值定理

2.技巧：加减中把极限存在（不管是否为0）的部分拆项先算出来

​ 乘除中把极限存在（必须不为0）的部分分离先算出来

​ **对$或且带x\to \infty (或+\infty ,-\infty )且带\frac{1}{x}$**的极限采用倒代换、抓大头、有理化、通分等

$$\begin{gathered} 等价无穷小 \\ x-sinx\sim \frac{1}{6}{x}^{3}x-arcsinx\sim -\frac{1}{6}{x}^{3}x-tanx\sim -\frac{1}{3}{x}^3 \\ x-arctanx\sim \frac{1}{3}{x}^{3}x-\ln (1+x)\sim \frac{1}{2}{x}^{2}tanx-sinx\sim \frac{1}{2}{x}^3 \\ 1-co{s}^{\alpha }x\sim \frac{\alpha }{2}{x}^2{e}^x-1-x\sim \frac{1}{2}{x}^2\sqrt{1+x}-1-\frac{1}{2}x\sim -\frac{1}{8}{x}^2 \\ f(x)\to 1时,\ln f(x)\sim f(x)-1 \end{gathered}$$

题目

$$\begin{array}{rl}& 1.\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1-\frac{x}{2}}{{e^x}^2-1}\\ & 2.\underset{x\to 0}{\lim }\frac{e^x+\ln (1-x)-1}{x-arctanx}\\ & 3.\underset{x\to 0}{\lim }\frac{(1+x{)}^{\frac{2}{x}}-e^2[1-\ln (1+x)]}{x}\\ & 4.\underset{x\to 0}{\lim }\frac{(1+x^2)(1-cos2x)-2x^2}{x^4}\\ & 5.\underset{x\to 0}{\lim }\frac{\sqrt{1-x^2}si{n}^{2}x-ta{n}^{2}x}{x^2[\ln (1+x){]}^2}\\ & 6.\underset{x\to 0}{\lim }\frac{(3+2tanx{)}^x-3^x}{3si{n}^{2}x+x^{3}cos\frac{1}{x}}\\ & 7.\underset{x\to \infty }{\lim }{e}^{-x}(1+\frac{1}{x}{)}^{x^2}\\ & 8.\underset{x\to \infty }{\lim }{x}^2(a^{\frac{1}{x}}+a^{-\frac{1}{x}}-2),其中常数a>0\\ & 9.\underset{x\to 0}{\lim }(\frac{cosx}{cos2x}{)}^{\frac{1}{x^2}}\\ & 10.\underset{x\to \infty }{\lim }(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5})\\ & 11.\underset{x\to +\infty }{\lim }[(x^3+\frac{x}{2}-tan\frac{1}{x})e^{\frac{1}{x}}-\sqrt{1+x^6}]\\ & 12.\underset{x\to 0}{\lim }\frac{sinx+x^{2}sin\frac{1}{x}}{(1+cosx)\ln (1+x)}\\ & 13.\underset{x\to 0}{\lim }[\frac{a}{x}-(\frac{1}{x^2}-a^2)\ln (1+ax)].其中a̸=0\\ & 14.\underset{x\to 0}{\lim }\frac{(1+x{)}^{\frac{1}{x}}-(1+2x{)}^{\frac{1}{2x}}}{sinx}\\ & 15.\underset{x\to 0}{\lim }\frac{1}{x^3}[(\frac{2+cosx}{3}{)}^x-1]\\ & 16.\underset{x\to 0}{\lim }\frac{\sqrt{cosx}-\sqrt[3]{cosx}}{si{n}^{2}x}\\ & 17.\underset{x\to 0}{\lim }\frac{1-cosx\cdot \sqrt{cos2x}\cdot \sqrt[3]{cos3x}}{x^2}\\ & 18.\underset{x\to 1}{\lim }\frac{(1-\sqrt[3]{x})(1-\sqrt[4]{x})\cdots (1-\sqrt[n]{x})}{(1-x{)}^{n-2}}\end{array}$$

