Conquer the limit of Higher Mathematics

2021-06-27 09:52:40

One 、 Common methods of finding function limit

1.1 Using rational operation algorithm

1. There is +- non-existent = non-existent
2. There is *÷ non-existent = not always
3. non-existent +-*÷ non-existent = not always

1.2 Using the basic limit to find the limit

\begin{aligned} & \lim_{n\to\infin}\sqrt[n]n=1 \\ & \lim_{n\to\infin}\sqrt[n]a=1 \\ \end{aligned}

1.3 Using the Equivalent Infinitesimal Substitution

\begin{aligned} & \alpha{x}-1\backsim{xln\alpha} \\ & x-\ln{(1+x)} \backsim\frac{x^2}{2} \\ & \arctan{x}<\sin{x}<x<\arcsin{x}<\tan{x} \qquad\text{ It's easy to remember the following infinitesimal substitutions } \\ & x-\arctan{x} = \frac{x^3}{3} \\ & x-\sin{x} = \frac{x^3}{6} \\ & \arcsin{x}-x = \frac{x^3}{6} \\ & \tan{x}-x = \frac{x^3}{3} \end{aligned}

1. When $$\frac{f(x)}{g(x)}=1$$,$$\int_0^x f(t)dt \backsim \int_0^x g(t)dt$$

2. The principle of Equivalent Infinitesimal Substitution ：

1. The multiplication division relationship can be changed at will
2. The relationship of addition and subtraction can be changed under certain conditions （ The principle is that the result of substitution can not be added or subtracted as 0）

if $$\alpha\backsim\alpha_1\,,\beta\backsim\beta_1$$, And $$\lim \frac{\alpha_1}{\beta_1}=A\neq{1}$$, be $$\alpha-\beta\backsim\alpha_1-\beta_1$$
if $$\alpha\backsim\alpha_1\,,\beta\backsim\beta_1$$, And $$\lim \frac{\alpha_1}{\beta_1}=A\neq{-1}$$, be $$\alpha+\beta\backsim\alpha_1+\beta_1$$

1. A little

1.5 Using Taylor's formula

\begin{aligned} & e^x = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+o(x^n) \\ & \sin{x} = x-\frac{x^3}{3!}+\cdots+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} \\ & \cos{x} = 1-\frac{x^2}{2!}+\cdots+(-1)^n\frac{x^n}{2n!}+o(n^2n) \\ & \ln{(x+1)} = x-\frac{x^2}{2}+\cdots+(-1)^{n-1}\frac{x^n}{n}+o(x^n) \\ & \frac{1}{1-x} = 1+x+x^2+\cdots+x^n+\cdots \\ & \frac{1}{1+x} = 1-x+x^2-x^3+\cdots+(-1)^nx^n \\ \end{aligned}

1. A little

1. A little

1. A little

1.9 Using Lagrange's middle finger Theorem

1. A little
Two 、 Common problems of finding function limit

2.1 $$\frac{0}{0}$$ （ a key ）

1. The core idea ： Eliminate... From the denominator 0 factor

2. Common operations

1. Lobida
2. Equivalent Infinitesimal Substitution
3. Taylor formula

2.2 $$\frac{\infin}{\infin}$$

1. Common methods ：

1. Lobida
2. Molecular denominator At the same time, divide by the infinity of the highest order of each term

2.3 $$0*\infin$$

1. Common methods ：

1. Into $$\frac{0}{0} or \frac{\infin}{\infin}$$
2. You can put 0 Making infinitesimal substitutions

2.4 $$\infin-\infin$$

1. Common methods ：

1. （ The difference between fractions ） Through differentiation into $$\frac{0}{0} or \frac{\infin}{\infin}$$
2. （ The radical difference ） The radical is rationalized into $$\frac{0}{0} or \frac{\infin}{\infin}$$
3. Extracting infinite factors , become $$(1+x)^\alpha-1$$ In the form of

2.5 $$1^\infin$$ （ a key ）

1. The best way ：

1. In standard form ：$$simple form =\lim{[1+\alpha(x)]^{\beta{(x)}}}$$
2. Find the limit ：$$\lim{\alpha(x)\beta(x)}=A$$
3. Write the result ：$$simple form =e^A$$

2.6 $$\infin^{0} and 0^0$$

1. Common methods ：

1. $$\lim{[f(x)]^{g(x)}}=\lim{{e}^{g(x)\ln{f(x)}}}$$
3、 ... and 、 sequence limit

3.1 The limit of the sequence of infinitives

1. Pay attention to the sequence In limit n Need to change to x To use lobita or other methods , because n It's a discrete point ,x It's a continuous point

3.2 n Xiang He

1. Pinch theorem
2. The definition of definite integral
3. notes ： When the changing part relative to the main part is Second order , With a pinch ; When the changing part is of the same magnitude as the main part , Define... With definite integral

3.3 n Xiang Liancheng

1. Pinch theorem
2. Take the logarithm Into n Xiang He

3.4 Recursive relations

3.4.1 Method 1

1. Proband $$\{x_n\}$$ convergence （ Monotone bounded criteria ）

2. Using the equation $$x_{n+1}=f(x_n)$$ Take the limit on both sides and find out A

3. prove $$\{x_n\}$$ Monotone method ：

1. $$x_{n+1}-x_n$$
2. $$\{x_n\}$$ Invariant number , And $$\frac{x_{n+1}}{x_n}\geq{1}(\leq{1})$$
3. adopt $$f(x_n)$$ Monotonicity judgment ：

$$f(x_n)$$ Monotonous increase ,$$x_1\leq{x_2}$$,$$\{x_n\}$$ Monotonous increase
$$f(x_n)$$ Monotonous increase ,$$x_1\geq{x_2}$$,$$\{x_n\}$$ Monotone minus
$$f(x_n)$$ Monotone minus ,$$\{x_n\}$$ Not monotonous , Only method two

3.4.2 Method 2

1. First, we use the recursive equation to find A, Then suppose $$\lim_{n\to\infin} x_n = A$$
2. Then we use the recurrence relation to prove that $$|x_n-A| < B|x_{n-1}-A|$$, As long as $$B\in(0,1)$$
3. Finally, we prove that $$0<|x_n-A|<B^{n-1}|x_1-A|<0$$, Obtain evidence
Four 、 The simplification technique of finding limit （ The accumulation of ）
1. notice $$\frac{1}{x}$$, We can consider the substitution
2. infinity + infinity , Low order infinity can be ignored ; An infinitesimal + An infinitesimal , Higher order infinitesimals can be ignored
3. $$x\to-\infin$$ when , Divide $$-x$$ It can avoid the positive and negative problems of factors
4. Encounter non 0 The factor has to be found first , And then bring it up
5. $${\frac{x}{x-a}}^x \implies {\frac{x-a}{x}}^{-x}$$
6. 



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