From 3305da1358bf681927611c2598d8a68238bfa833 Mon Sep 17 00:00:00 2001 From: Didnelpsun <48906416+Didnelpsun@users.noreply.github.com> Date: Wed, 10 Feb 2021 23:15:27 +0800 Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E4=B8=8E=E7=BB=93=E6=9E=84?= =?UTF-8?q?=E8=B0=83=E6=95=B4?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../{ => exercise}/1-limit/limit.tex | 36 +++++- .../continuity-and-discontinuity.tex | 119 ++++++++++++++++++ .../{ => knowledge}/0-perpare/perpare.tex | 0 .../function-and-limit.tex | 2 +- .../derivatives-and-differential.tex | 0 ...heorem-and-applications-of-derivatives.tex | 0 model/model.tex | 2 +- 7 files changed, 155 insertions(+), 4 deletions(-) rename advanced-math/{ => exercise}/1-limit/limit.tex (93%) create mode 100644 advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex rename advanced-math/{ => knowledge}/0-perpare/perpare.tex (100%) rename advanced-math/{ => knowledge}/1-function-and-limit/function-and-limit.tex (99%) rename advanced-math/{ => knowledge}/2-derivatives-and-differential/derivatives-and-differential.tex (100%) rename advanced-math/{ => knowledge}/3-differential-mean-value-theorem-and-applications-of-derivatives/differential-mean-value-theorem-and-applications-of-derivatives.tex (100%) diff --git a/advanced-math/1-limit/limit.tex b/advanced-math/exercise/1-limit/limit.tex similarity index 93% rename from advanced-math/1-limit/limit.tex rename to advanced-math/exercise/1-limit/limit.tex index 74d9735..b9911a4 100644 --- a/advanced-math/1-limit/limit.tex +++ b/advanced-math/exercise/1-limit/limit.tex @@ -28,7 +28,7 @@ \usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref} % 超链接 \author{Didnelpsun} -\title{极限练习题} +\title{极限} \date{} \begin{document} \maketitle @@ -485,11 +485,43 @@ $=\lim\limits_{x\to 0}\dfrac{(1-a)x-\left(\dfrac{1}{2}+b\right)x^2+o(x^2)}{x^2}= $1-a=0;-\left(\dfrac{1}{2}+b\right)=2$\medskip -$\therefore a=1;b=-\dfrac{5}{2}$ +$\therefore a=1;b=-\dfrac{5}{2}$。 \subsection{极限存在性} +一般会给出带有参数的例子，并给定一个点指明在该点极限存在，求参数。 +若该点极限存在，则该点两侧的极限都相等。\medskip + +\textbf{例题：}设函数$f(x)=\left\{\begin{array}{lcl} + \dfrac{\sin x(b\cos x-1)}{e^x+a}, & & x>0 \\ + \dfrac{\sin x}{\ln(1+3x)}, & & x<0 +\end{array} +\right.$在$x=0$处极限存在，则$a$，$b$分别为。 + +解：首先根据极限在$x=0$存在，且极限的唯一性。分段函数在0两侧的极限值必然相等。 + +$\because\lim\limits_{x\to 0^-}\dfrac{\sin x}{\ln(1+3x)}=\lim\limits_{x\to 0^-}\dfrac{\sin x}{3x}=\dfrac{1}{3}=\lim\limits_{x\to 0^+}\dfrac{\sin x(b\cos x-1)}{e^x+a}$。 + +\medskip + +又$\lim\limits_{x\to 0^+}\dfrac{\sin x(b\cos x-1)}{e^x+a}$的分母的$e^x$当$x\to 0^+$时$e^x\to 1$，假如$a\neq-1$，则$e^x+a\neq 0$，则为一个常数。 + +从而提取常数因子：$\lim\limits_{x\to 0^+}\dfrac{\sin x(b\cos x-1)}{e^x+a}=\dfrac{1}{1+a}\lim\limits_{x\to 0^+}\sin x(b\cos x-1)$，这时候$\sin x$是趋向0的，而$b\cos x-1$无论其中的$b$为何值都是趋向一个常数或0，这时候他们的乘积必然为无穷小，从而无法等于$\dfrac{1}{3}$这个常数。 + +$\therefore a=-1$，从而让极限式子变为一个商的形式：\medskip + +$\lim\limits_{x\to 0^+}\dfrac{\sin x(b\cos x-1)}{e^x+a}$\medskip + +$=\lim\limits_{x\to 0^+}\dfrac{\sin x(b\cos x-1)}{e^x-1}$\medskip + +$=\lim\limits_{x\to 0^+}\dfrac{\sin x(b\cos x-1)}{x}$\medskip + +$=\lim\limits_{x\to 0^+}b\cos x-1$\medskip + +$=b-1=\dfrac{1}{3}$\medskip + +$\therefore a=-1,b=\dfrac{4}{3}$。 \subsection{极限唯一性} diff --git a/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex b/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex new file mode 100644 index 0000000..34343d8 --- /dev/null +++ b/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex @@ -0,0 +1,119 @@ +\documentclass[UTF8, 12pt]{ctexart} +% UTF8编码，ctexart现实中文 +\usepackage{color} +% 使用颜色 +\usepackage{geometry} +\setcounter{tocdepth}{4} +\setcounter{secnumdepth}{4} +% 设置四级目录与标题 +\geometry{papersize={21cm,29.7cm}} +% 默认大小为A4 +\geometry{left=3.18cm,right=3.18cm,top=2.54cm,bottom=2.54cm} +% 默认页边距为1英尺与1.25英尺 +\usepackage{indentfirst} +\setlength{\parindent}{2.45em} +% 首行缩进2个中文字符 +\usepackage{setspace} +\renewcommand{\baselinestretch}{1.5} +% 1.5倍行距 +\usepackage{amssymb} +% 因为所以 +\usepackage{amsmath} +% 数学公式 +\author{Didnelpsun} +\title{连续与间断} +\date{} +\begin{document} +\maketitle +\pagestyle{empty} +\thispagestyle{empty} +\tableofcontents +\thispagestyle{empty} +\newpage +\pagestyle{plain} +\setcounter{page}{1} +\section{连续} + +连续则极限值等于函数值。 + +\subsection{求连续区间} + +若要考察一个函数的连续区间，必须要了解函数的所有部分，一般会给出分段函数，所以要了解分段函数的每段函数的性质。\medskip + +\textbf{例题：}$f(x)=\lim\limits_{n\to\infty}\dfrac{x+x^2e^{nx}}{1+e^{nx}}$，求函数连续区间。\medskip + +注意到函数的形式为一个极限值，其极限趋向的变量为$n$（$n\to\infty$指$n\to+\infty$）。 + +\subsection{已知连续区间求参数} + +一般会给出带有参数的分段函数，要计算参数就必须了解连续区间与函数之间的关系。 + +\textbf{例题：}$f(x)=\left\{\begin{array}{lcl} + 6, & & x\leqslant 0 \\ + \dfrac{e^{ax^3}-1}{x-\arcsin x}, & & x>0 +\end{array} +\right.$，$g(x)=\left\{\begin{array}{lcl} + \dfrac{3\sin(x-1)}{x-1}, & & x<1 \\ + e^{bx}+1, & & x\geqslant 1 +\end{array} +\right.$，\smallskip \\ 若$f(x)+g(x)$在$R$上连续，则求$a,b$。 + +解：已知$f(x)+g(x)$在$R$上连续，但是不能判断$f(x)$与$g(x)$的连续性。 + +所以分开讨论。 + +对于$f(x)$因为左侧为常数函数，所以若是$f(x)$连续，则必然：\medskip + +$\lim\limits_{x\to 0^+}\dfrac{e^{ax^3}-1}{x-\arcsin x}=6$\medskip + +$\therefore\lim\limits_{x\to 0^+}\dfrac{e^{ax^3}-1}{x-\arcsin x}$\medskip + +$=\lim\limits_{x\to 0^+}\dfrac{ax^3}{x-\arcsin x}$\medskip + +$\text{令}t=\arcsin x\Rightarrow=\lim\limits_{x\to 0^+}\dfrac{a\sin^3t}{\sin t-t}$ + +$=a\lim\limits_{x\to 0^+}\dfrac{t^3}{\sin t-t}$\medskip + +$=a\lim\limits_{x\to 0^+}\dfrac{3t^2}{\cos t-1}=-6a=6$。\medskip + +$\therefore a=-1$时$f(x)$在$R$上连续。\medskip + +对于$g(x)$，当$x<1$时，$\lim\limits_{x\to 1^-}\dfrac{3\sin(x-1)}{x-1}=\lim\limits_{t\to 0^-}\dfrac{3\sin t}{t}=3$。\medskip + +$\therefore\lim\limits_{x\to 1^+}e^{bx}+1=e^b+1=3$。\medskip + +$\therefore b=\ln 2$时$g(x)$在$R$上连续。\medskip + +$\therefore a=-1,b=\ln 2$时$f(x)+g(x)$在$R$上连续。而$a\neq -1$时$f(x)+g(x)$在$x=0$时不连续，$b\neq\ln 2$时$f(x)+g(x)$在$x=1$时不连续。 + +\section{间断} + +\subsection{求间断点} + +\subsection{已知间断点求参数} + +这种题目已知间断点，而未知式子中的参数，只用将间断点代入式子并利用极限计算间断点的类型就可以了。 + +\textbf{例题：}$f(x)=\dfrac{e^x-b}{(x-a)(x-b)}$有无穷间断点$x=e$，可去间断点$x=1$，求$ab$的值。 + +已知有两个间断点$x=a,x=b$，其中无穷间断点指极限值为无穷的点，可去间断点表示极限值存在且两侧相等，但是与函数值不相等的点。 + +已经给出两个间断点的值为$x=1$和$x=e$，所以$ab$必然对应其中一个，但是不清楚到底谁是谁。 + +当$a=1,b=e$时，$f(x)=\dfrac{e^x-e}{(x-1)(x-e)}$。