答案

$$\begin{array}{rl}1.I& =\underset{x\to 0}{\lim }\frac{-\frac{1}{8}{x}^2}{x^2}=-\frac{1}{8}\\ 2.I& =\underset{x\to 0}{\lim }\frac{e^x+\ln (1+x)-1}{\frac{1}{3}{x}^3}\\ & =\underset{x\to 0}{\lim }\frac{1+\frac{1}{2}{x}^2+x+\frac{1}{6}{x}^3+\circ (x^3)-x-\frac{1}{2}{x}^2-\frac{1}{3}{x}^3+\circ (x^3)-1}{\frac{1}{3}{x}^3}\\ & =\underset{x\to 0}{\lim }\frac{-\frac{1}{6}{x}^3}{\frac{1}{3}{x}^3}=-\frac{1}{2}\\ 3.I& =\underset{x\to 0}{\lim }\frac{(1+x{)}^{\frac{2}{x}}-e^2}{x}+e^2\underset{x\to 0}{\lim }\frac{\ln (1+x)}{x}\\ & =\underset{x\to 0}{\lim }\frac{e^{2\ln (1+x)/x}-e^2}{x}+e^2\\ & =\underset{x\to 0}{\lim }\frac{e^2(e^{\frac{2\ln (1+x)}{x}-2}-1)}{x}+e^2=\underset{x\to 0}{\lim }\frac{e^{\xi }(\frac{2\ln (1+x)}{x}-2)}{x}+e^2\\ & =e^2\underset{x\to 0}{\lim }\frac{\frac{2\ln (1+x)}{x}-2}{x}+e^2=2e^2\underset{x\to 0}{\lim }\frac{\frac{\ln (1+x)}{x}-1}{x}+e^2\\ & =2e^2\underset{x\to 0}{\lim }\frac{\ln (1+x)-x}{x^2}+e^2=2e^2\underset{x\to 0}{\lim }\frac{\ln (1+x)-x}{x^2}+e^2\\ & =(2e^2\cdot (-\frac{1}{2}))+e^{2}d=2e^2\underset{x\to 0}{\lim }\frac{-\frac{1}{2}{x}^2}{x^2}+e^2\\ & =-e^2+e^2=0=0\\ 4.I& =\underset{x\to 0}{\lim }\frac{1-\cos 2x-2x^2}{x^4}+\underset{x\to 0}{\lim }\frac{x^2(1-cos2x)}{x^4}\\ & =\underset{x\to 0}{\lim }\frac{1-(1-\frac{1}{2}(2x{)}^2+\frac{1}{24}(2x{)}^4+\circ (x^4))-2x^2}{x^4}+2\\ & =\underset{x\to 0}{\lim }\frac{-\frac{16}{24}{x}^4}{x^4}+2=\frac{4}{3}\\ 5.I& =\underset{x\to 0}{\lim }\frac{\sqrt{1-x^2}{\sin }^{2}x-{\sin }^{2}x+{\sin }^{2}x-{\tan }^{2}x}{x^4}\\ & =\underset{x\to 0}{\lim }\frac{{\sin }^x(\sqrt{1-x^2}-1)}{x^4}+\underset{x\to 0}{\lim }\frac{{\sin }^{2}x-{\tan }^{2}x}{x^4}\\ & =-\frac{1}{2}+\underset{x\to 0}{\lim }\frac{(\sin x+\tan x)(\sin x+\tan )}{x^4}\\ & =-\frac{1}{2}+\underset{x\to 0}{\lim }\frac{(\sin x+\tan x)(-\frac{1}{2}{x}^3)}{x^4}\\ & =-\frac{1}{2}+(-\frac{1}{2})(1+1)=-\frac{3}{2}\\ 6.& 由\frac{x^3\cos \frac{1}{x}}{3{\sin }^x}=\frac{1}{3}\underset{x\to 0}{\lim }x\cos \frac{1}{x}=0\\ & 则I=\underset{x\to 0}{\lim }\frac{3^x[(1+\frac{2}{3}\tan x{)}^x-1]}{3x^2}\\ & =\underset{x\to 0}{\lim }\frac{e^{x\ln (1+\frac{2}{3}\tan x)}-1}{3x^2}\\ & =\underset{x\to 0}{\lim }\frac{x\ln (1+\frac{2}{3}\tan x)}{3x^2}=\frac{2}{9}\\ 7.I& =\underset{x\to \infty }{\lim }{e}^{-x}{e}^{x^2\ln (1+\frac{1}{x})}\\ & =\underset{x\to \infty }{\lim }{e}^{x^2\ln (1+\frac{1}{x})-x}\\ & 且\underset{x\to \infty }{\lim }(x^2\ln (1+\frac{1}{x})-x)\\ & \underrightarrow{t=\frac{1}{x}}\underset{t\to 0}{\lim }\frac{\ln (1+t)}{t^2}-\frac{1}{t}=\underset{x\to \infty }{\lim }{x}^2(\ln (1+\frac{1}{x})-\frac{1}{x})\\ & =\underset{t\to 0}{\lim }\frac{\ln (1+t)-t}{t^2}=-\frac{1}{2}=\underset{x\to \infty }{\lim }{x}^2(-\frac{1}{2})(\frac{1}{x}{)}^2=-\frac{1}{2}\\ & \therefore I=e^{-\frac{1}{2}}\\ 8.& 令\frac{1}{x}=t\\ I& =\underset{t\to 0}{\lim }\frac{1}{t^2}(a^t+a^{-t}-2)\\ & =\underset{t\to 0}{\lim }\frac{a^t\ln a+a^{-t}\cdot \ln a^{-1}}{2t}\\ & ={\ln }^{2}a\\ 9.I& =e^A,其中A=\underset{x\to 0}{\lim }\frac{1}{x^2}(\frac{\cos x}{cos2x}-1)\\ & =\underset{x\to 0}{\lim }\frac{\cos x-\cos 2x}{x^2\cos 2x}\\ & =\underset{x\to 0}{\lim }\frac{\cos x-\cos 2x}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\cos x-1+1-\cos 2x}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\cos x-1}{x^2}+\underset{x\to 0}{\lim }\frac{1-\cos 2x}{x^2}\\ & =-\frac{1}{2}+\underset{x\to 0}{\lim }\frac{\frac{1}{2}(2x{)}^2}{x^2}\\ & =\frac{3}{2}故I=e^{\frac{3}{2}}\\ 10.I& =\underset{x\to \infty }{\lim }{f}^{\prime }(\xi )(2x^5)\\ & =\underset{x\to \infty }{\lim }\frac{1}{6}{\xi }^{-\frac{5}{6}}(2x^5)\\ & =\underset{x\to \infty }{\lim }\frac{1}{6}\cdot x^{6(-\frac{5}{6})}(2x^5)\\ & =\underset{x\to \infty }{\lim }\frac{1}{6}{x}^{-5}(2x^5)=\frac{1}{3}\\ 11.I& =\underset{x\to +\infty }{\lim }[(x^3+\frac{x}{2})e^{\frac{1}{x}}-\sqrt{1+x^6}]-\underset{x\to +\infty }{\lim }\tan \frac{1}{x}\cdot e^{\frac{1}{x}}\\ & \underrightarrow{t=\frac{1}{x}}\underset{t\to 0^{+}}{\lim }[(\frac{1}{t^3}+\frac{1}{2t})e^t-\sqrt{1+\frac{1}{t^6}}]-0\\ & =\underset{t\to 0^{+}}{\lim }\frac{(2+t^2)e^t-2\sqrt{1+t^6}+2-2}{2t^3}\\ & =\underset{t\to 0^{+}}{\lim }\frac{(2+t^2)e^t-2}{2t^3}-2\underset{t\to 0^{+}}{\lim }\frac{\sqrt{1+t^6}-1}{2t^3}\\ & =\underset{t\to 0^{+}}{\lim }\frac{(2+t^2)(1+t+\frac{1}{2}{t}^2+\frac{1}{6}{t}^3+\circ (t^3))-2}{2t^3}-0\\ & =\underset{t\to 0^{+}}{\lim }\frac{2t+2t^2+\frac{4}{3}{t}^3}{2t^3}=\infty \\ 12.I& =\underset{x\to 0}{\lim }\frac{x}{2x}=\frac{1}{2}\\ 13.I& =\underset{x\to 0}{\lim }[\frac{a}{x}-\frac{1}{x^2}\ln (1+ax)]+a^2\underset{x\to 0}{\lim }\ln (1+ax)\\ & =\underset{x\to 0}{\lim }\frac{ax-\ln (1+ax)}{x^2}+0\\ & =\underset{x\to 0}{\lim }\frac{\frac{1}{2}(ax{)}^2}{x^2}=\frac{1}{2}{a}^2\\ 14.I& =\underset{x\to 0}{\lim }\frac{e^{\ln (1+x)/x}-e^{\ln (1+2x)/2x}}{x}\\ & =\underset{x\to 0}{\lim }\frac{e^{\ln (1+2x)/2x}(e^{\ln (1+x)/x-\ln (1+2x)/2x}-1)}{x}\\ & =e\underset{x\to 0}{\lim }\frac{\ln (1+x)/x-\ln (1+2x)/2x}{x}\\ & =e\underset{x\to 0}{\lim }\frac{2\ln (1+x)-\ln (1+2x)}{2x^2}\\ & =e\cdot \frac{2(x-\frac{1}{2}{x}^2)+\circ (x^2)-(2x-\frac{1}{2}(2x{)}^2)-\circ (x^2)}{2x^2}\\ & =e\cdot \frac{1}{2}=\frac{e}{2}\\ 15.I& =\underset{x\to 0}{\lim }\frac{1}{x^3}[e^{x\ln \frac{2+\cos x}{3}}-1]I=\underset{x\to 0}{\lim }\frac{1}{x^3}[(1+\frac{\cos x-1}{3}{)}^x-1]\\ & =\underset{x\to 0}{\lim }\frac{1}{x^3}\cdot x\ln \frac{2+\cos x}{3}=\underset{x\to 0}{\lim }\frac{1}{x^3}\cdot x\cdot \frac{\cos x-1}{3}\\ & =\underset{x\to 0}{\lim }\frac{1}{x^2}\ln \frac{2+\cos x}{3}=-\frac{1}{6}\\ & =\underset{x\to 0}{\lim }\frac{1}{x^2}\ln (1+\frac{2+\cos x}{3}-1)\\ & =\underset{x\to 0}{\lim }\frac{1}{x^2}(\frac{2+\cos x}{3}-1)\\ & =\underset{x\to 0}{\lim }\frac{\cos x-1}{3x^2}=-\frac{1}{6}\\ 16.I& =\underset{x\to 0}{\lim }\frac{\sqrt{\cos x}-1+1-\sqrt[3]{\cos x}}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\sqrt{\cos x}-1}{x^2}+\underset{x\to 0}{\lim }\frac{1-\sqrt[3]{\cos x}}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{-\frac{1}{4}{x}^2}{x^2}+\underset{x\to 0}{\lim }\frac{\frac{1}{6}{x}^2}{x^2}\\ & =-\frac{1}{12}\\ 17.I& =\underset{x\to 0}{\lim }\frac{1-\cos x+\cos x-\cos x\sqrt{\cos 2x}\sqrt[3]{cos3x}}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}+\frac{\cos x-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}\\ & =\frac{1}{2}+\underset{x\to 0}{\lim }\frac{\cos x(1-\sqrt{\cos 2x}\sqrt[3]{\cos 3x})}{x^2}\\ & =\frac{1}{2}+\underset{x\to 0}{\lim }\frac{1-\sqrt{\cos 2x}+\sqrt{\cos 2x}(1-\sqrt[3]{\cos 3x})}{x^2}\\ & =3\\ 18.& \underset{x\to 1}{\lim }\frac{1-\sqrt[t]{x}}{1-x}(t=3,4,5,\cdots ,n)\\ & =\underset{x\to 1}{\lim }\frac{-\frac{1}{t}{x}^{\frac{1}{t}-1}}{-1}=\frac{1}{t}\\ I& =\frac{1}{3}\cdot \frac{1}{4}\cdots \frac{1}{n}=\frac{2}{n!}\end{array}$$

