Chapter 1: Functions, Limits, Continuity

1.1 Basic Properties of Functions

Basic Elementary Functions

Power functions, exponents, logarithms, trigonometric functions, inverse trigonometric functions;

Elementary Functions

1. Constants and five classes of basic elementary functions;
2. Addition, subtraction, multiplication, division, and combination;
3. The resulting function expressed by an analytical expression;

Inverse and Composite Functions

Scenario One: $f^{-1}[f(x)]=x$
Scenario 2: $f[f^{-1}(x)]=x$

Parity

Parity Judgment:
Common odd functions: $\sin x, \tan x, \arcsin x, \arctan x, \ln\frac{1-x}{1+x}, {\ln(x+\sqrt{1+x^2})}, \frac{e^x-1}{e^x+1}, $$f(-x)=-f(x)$
Common even functions: $x^2, |x|, \cos x, f(x)=f(-x)$
Operation rules:
odd function + odd function =odd function
even function + even function = even function
odd function + even function =neither odd nor even
Odd function × odd function =Even function
Even function × even function = even function
odd function × even function = odd function
composite function:
What is odd inside is the same outside, what is even inside is even.
If the inner function is an odd function, the parity of the composite function is the same as that of the outer function;
If the inner function is an even function, then the composite function is an even function;
derivative relationship:
Derivatives of odd functions: even functions;
Derivatives of even functions: odd functions;

Periodic

1. $\sin x,\cos x\text\text{ The period is }2\pi;\,\sin2x,|\sin x|\text\text{ The period is }\pi$
2. $\text{ If }f(x)\textit{\text{ With }}T\text\text{ This is the cycle },\text{ then }f(a\text{x}+b)\textit{ \text{ With }}\frac T{|a|}\text{ This is the cycle }.$

1.2 limit, continuity

Infinitesimal

Same-order infinitesimals: $\frac{\alpha(x)}{\beta(x)}\to C$, where $C\neq0$.
Equivalent infinitesimals: $\frac{\alpha(x)}{\beta(x)}\to1$.
Higher-order infinitesimal: $\frac{\alpha(x)}{\beta(x)}\to0$; equivalently, $\alpha(x)=o(\beta(x))$.
Lower-order infinitesimal: $\left|\frac{\alpha(x)}{\beta(x)}\right|\to\infty$.
$$\text{ If }\lim\frac{\alpha (x)^{\color{red}{}}}{\left[\beta (x)\right]^{k}\color{red}}=C\neq 0,\text{ α is called the k-th order infinitesimal of β; }$$

continuity

Definition:
Definition 1: $y=f(x)$ is continuous at $x_0$ if $\lim_{\Delta x\to0}[f(x_0+\Delta x)-f(x_0)]=0$.
Definition 2: $y=f(x)$ is continuous at $x_0$ if $\lim_{x\to x_0}f(x)=f(x_0)$.
$f(x)$ Necessary and sufficient conditions for continuity: $f(x)$left continuity and right continuity,left and right continuity are equal;

Discontinuity

1. The first type of discontinuity point: both left and right limits exist;
Can remove discontinuous points: and left limit = right limit;
Jump break point: left limit is not equal to right limit;
2. The second type of discontinuity point: at least one of the left and right limits does not exist;
Infinite discontinuity point: for example, in $1/x$, when x=0;
Oscillation discontinuity point: For example, when x tends to 0, $sin1/x$ is the oscillation discontinuity point;
Infinity and oscillation are two common types in the second category, but they are not necessarily the only two types, and it may be neither infinity nor oscillation;
Common situations where it is necessary to divide left and right limits:
1. Piecewise function dividing point;
2. E infinity: $e^{\infty}$;
3. Arctan infinity;

1.3 Method to find limit

1.3.1 Use basic limit to find limit

Commonly used basic limit

$$\lim_{x\to0}\frac{\sin x}x=1;$$

Power exponent and root

$$\lim_{x\to0}(1+x)^{\frac1x}=e;$$
It is of type $1^\infty$, and the base of the power in the brackets is greater than 0;
$\lim_{x\to0}(1+\frac{1}{x})^{x}$ does not exist, because when x tends to 0, it can also tend to be negative toward 0;
Common mistakes: Note that the following limits do not equal a; $$\lim_{n\to \infty} a(\frac{n}{{n+1}})^n=\frac{a}{e}$$
$\lim_{x\to0^+}(1+\frac{1}{x})^{x}=1$
$\lim_{x\to\infty}(1+\frac1x)^x=e;$
It is of type $1^\infty$, and the base of the power in the brackets is greater than 0
$\lim_{x\to\infty}(1+x)^{\frac{1}{x}}$
$\lim_{x\to+\infty}(1+x)^{\frac{1}{x}}=1$
$\lim_{x\to0}\frac{a^x-1}x=\ln a;$
$\lim_{n\to\infty}\sqrt[n]{n}=1.$
Note that here is n, so it tends to positive infinity;
$\lim_{n\to\infty}\sqrt[n]{a}=1,(a>0).$
Note that here is n, so it tends to positive infinity;

