Calculate the limit of a function. If the numerator and denominator are all 0, what should we do next?

Calculate the limit of a function. If the numerator and denominator are all 0, what should we do next?

Taylor expansion formula of common functions

If a function is n-order differentiable, then the function can be expanded n-order by Taylor formula
That is, f (x) = f (x0) + F '(x0) (x-x0) + F' (x0) (x-x0) / 2! +... + F ＾ (n) (x0) (x-x0) ＾ (n) / N! + 0x
F ＾ (n) (x0) denotes the n-th derivative of F (x) at x0. 0x denotes infinitesimal of higher order than (x-x0) ＾ (n)
If expressed by Lagrangian remainder, then 0x = f ＾ (n + 1) (zeta) (x - zeta) ＾ (n + 1) / N + 1!
McLaurin's formula is a special case of Taylor's formula expanded at 0
Taylor's formula can easily let you get the coefficient of the power term of X in the expansion of F (x). It can also deduce the original function from the derivative value of the known function. It is often used to solve limit problems
For example, find the limit of LIM (E ＾ x-x-1) / X when x approaches 0
F (x) = e ＾ x quadratic expansion at x = 0 = e ＾ (0) + e ＾ (0) * (x-0) + e ＾ (0) (x-0) / 2! + 0x
=1+x+x/2;
Then LIM (E ＾ x-x-1) / x = LIM (1 + X + X / 2-x-1) / x = 1 / 2
F '(x) = Lim [f (x) - f (x0)] / (x-x0) where X - > X0
If X - > x0, Lim f (x) - f (x0) = f '(x) (x-x0)
Lim f (x) whose error Lagrangian remainder in F (x) is f ＾ (2) (zeta) (x - zeta) ＾ (2) / 2! Is a higher order infinitesimal of (x-x0), which is generally used in proving problems

How to expand complex function into Taylor function?
The title is: F (z) = Z-1 / Z + 1, at z = 1
The expansion is Taylor series and its convergence region is pointed out

f(z)=1-(2/z+1)=1-(2/z-1+2)=1-(1/1+(z-1/2))=1-E(z-1/2)^n*(-1)^n
The convergence domain with convergence domain of / Z-1 / 2 / 1 {only the numerator and denominator divide Z-1}

Taylor expansion of two functions
Find the Taylor expansion of function f (x) = (x + 2) ^ (1 / 2) in x = 2
Find the Taylor expansion of F (x) = cos (2x) in x = Pi

Let t = X-2, then x = t + 2,
F (x) = (T + 4) ^ (1 / 2), which can be expanded into an equation about t
f(x)=2(1+t/4)^(1/2)
Because (1 + x) ^ μ = 1 + μ x + (μ (μ - 1) / 2!) x ^ 2 + (μ (μ - 1) (μ - 2) / 3!) x ^ 3 +
(1+x)^(1/2)=1+x/2-x^2/8+x^3/16-.
So f (x) = 2 [1 + T / 8-t ^ 2 / 128 + T ^ 3 / 1024 -.], the convergence domain is | T / 4|

What is the Taylor expansion of COS function?

See figure

How to prove Taylor expansion of analytic function

In mathematics, Taylor's formula is a formula that uses the information of a function at a certain point to describe the value of a function nearby. If the function is smooth enough and the derivative values of the function at a certain point are known, the Taylor's formula can be used to describe the value of a function, Taylor's formula can use these derivative values as coefficients to construct a polynomial to approximate the value of the function in the neighborhood of this point. Taylor's formula also gives the deviation between the polynomial and the actual value of the function. Taylor's formula is named after British mathematician Brooke Taylor. He first described this formula in a letter in 1712, Although James Gregory discovered its special case in 1671, in mathematics, Taylor formula is a formula that uses the information of a function at a certain point to describe its value nearby. If the function is smooth enough, and the derivative values of the function at a certain point are known, the Taylor formula can be used to describe the value of the function, Taylor's formula can use these derivative values as coefficients to construct a polynomial to approximate the value of the function in the neighborhood of this point. Taylor's formula also gives the deviation between the polynomial and the actual value of the function. Taylor's formula is named after British mathematician Brooke Taylor. He first described this formula in a letter in 1712, Although in 1671 James Gregory had discovered its special case

High number commonly used in the McLaurin formula list, there is to ask experts to help how to remember these formulas

e^x=1+x+x^2/2!+x^3/3!+.
There are just a few formulas, which are easy to remember. The important thing is that you know how to get them. In fact, they are the expansion of Taylor series

Why do we use Taylor's formula to make x zero = 0 (that is, McLaughlin's formula)?
Same as above, please do me a favor,

It's not necessary to make x0 = 0. Otherwise, with McLaughlin's formula, why return Taylor's formula?
Let x0 = 0 be the meaning of the problem, which requires the expansion to be found in the neighborhood of 0
In addition, the expansion is the simplest polynomial, which is convenient for analysis and calculation
Sometimes for special needs, we have to make x0 other values

What Taylor formula can approximately Express function and derivation process
How does Taylor's formula come from and why can it approximately express functions

Function is a curve intuitively. The purpose of Taylor's formula is to express some formulas that are not easy to operate, such as sine and cosine, with polynomials. (later you will realize that polynomials are the simplest function expressions). The approximation method of Taylor's formula is to use the n-order derivative of multinomial formula and the n-order derivative of F (x)

How to deduce the second order Taylor formula of Taylor Theorem f (x + H)

Mathematical analysis books are available