数列极限

1.方法：转化为函数极限（求导、洛必达、拉格朗日中值定理），先求和（积）、夹逼准则、定积分定义、单调有界法则

2.技巧：夹逼准则的套路，单调有界的套路，先求极限再证明的套路

3.总结：数列的构造法

题目

部分题目之前已总结，请去数列极限查看。
$$\begin{array}{rl}& 1.\underset{n\to \infty }{\lim }{n}^3(sin\frac{1}{n}-\frac{1}{2}sin\frac{2}{n})\\ & 2.\underset{n\to \infty }{\lim }{n}^2(arctan\frac{a}{n}-arctan\frac{a}{n+1}),a>0\\ & 3.\underset{n\to \infty }{\lim }\frac{n^{99}}{n^k-(n-1{)}^k}存在且不为0，则常数k=\underline{}\end{array}$$

答案

$$\begin{array}{rl}1.I& =\underset{n\to \infty }{\lim }{n}^3(\sin \frac{1}{n}-\frac{1}{2}\sin \frac{2}{n})\\ & =\underset{n\to \infty }{\lim }{n}^3(\sin \frac{1}{n}-\frac{1}{n})+\underset{n\to \infty }{\lim }{n}^3(\frac{1}{n}-\frac{1}{2}\sin \frac{2}{n})\\ & =-\frac{1}{6}+\underset{n\to \infty }{\lim }\frac{1}{2}\cdot n^3(\frac{2}{n}-\sin frac2n)\\ & =\frac{1}{2}\\ 2.I& =\underset{n\to \infty }{\lim }{n}^2(f^{\prime }(\xi )\cdot (\frac{a}{n}-\frac{a}{n+1}))\\ & =\underset{n\to \infty }{\lim }{n}^2\cdot \frac{1}{1+{\xi }^2}(\frac{a}{n}-\frac{a}{n+1})\\ & =\underset{n\to \infty }{\lim }\frac{a(n+1)-an}{n(n+1)}\cdot n^2=a\\ 3.I& =\underset{n\to \infty }{\lim }\frac{n^{99}}{n^k(1-(1-\frac{1}{n}{)}^k)}\\ & =\underset{n\to \infty }{\lim }\frac{n^{99}}{n^k(-k)(-\frac{1}{n})}\\ & =\underset{n\to \infty }{\lim }\frac{n^{99}}{k{n}^{k-1}}=c̸=0⟹k-1=99\end{array}$$

导数计算

题目

$$\begin{array}{rl}& 1.设y=e^{x^2},求\frac{dy}{dx},\frac{dy}{d(x^2)},\frac{d^{2}y}{d{x}^2}\\ & 2.设f(x)=(cosx-4)sinx+3x,求\frac{df(x)}{d(x^2)}\\ & 3.设f^{\prime }(0)=1,f^{\prime \prime }(0)=0,求证：在x=0处，有\frac{d^2}{d{x}^2}f(x^2)=\frac{d^2}{d{x}^2}{f}^2(x)\\ & 4.已知可微函数y=y(x),由方程y=-y{e}^x+2e^{y}sinx-7x所确定，求y^{\prime \prime }(0)\\ & 5.设函数y=y(x)由参数方程\{\begin{array}{l}x=1+t^2\\ y=cost\end{array}所确定，求\frac{dy}{dx}和\frac{d^{2}y}{d{x}^2}\\ & 6.设y=x^3+3x+1,则\frac{dx}{dy}{|}_{y=1}=\underline{}\\ & 7.x=f(y)是函数y=x+\ln x的反函数，求\frac{d^{2}f}{d{y}^2}\\ & 8.设f(x)=\sqrt{\frac{(1+x)\sqrt{x}}{e^{x-1}}}+arcsin\frac{1-x}{\sqrt{1+x^2}},求f^{\prime }(1)\end{array}$$

答案

$$\begin{array}{rl}1.& \frac{dy}{dx}=y^{\prime }=e^{x^2}\cdot 2x\\ & \frac{d^{2}y}{d{x}^2}=y^{{}^{\prime \prime }}=e^{x^2}\cdot 4x^2+e^{x^2}\cdot 2\\ & \frac{dy}{d(x^2)}=\frac{e^{x^2}\cdot 2xdx}{2xdx}=e^{x^2}\frac{dy}{2xdx}=\frac{1}{2x}\cdot e^{x^2}\cdot 2x=e{x}^2\\ 2.& df(a)=f^{\prime }(x)dx=(-{\sin }^{2}x+(\cos x-4)\cdot \cos x+3)dx\\ & d(x^2)=2xdx\\ \therefore & \frac{df(x)}{d(x^2)}=\frac{-{\sin }^{2}x+(\cos x-4)\cos x+3}{2x}=\frac{(\cos x-1{)}^2}{x}\\ 3.& y_1^{\prime }=f^{\prime }(x^2)\cdot 2x,y_1^{{}^{\prime \prime }}{|}_0=f^{\prime \prime }(x^2)\cdot 2x\cdot 2x+f^{\prime }(x^2)\cdot 2{|}_{x=0}=2\\ & y_2^{\prime }=2f(x)\cdot f^{\prime }(x),y_2^{{}^{\prime \prime }}{|}_0=2f^{\prime }(x)f^{\prime }(x)+2f(x)f^{\prime \prime }(x){|}_{x=0}=2\\ 4.& y=-y{e}^x+2e^y\sin x-7x\\ & ⟹y^{\prime }=-y^{\prime }{e}^x-y{e}^x+2e^y\sin x\cdot y^{\prime }+2e^y\cdot \cos x-7\\ & ⟹y^{\prime \prime }=-y^{\prime \prime }{e}^x-y^{\prime }{e}^x-y^{\prime }{e}^x-y{e}^x+2e^y\cdot (y^{\prime }{)}^2\sin x+\\ & 2e^y\cos x\cdot y^{\prime }+2e^y\sin x\cdot y^{\prime \prime }+2e^y\cdot y^{\prime }\cos x-2e^y\sin x\\ & 由x=0代入，分别得：\{\begin{array}{l}y=0\\ y^{\prime }=-\frac{5}{2}\\ y^{\prime \prime }=-\frac{5}{2}\end{array}\\ 5.& \frac{dy}{dx}=\frac{y_t^{\prime }}{x_t^{\prime }}=\frac{-\sin t}{2t}\\ & \frac{d^{2}y}{d{x}^2}=\frac{d}{dt}(\frac{-\sin t}{2t})\frac{dt}{dx}=\frac{-\cos t\cdot 2t+\sin t\cdot 2}{4t^2}\cdot \frac{1}{2t}\\ 6.& \frac{dx}{dy}{|}_{y=1}=\frac{1}{y_x^{\prime }}{|}_{x=0}=\frac{1}{3x^2+3}{|}_{x=0}=\frac{1}{3}\\ 7.& x_y^{\prime }={\frac{1}{y}}_x^{\prime },x_y^{\prime \prime }=-\frac{y_x^{\prime \prime }}{(y_x^{\prime }{)}^3}\\ & 有y_x^{\prime }=1+\frac{1}{x},y_x^{\prime \prime }=-\frac{1}{x^2},x_y^{\prime \prime }=-\frac{y_x^{\prime \prime }}{(y_x^{\prime }{)}^3}=-\frac{-1/x^2}{(1+\frac{1}{x}{)}^3}=\frac{x}{(1+x{)}^3}\\ 8.& 令y_1=\sqrt{\frac{(1+x)\sqrt{x}}{e^{x-1}}}⟹\ln y_1=\frac{1}{2}(\ln (1+x)+\frac{1}{2}\ln x-(x-1))\\ & ⟹\frac{1}{y_1}\cdot y_1^{\prime \prime }=\frac{1}{2}(\frac{1}{1+x}+\frac{1}{2x}-1)\\ & ⟹y_1^{\prime }(1)\underrightarrow{代入}0\\ & 令y_2=\arcsin \frac{1-x}{\sqrt{1+x^2}}⟹y_2^{\prime }(1)=\underset{x\to 1}{\lim }\frac{y_2(x)-y_2(1)}{x-1}\\ & =\underset{x\to 1}{\lim }\frac{\arcsin \frac{1-x}{\sqrt{1+x^2}}-0}{x-1}=\underset{x\to 1}{\lim }\frac{\frac{1-x}{\sqrt{1+x^2}}}{x-1}=-\frac{\sqrt{2}}{2}\\ & 故f^{\prime }(1)=-\frac{\sqrt{2}}{2}\end{array}$$