piecewise function

$\left.\lim_{{x\to\infty}}\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}{{b}_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0}=\left|\begin{array}{ll}\frac{a_n}{b_m},&n=m,\\0,&n<m,\\\infty,&n>m.\end{array}\right.\right.$
With so many terms in the numerator and denominator, the result mainly depends on the largest number (the highest degree);
Note that x goes to infinity.
When x tends to 0, the value of the function should depend on the smallest number (lowest degree);
$\lim_{{n\to\infty}}x^n=\begin{cases}0,&|x|<1,\\1,&x=1,\\+\infty,&x>1,\\\text{does not exist},&x\leq-1.\end{cases}$
$\lim_{n\to\infty}e^{{nx}}=\begin{cases}0,&x<0,\\+\infty,&x>0\\1,&x=0.&\end{cases}$

1 to the power of infinity

For $1^{\infty}$, the result is an infinitive, not necessarily equal to 1;
$\text{ If }\lim\alpha(x)=0,\lim\beta(x)=\infty,\text{ and }\lim\alpha(x)\beta(x)=A.\text{ then }\lim(1+\alpha(x))^{\beta(x)}=e^A$
that is, the core is in $\lim\alpha(x)\beta(x)=A$, see if it is established;
Method Trilogy
1. Written in standard form: Original form $=\mathbf{lim}[1+\alpha(x)]^{\beta(x)}$;
2. Find limit: $\lim\alpha(x)\beta(x)=A; $
3. Write the result: Original $=e^A.$

1.3.2 equivalent infinitesimal substitution

Principle of equivalent substitution

1. Multiplication and division relationships can be exchanged
$\text{ If }\alpha\sim\alpha_1,\beta\sim\beta_1,\text{ then }\lim\frac\alpha\beta=\lim\frac{\alpha_1}\beta=\mathbf{lim}\frac\alpha{\beta_1}=\mathbf{lim}\frac{\alpha_1}{\beta_1}$
2. Addition and subtraction relationships can be exchanged under certain conditions
Subtraction
$\text{ If }\alpha{\sim}\alpha_1,\beta{\sim}\beta_1,\text{ and }\lim\frac{\alpha_1}{\beta_1}=A\neq1.\text{ then }\alpha{-}\beta{\sim}\alpha_1{-}\beta_1$
Conditions for subtraction and substitution:Two subtraction terms are not equivalent, A is not equal to 1;
For example: $sin2x - tanx$ ~ $2x - x$
Addition
$\text{ If }\alpha{\sim}\alpha_1,\beta{\sim}\beta_1,\text{ and }\lim\frac{\alpha_1}{\beta_1}=A\neq-1.\text{ then }\alpha{-}\beta{\sim}\alpha_1{+}\beta_1$
Conditions for subtraction and substitution: two addition terms are not equivalent, A is not equal to -1;

commonly equivalent infinitesimal

when x tends to 0;
${(1+x)^\alpha-1}\sim{\alpha x}$
$e^{x}-1\sim x$
Substitution of power exponential functions;
$$x\sim\sin x\sim\tan x\sim\arcsin x\sim\arctan x\sim\ln(1+x)\thicksim e^x-1;$$
Variant: $\ln q(x)\sim q(x)-1, q(x)\to1$
$$a^x-1\sim x\ln a,\quad{(1+x)^\alpha-1\sim\alpha x},\quad1-\cos x\sim\frac12x^2$$

Higher order equivalent infinitesimal

$x-\sin x\sim\frac16x^3$,\,$\tan x-x\sim\frac13x^3$
$\arcsin x-x\sim\frac16x^3\quad x-\arctan x\sim\frac13x^3$
$x-\ln(1+x)\sim\frac12x^2$

More general form of two equivalent infinitesimal

$\log^{a}(1+2x)\sim(2x)^{a}=2^{a}x^{a}$
$(1-ax)^{\frac{1}{\alpha}}\sim(\frac{1}{2}x^{2})^{\frac{1}{\alpha}}$

1.3.3 evaluating limits by algebraic manipulation

Calculation premise

The addition, subtraction, multiplication and division operations of limit can be calculated separately, provided that:The limit of each split part needs to exist;

Operation Rules

Rules:

lim(f(x)\pm g(x))=\lim f(x)\pm\lim g(x) \\\quad\quad\quad\quad\quad\quad lim(f(x)\cdot g(x))=\lim f(x)\cdot\lim g(x) \\\quad\quad\quad\quad\quad\quad lim\left(\frac{f(x)}{g(x)}\right)=\frac{\lim f(x)}{\lim g(x)}\quad(B\neq0)