\medskip + +当$x\to 1$时，$\lim\limits_{x\to 1}\dfrac{e^x-e}{(x-1)(x-e)}$$=\dfrac{1}{1-e}\lim\limits_{x\to 1}\dfrac{e^x-e}{x-1}$$=\dfrac{e}{1-e}\lim\limits_{x\to 1}\dfrac{e^{x-1}-1}{x-1}$$=\dfrac{e}{1-e}\lim\limits_{x\to 1}\dfrac{x-1}{x-1}$$=\dfrac{e}{1-e}$。\medskip + +$\therefore x=1$为可去间断点。\medskip + +当$x\to e$时，$\lim\limits_{x\to e}\dfrac{e^x-e}{(x-1)(x-e)}$$=\dfrac{1}{e-1}\lim\limits_{x\to e}\dfrac{e^x-e}{x-e}$$=\dfrac{e}{e-1}\lim\limits_{x\to e}\dfrac{e^{x-1}-1}{x-e}$\medskip$=\dfrac{e}{e-1}\lim\limits_{x\to e}\dfrac{x-1}{x-e}$$=\dfrac{e(e-1)}{e-1}\lim\limits_{x\to e}\dfrac{1}{x-e}=\infty$。\medskip + +$\therefore x=e$为无穷间断点。\medskip + +当$a=e,b=1$时，$f(x)=\dfrac{e^x-1}{(x-e)(x-1)}$。\medskip + +而作为分子的$e^x-1$必然为一个常数，当式子趋向$1$或$e$的时候分母两个不等式中的一个不等式必然为一个常数，从而另一个不等式则变为了无穷小，所以$\lim\limits_{x\to 1}f(x)=\lim\limits_{x\to e}f(x)=\infty$。 + +$\therefore a=1,b=e$。 + +\end{document} diff --git a/advanced-math/0-perpare/perpare.tex b/advanced-math/knowledge/0-perpare/perpare.tex similarity index 100% rename from advanced-math/0-perpare/perpare.tex rename to advanced-math/knowledge/0-perpare/perpare.tex diff --git a/advanced-math/1-function-and-limit/function-and-limit.tex b/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex similarity index 99% rename from advanced-math/1-function-and-limit/function-and-limit.tex rename to advanced-math/knowledge/1-function-and-limit/function-and-limit.tex index 9aadd25..413a917 100644 --- a/advanced-math/1-function-and-limit/function-and-limit.tex +++ b/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex @@ -1074,7 +1074,7 @@ $\lim\limits_{x\to 0}f(x)=\lim\limits_{x\to 0}\ln\vert x\vert\cdot\sin x=\lim\li \subsubsection{连续函数四则运算的连续性} -若两个函数在某点连续，则这两个函数的和差积商在该点都连续。 +若两个函数在某点连续，则这两个函数的和差积商在该点都连续。但是如果两个在某点不连续的函数，其和差积商在某点的连续性都是不一定的，所以反过来，如果一个函数的和差积商是在某点连续的，不能说明这个组成的多个函数在该点是连续的。 \subsubsection{反函数的连续性} diff --git a/advanced-math/2-derivatives-and-differential/derivatives-and-differential.tex b/advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex similarity index 100% rename from advanced-math/2-derivatives-and-differential/derivatives-and-differential.tex rename to advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex diff --git a/advanced-math/3-differential-mean-value-theorem-and-applications-of-derivatives/differential-mean-value-theorem-and-applications-of-derivatives.tex b/advanced-math/knowledge/3-differential-mean-value-theorem-and-applications-of-derivatives/differential-mean-value-theorem-and-applications-of-derivatives.tex similarity index 100% rename from advanced-math/3-differential-mean-value-theorem-and-applications-of-derivatives/differential-mean-value-theorem-and-applications-of-derivatives.tex rename to advanced-math/knowledge/3-differential-mean-value-theorem-and-applications-of-derivatives/differential-mean-value-theorem-and-applications-of-derivatives.tex diff --git a/model/model.tex b/model/model.tex index b851cd3..a54ae48 100644 --- a/model/model.tex +++ b/model/model.tex @@ -1,4 +1,4 @@ -\documentclass[UTF8]{ctexart} +\documentclass[UTF8, 12pt]{ctexart} % UTF8编码，ctexart现实中文 \usepackage{color} % 使用颜色