导数应用

1.切线问题——曲线的构造（显式、隐式、参数方程、极坐标）

2.性态问题——函数的构造（常规函数如显隐参极分变限），用极限${\lim }_{n\to \infty }$,用定积分定义，用微分方程的解，用行列式

3.渐近线、曲率、变化率问题——记好公式

题目

$$\begin{array}{rl}& 1.若曲线C:f(x)由方程2x-y=2arctan(y-x)确定，则曲线在点(1+\frac{2}{\pi },2+\frac{\pi }{2})的切线方程是\underline{}\\ & 2.已知两条曲线由y=f(x)与xy+e^{x+y}=1所确定，且在点(0,0)处的切线相同，写出此切线方程，求极限\underset{n\to 0}{\lim }nf(\frac{2}{n}）\\ & 3.使曲线f(x)=x^n在点(1,1)处的切线与x轴的交点为(x_n,0),n=1,2,\cdots ,求\underset{n\to \infty }{\lim }f(x_n)\\ & 4.求双曲线y_1=\frac{1}{x}与抛物线y_2=\sqrt{x}的交角\\ & 5.求函数f(x)=|x|e^{-|x-1|}的极值\\ & 6.设正值函数f(x)在(1,+\infty )内连续，求函数F(x)={\int }_1^x[(\frac{2}{x}+\ln x)-(\frac{2}{t}+\ln t)]f(t)dt的最小值点\\ & 7.设f(x)=\{\begin{array}{l}{\lim }_{n\to \infty }\frac{1}{n}(1+cos\frac{x}{n}+cos\frac{2x}{n}+\cdots +cos\frac{n-1}{n}x),x>0\\ 1,x=0\\ f(-x),x<0\end{array}\\ & (1)求f^{\prime }(0)(2)求f(x)在[-\pi ,\pi ]上的最大值\\ & 8.已知f^{\prime }(-x)=x[f^{\prime }(x)+1],求f(x)的极值点，并说明是极大值点还是极小值点\\ & 9.讨论常数a的值，确定曲线y=a{e}^x与y=1+x的公共点的个数\\ & 10.设f(x)可导，证明：f(x)的两个零点之间一定有f(x)+f^{\prime }(x)的零点\\ & 11.设x\in (0,1),证明下面不等式：（1）(1+x){\ln }^2(1+x)<x^2(2)\frac{1}{\ln 2}-1<\frac{1}{\ln (1+x)}-\frac{1}{x}<\frac{1}{2}\\ & 12.已知f(x)二阶可导，且f(x)>0,f(x)\cdot f^{\prime \prime }(x)-[f^{\prime }(x){]}^2\ge 0(x\in R)\\ & (1)证明f(x_1)\cdot f(x_2)\ge f^2(\frac{x_1+x_2}{2})(x_1,x_2\in R)\\ & (2)若f(0)=1,证明f(x)\ge e^{f^{\prime }(0)x}(x\in R)\end{array}$$