Addition:
presence + presence: detachable
presence + absence: not removable
does not exist + does not exist: not necessarily

Common conclusions

1. $\lim f(x)=A\neq0\Rightarrow\lim f(x)g(x)=A\lim g(x);$
The limit of non-zero factors can be found first;
2. $\lim\frac{f(x)}{g(x)}$ exists, $\lim g(x)=0\Rightarrow\lim f(x)=0; $
3. $\lim\frac{f(x)}{g(x)}=A\neq0,\lim f(x)=0\Rightarrow\lim g(x)=0;$

1.3.4 L'Hôpital's rule

L'Hôpital's rule

$$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$

Prerequisites

1. $\lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)=0\left(\infty\right)$
2. $f(x)$ and $g(x)$ are differentiable in a punctured neighborhood of $x_0$, and $g'(x)\neq0$.
3. $\lim_{x\to x_0}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists;
Example: If the condition is that the function f (x) is second-order differentiable, use L'Hôpital's rule to use at most first-order derivatives

Usage environment

Type: $$\frac00;\quad\frac\infty\infty,\quad0\cdot\infty;\quad\infty-\infty;\quad1^\infty;\quad\infty^0;\quad0^0$$
Solution idea: $\frac00, \frac\infty\infty\quad\Leftarrow\begin{cases}\quad ∞-∞\\0\cdot\infty\quad\Leftarrow\begin{cases}1^\infty\\\infty^0\\0^0&\end{cases}&\end{cases}$
Conversion method:
$0\cdot\infty\quad$
Idea: Put one of 0 or infinity above (numerator) OR below (denominator);
$1^\infty;\quad\infty^0;\quad0^0$:\,
Idea: All three must be rewritten as 0 times infinity;
Method: rewrite $[f(x)]^{g(x)}$ as $e^{g(x)\ln f(x)}$.
Core: Find the infinity of the multiplied part above e;

1.3.5 Taylor's formula

Commonly used Taylor's formula

$$\mathrm{e^x=\sum_{n=0}^\infty\frac1{n!}x^n=1+x+\frac1{2!}x^2+\cdots\in(-\infty,+\infty)}$$
$$\sin\mathrm{x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac1{3!}x^3+\frac1{5!}x^5+\cdots,x\in(-\infty,+\infty)}$$
$$\cos\mathrm{x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac1{2!}x^2+\frac1{4!}x^4+\cdots,x\in(-\infty,+\infty)}$$
$$\mathrm{ln(1+x)=\sum_{n=0}^\infty\frac{(-1)^n}{n+1}x^{n+1}=x-\frac12x^2+\frac13x^3+\cdots,x\in(-1,1]}$$
$$\begin{aligned}&\frac1{1-\mathrm{x}}=\sum_{\mathrm{n=0}}^\infty\mathrm{x^n}=1+\mathrm{x+x^2+x^3+\cdots,x\in(-1,1)}\\&\frac1{1+\mathrm{x}}=\sum_{\mathrm{n=0}}^\infty(-1)^\mathrm{n}\mathrm{x}^\mathrm{n}=1-\mathrm{x}+\mathrm{x}^2-\mathrm{x}^3+\cdots,\mathrm{x}\in(-1,1)\end{aligned}$$
$$\mathrm{(1+x)^a=1+\sum_{n=1}^\infty\frac{a(a-1)\cdots(a-n+1)}{n!}x^n=1+ax+\frac{a(a-1)}{2!}x^2+\cdots,x\in(-1,1)}$$

Taylor's formula -Expand conclusion

Formula:

\begin{aligned} &\sin x=x-\frac1{3!}x^3+o\left(x^3\right) \\ &\arcsin x=x+\frac16x^3+o\left(x^3\right) \\ &\tan x=x+\frac13x^3+o\left(x^3\right) \\ &\arctan x=x-\frac13x^3+o\left(x^3\right) \\ &\cos x=1-\frac1{2!}x^2+\frac1{4!}x^4+o\left(x^4\right) \\ &\mathrm{e}^x=1+x+\frac1{2!}x^2+\frac1{3!}x^3+o\left(x^3\right) \\ &\ln\left(1+x\right)=x-\frac12x^2+\frac13x^3+o\left(x^3\right) \\ &\begin{aligned}\left(1+x\right)^\alpha=1+\alpha x+\frac{\alpha\left(\alpha-1\right)}{2!}x^2+\end{aligned}\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+o\left(x^3\right) \end{aligned}

Recitation skills:
Explanation:
Change: Change symbol, + - + - + -
order: factorial, such as 3
Formula:
sin: variable order 135
cox: level change 124
arcsin: positive 13 with 6
tanx: Positive 13
arctanx: change 135
ex: factorial 123
In (1+x): starting from x, changes to 123