当前位置:网站首页>Conquer the limit of Higher Mathematics

Conquer the limit of Higher Mathematics

2021-06-27 09:52:40 wx5b1fd43180419

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 \(\alpha+\beta\backsim\alpha_1+\beta_1\)

1.4 Using lobida

  1. A little

1.5 Using Taylor's formula

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

1.6 Using the pinch rule

  1. A little

1.7 Use definite integral to define

  1. A little

1.8 Using the monotone bounded criterion

  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. \[\]

\[\]

版权声明
本文为[wx5b1fd43180419]所创,转载请带上原文链接,感谢
https://chowdera.com/2021/05/20210520194604511g.html