答案

$$\begin{array}{rl}1.& 2-y^{\prime }=\frac{2}{1+(y-x{)}^2}(y^{\prime }-1)⟹k=y^{\prime }{|}_p=\frac{3}{2}\\ & ⟹y-(2+\frac{\pi }{2})=\frac{3}{2}(x-(1+\frac{\pi }{2}))\\ 2.& 由xy+e^{x+y}=1,知y+x{y}^{\prime }+e^{x+y}(1+y^{\prime })=0\\ & ⟹y^{\prime }(0)=-1=k,\therefore y-0=-x\\ & ⟹I=\underset{n\to \infty }{\lim }\frac{f(\frac{2}{n})}{\frac{1}{n}}=\underset{n\to \infty }{\lim }\frac{f(0+\frac{2}{n})-f(0)}{\frac{2}{n}}\cdot 2=2f^{\prime }(0)=-2\\ 3.& f^{\prime }(x)=n{x}^{n-1}⟹k=n,故y-1=n(x-1)⟹x_n=1-\frac{1}{n}\\ & 故I=\underset{n\to \infty }{\lim }f(x_n)=\underset{n\to \infty }{\lim }(1-\frac{1}{n}{)}^n=e^A=e^{-1}\\ & 其中A=\underset{n\to \infty }{\lim }n(1-\frac{1}{n}-1)=-1\\ 4.& 交点(1,1),y_1^{\prime }(1)=(-\frac{1}{x^2}){|}_{x=1}=-1=\tan \alpha \\ & y_2^{\prime }(1)=(\frac{1}{2\sqrt{x}}){|}_{x=1}=\frac{1}{2}=\tan \beta \\ & ⟹r=\alpha -\beta =\frac{3}{4}\pi -\arctan \frac{1}{2}\\ 5.& f(x)=\{\begin{array}{l}-x{e}^{x-1},x<0\\ 0,x=0\\ x{e}^{x-1},0<x<1\\ 1,x=1\\ x{e}^{1-x},x>1\end{array}{f}^{\prime }(x)=\{\begin{array}{l}-e^{x-1}-x{e}^{x-1},x<0\\ e^{x-1}+x{e}^{x-1},0<x<1\\ e^{1-x}-x{e}^{1_x},x>1\end{array}\\ & f_{(}^{\prime }0)=-e^{-1},f_{+}^{\prime }(0)=e^{-1},\therefore f^{\prime }(0)不存在\\ & f_{(}^{\prime }1)=2,f_{+}^{\prime }(1)=0,\therefore f^{\prime }(1)不存在\\ & 知x_1=-1,x_2=0,x_3=1,则x_1=-1为极大点，x_2=0为极小点，x_3=1为极大点\\ 6.& F(x)={\int }_1^x(\frac{2}{x}+\ln x)f(t)dt-{\int }_1^x(\frac{2}{t}+\ln t)f(t)dt\\ & =(\frac{2}{x}+\ln x){\int }_1^{x}f(t)dt-{\int }_1^x(\frac{2}{t}+\ln t)f(t)dt\\ & ⟹F^{\prime }(x)=(-\frac{2}{x^2}+\frac{1}{x}){\int }_1^{x}f(t)dt+(\frac{2}{x}+\ln x)F(x)-(\frac{2}{x}+\ln x)f(x)\\ & 由F^{\prime }(x)=0知x=2是唯一极小值点,\therefore x=2是最小值点\\ 7.(1)& x>0时,f(x)=\underset{n\to \infty }{\lim }\sum _{i=0}^{n-1}\cos \frac{i}{n}x\cdot \frac{1}{n}=\underset{n\to \infty }{\lim }\sum _{i=0}^{n-1}\cos \frac{x}{n}i\cdot \frac{x}{n}\cdot \frac{1}{x}\\ & =\frac{1}{x}{\int }_0^x\cos tdt=\frac{\sin x}{x}⟹f(x)=\{\begin{array}{l}\frac{\sin x}{x},x>0\\ 1,x=0\\ \frac{\sin x}{,}x<0\end{array}\\ 8.& f^{\prime }(-1x)=x[f^{\prime }(x)+1]⟹f^{\prime }(x)=-x[f^{\prime }(x)+1]\\ & 代入,得f^{\prime }(x)=-x[x[f^{\prime }(x)+1]+1]⟹f^{\prime }(x)=\frac{-x^2-x}{1+x^2}\\ & 由f^{\prime }(x)=0,知x_1=0,x_2=-1\\ & \therefore x_1=0是极大值点，x_2=-1是极小值点\\ 9.& a{e}^x=1+x,令f(x)=a{e}^x-1-x\\ & ⟹a=\frac{1+x}{e^x},令f(x)=\frac{1+x}{e^x},f^{\prime }(x)=\frac{e^x-(1+x)e^x}{(e^x{)}^2}=\frac{-x}{e^x}\\ & ⟹x<0时,f(x)单调递增，x>0时,f(x)单调递减，由f(-\infty )=\underset{n\to \infty }{\lim }\frac{1+x}{e^x}=-\infty \\ & f(0)=1,且f(+\infty )=\underset{x\to +\infty }{\lim }\frac{1+x}{e^x}=0\\ & \{\begin{array}{l}a\ge 0,1\\ 0<a<1,2\\ a=1,1\\ a>1,nan\end{array}\\ 10.& f(x_1)=f(x_2)=0(x_1<x_2)\\ & ⟹f(\xi )+f^{\prime }(\xi )=0(x_1<\xi <x_2)\\ & F(x)=e^{x}f(x),则F(x_1)=0=F(x_2)⟹\exists \xi \in (x_1,x_2),使F^{\prime }(\xi )=e^{x}f(x)+e^x{f}^{\prime }(x){|}_{\xi }=0\\ & 即f(\xi )+f^{\prime }(\xi )=0\\ 11.(1)& 令f(x)=(1+x){\ln }^2(1+x)-x^2,0<x<1,\therefore f^{\prime }(x)={\ln }^{(}1+x)+2\ln (1+x)-2x,0<x<1\\ & 且f^{\prime \prime }(x)=\frac{2\ln (1+x)}{1+x}+\frac{2}{1+x}-2=\frac{2[\ln (1+x)-x]}{1+x}<0\\ & ⟹f^{\prime }(x)单调递减且f^{\prime }(0)=0,故f^{\prime }(x)<0,0<x<1\\ & ⟹f(x)单调递减且f(0)=0,故f(x)<0,0<x<1\\ (2)& 令g(x)=\frac{1}{\ln (1+x)}-\frac{1}{x},0<x<1⟹g^{\prime }(x)=\frac{-\frac{1}{1+x}}{{\ln }^2(1+x)}+\frac{1}{x^2}=\frac{(1+x){\ln }^2(1+x)-x^2}{x^2(1+x){\ln }^2(1+x)}\\ & 由(1)可知，g^{\prime }(x)<0,故g(x)单调递减，g(1)=\frac{1}{\ln 2}-1最小\\ & g(0)=\underset{x\to 0}{\lim }(\frac{1}{\ln (1+x)}-\frac{1}{x})=\underset{x\to 0}{\lim }\frac{x-\ln (1+x)}{x\ln (1+x)}=\frac{1}{2}最大\\ & 即\frac{1}{\ln 2}-1<\frac{1}{\ln (1+X)}-\frac{1}{x}<\frac{1}{2}\\ 12.(1)& \ln f(x_1)+\ln f(x_2)\ge 2\ln f(\frac{x_1+x_2}{2})⟹\frac{\ln f(x_1)+\ln f(x_2)}{2}\ge \ln f(\frac{x_1+x_2}{2})\\ & 令g(x)=\ln f(x),g^{\prime }(x)=\frac{f^{\prime }(x)}{f(x)}⟹g^{\prime \prime }(x)=\frac{f^{\prime \prime }(x)f(x)-f^{\prime }(x)f^{\prime }(x)}{f^2(x)}\ge 0\\ & 故g(x)是凹曲线，于是\frac{g(x_1)+g(x_2)}{2}\ge g(\frac{x_1+x_2}{2})\\ & 即\frac{\ln f(x)+\ln f(x_2)}{2}\ge \ln f(\frac{x_1+x_2}{2})\\ (2)& 由(1)得g(x)=\ln f(x)是凹曲线\\ & y-g(0)=g^{\prime }(0)(x-0)即y=f^{\prime }(0)x\\ & ⟹g(x)\ge y即\ln f(x)\ge f^{\prime }(0)x⟹f(x)\ge e^{f^{\prime }(0)x}\end{array}$$

积分

一元积分比大小

题目

$$\begin{array}{rl}& 1.设F(x)={\int }_x^{x+2\pi }{e}^{sint}sintdt,则F(x)=\underline{}\\ & 2.设I_k={\int }_0^{k\pi }{e}^{x^2}sinxdx(k=1,2,3),则有:()\\ & A.I_1<I_2<I_{3}B.I_3<I_2<I_{1}C.I_2<I_3<I_{1}D.I_2<I_1<I_3\\ & 3.设常数a>0,积分I_1={\int }_0^{\frac{2}{\pi }}\frac{cosx}{1+x^{\alpha }}dx,I_2={\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+x^{\alpha }}dx,则()\\ & A.I_1>I_{2}B.I_1<I_{2}C.I_1=I_{2}D.大小与\alpha 有关\\ & 4.证明{\int }_0^1\frac{x\cdot sin\frac{\pi }{2}x}{1+x}dx>{\int }_0^1\frac{x\cdot cos\frac{\pi }{2}x}{1+x}dx\end{array}$$

答案

$$\begin{array}{rl}1.& F^{\prime }(x)=e^{\sin (x+2\pi )}\sin (x+2\pi )-e^{\sin x}\sin x=0\\ & ⟹F(x)=c=F(0)={\int }_0^{2\pi }{e}^{\sin t}\sin tdt>0\\ & 故F(x)为正常数\\ 2.& I_1={\int }_0^{\pi }{e}^{x^2}\sin xdx,I_2={\int }_0^{2\pi }{e}^{x^2}\sin xdx,I_3={\int }_0^{3\pi }{e}^{x^2}\sin xdx\\ & 画图知该函数随着横坐标x的增大，其因式e^{x^2}也会增大，故其凹或凸的区间会增大\\ & I_1>0,I_2<0,I_3>I_1>0\\ 3.& I_1-I_2={\int }_0^{\frac{\pi }{2}}\frac{1}{1+x^{\alpha }}(\cos x-\sin x)dx\\ & ={\int }_0^{\frac{\pi }{4}}\frac{1}{1+x^{\alpha }}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1}{1+x^{\alpha }}(\cos x-\sin x)dx>0(前者大，后者小)\\ 4.& I_{左}-I_{右}={\int }_0^1\frac{x}{1+x}(\sin \frac{\pi }{2}x-\cos \frac{\pi }{2}x)dx\\ & ={\int }_0^{\frac{1}{2}}(\sin \frac{\pi }{2}x-\cos \frac{\pi }{2}x)dx+{\int }_{\frac{1}{2}}^1\frac{x}{1+x}(\sin \frac{\pi }{2}x-\cos \frac{\pi }{2}x)dx>0\end{array}$$

定积分定义

部分题在36讲配套视频中出现过，故请至相关笔记页面查看。
数列极限

题目

$$\begin{array}{rl}& 1.\underset{n\to \infty }{\lim }\frac{\sqrt[n]{(n+1)(n+2)\cdots (n+n)}}{n}=\underline{}\\ & 2.f(x)=\{\begin{array}{l}{e}^{-x},x̸=0\\ {\lim }_{n\to \infty }2[\frac{n}{(n+1{)}^2}+\frac{n}{(n+2{)}^2}+\cdots +\frac{n}{(n+n{)}^2}],x=0\end{array},求f^{\prime }(0)\\ & 3.(1)证明:当x\to 0^{+}时，不等式0<{\tan }^x-x^x<x^4成立。(2)设x_n=\sum _{k=1}^n{\tan }^2\frac{1}{\sqrt{n+k}},求\underset{n\to \infty }{\lim }{x}_n\end{array}$$

答案

$$\begin{array}{rl}1.I& =\underset{n\to \infty }{\lim }\sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})}\\ & =e^{\ln \sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})}}\\ & =e^{\frac{1}{n}{\sum }_{i=1}^n}\ln (1+\frac{i}{n})\\ & =e^{{\int }_0^1\ln (1+x)dx}=e^{2\ln 2-1}=\frac{4}{e}\\ 2.& x=0时，f(x)=\underset{n\to \infty }{\lim }2\sum _{i=1}^n\frac{n}{(n+i{)}^2}=2\underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{1}{(1+\frac{i}{n}{)}^2}\cdot \frac{1}{n}\\ & =2{\int }_0^1\frac{1}{(1+x{)}^2}dx=2(-\frac{1}{1+x}){|}_0^{\prime }=2(-\frac{2}{x}+1)=1\\ & f(x)=\{\begin{array}{l}{e}^{-x},x̸=0\\ 1,x=0\end{array},f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x}=\underset{x\to 0}{\lim }\frac{e^{-x}-1}{x}=-1\\ 3.(1)& \underset{x\to 0^{+}}{\lim }\frac{{\tan }^{2}x-x^2}{x^4}=\frac{2}{3}<1⟹x\to 0^{+}时，0<{\tan }^{2}x-x^2<x^4\\ (2)& 由(1)知x^2<{\tan }^{2}x<x^2+x^4\\ & ⟹\sum _{k=1}^n\frac{1}{n+k}<\sum _{k=1}^n{\tan }^2\frac{1}{\sqrt{n+k}}<\sum _{k=1}^n\frac{1}{n+k}+\sum _{k=1}^n\frac{1}{(n+k{)}^2}\\ & 由\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{1}{n+k}=\underset{n\to \infty }{\lim }\frac{1}{n}\cdot \frac{1}{1+\frac{k}{n}}=\int +0^1\frac{1}{1+x}dx\\ & 且\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{1}{(n+k{)}^2}<\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{1}{n^2}=\frac{1}{n}\\ & \therefore \underset{n\to \infty }{\lim }{x}_n=0\end{array}$$

变现积分（导数）

直接求导

题目

$$\begin{array}{rl}& 1.设f(x)为连续函数，且F(x)={\int }_{\frac{1}{x}}^{\ln x}f(t)dt,则F(x)=\\ & 2.\underset{x\to 0}{\lim }{\int }_0^x\frac{sin2t}{\sqrt{4+t^2}{\int }_0^x(\sqrt{t+1}-1)dt}dt\\ & 3.\underset{x\to 0}{\lim }\frac{{\int }_0^x[{\int }_0^{u^2}\arctan (1+t)dt]du}{x(1-\cos x)}\end{array}$$

答案

$$\begin{array}{rl}1.& F^{\prime }(x)=f(\ln x)\cdot \frac{1}{x}-f(\frac{1}{x})(\frac{-1}{x^2})\\ 2.& I=\underset{x\to 0}{\lim }\frac{{\int }_0^x\frac{\sin 2t}{\sqrt{4+t^2}}dt}{{\int }_0^x(\sqrt{t+1}-1)dt}=\underset{x\to 0}{\lim }\frac{\frac{\sin 2x}{\sqrt{4+x^2}}}{\sqrt{x+1}-1}\\ & =\frac{1}{2}\underset{x\to 0}{\lim }\frac{\sin 2x}{\sqrt{1+x}-1}=2\\ 3.& 令\phi (u)={\int }_0^{u^2}\arctan (1+t)dt\\ & 则I=\underset{x\to 0}{\lim }\frac{{\int }_0^x\phi (u)du}{x\cdot \frac{1}{2}{x}^2}=\underset{x\to 0}{\lim }\frac{\phi (x)}{\frac{3}{2}{x}^2}\\ & \underset{x\to 0}{\lim }\frac{{\int }_0^{x^2}\arctan (1+t)dt}{\frac{3}{2}{x}^2}\\ & \underset{x\to 0}{\lim }\frac{\arctan (1+x^2)\cdot 2x}{3x}=\frac{\pi }{6}\end{array}$$

拆分后再求导

题目

$$\begin{gathered} 设\phi (x)在[a,b]上连续，且\phi (x)>0,则函数y={\int }_a^b|x-t|\phi (t)dt,则() \\ A.在(a,b)内为凸B.在(a,b)内为凹C.在(a,b)内有拐点D.在(a,b)内有间断点 \end{gathered}$$

答案

$$\begin{array}{rl}y& ={\int }_a^b|x-t|\phi (t)dt\\ & ={\int }_a^x(x-t)\phi (t)dt+{\int }_x^b(t-x)\phi (t)dt\\ & ={\int }_a^{x}x\phi (t)dt-{\int }_a^{x}t\cdot \phi (t)dt+{\int }_x^{b}t\phi (t)dt-{\int }_x^{b}x\phi (t)dt\\ & =x{\int }_a^x\phi (t)dt-{\int }_a^{x}t\phi (t)dt+{\int }_x^{b}t\phi (t)dt-x{\int }_x^b\phi (t)dt\\ & ⟹y^{\prime }={\int }_a^x\phi (t)dt+x\phi (x)-x\phi (x)-x\phi (x)-{\int }_x^b\phi (t)dt+x\phi (x)\\ & ={\int }_a^x\phi (t)dt-{\int }_x^b\phi (t)dt⟹y^{\prime \prime }=\phi (x)+\phi (x)=2\phi (2)>0\end{array}$$

换元后再求导

题目

$$设f(x)在[0,+\infty )上可导，f(0)=0,其反函数为g(x),若{\int }_x^{x+f(x)}g(t-x)dt=x^2\ln (1+x),求f(x)$$

答案

$$\begin{array}{rl}& {\int }_x^{x+f(x)}g(t-x)dt\underrightarrow{令u=t-x}{\int }_0^{f(x)}g(u)du\\ & ⟹{\int }_0^{f(x)}g(u)du=x^2\ln (1+x)⟹g(f(x))\cdot f^{\prime }(x)=2x\ln (1+x)+\frac{x^2}{1+x}\\ & ⟹x\cdot f^{\prime }(x)=2x\ln (1+x)+\frac{x^2}{1+x}\\ & ⟹f^{\prime }(x)=2\ln (1+x)+\frac{x}{1+x}\\ & ⟹f(x)=2\int \ln (1+x)dx+\int \frac{x}{1+x}dx\\ & =2(x+1)\ln (1+x)-2\int (x+1)\cdot \frac{1}{1+x}dx\\ & =2(x+1)\ln (1+x)-2x+x-\ln (1+x)+c\\ & 又f(0)=0,故c=0\\ & 则f(x)=(2x+1)\ln (1+x)-x\end{array}$$