Finding limit limx → 0 x-arcsinx / xsinxtan2x

Finding limit limx → 0 x-arcsinx / xsinxtan2x

When X - > 0, SiNx ~ x, tan2x ~ 2x
lim x-arcsinx/xsinxtan2x
=lim x-arcsinx/2x^3
=lim [1-1/√(1-x²)]/6x²
=lim [-x(1-x²)^(-3/2)]/12x
=lim [-(1-x²)^(-3/2)]/12
=-1/12

Find that limx tends to infinity (1 / 2 of xsinx)

LIM (x →∞) (1 / 2 of xsinx)
=lim（x→∞）sin(1/x)/(1/x)
=lim（t→0）sin(t)/t
=1

F (x) = x (x + 1) (x + 2). (x + n) finding the n-th derivative of F (0)
To be exact, it is the value when the n-th derivative of F (x) is equal to 0

On the right is a polynomial of degree n + 1, which can be extended to the following form: ax ^ (n + 1) power + BX ^ (n) power + Omit in the middle + cX.abc Let's not consider the coefficient for the moment. After n times of derivation, the rest must be ax + B. first of all, let's see that a, a is the coefficient of N + 1 power, we can see that it is 1. Then let's see that B, B are the coefficient of n power

Finding the n-order derivative of F (x) = 1 / (1 + X + x ^ 2) at x = 0

F (x) * (1 + X + x ^ 2) = 1, use Leibniz formula to obtain n-order derivative f ^ n (x) * (1 + X + x ^ 2) + NF ^ (n-1) (x) * (1 + 2x) + n (n-1) f ^ (n-2) (x) = 0, let x = 0 substitute an + Na (n-1) + n (n-1) a (n-2) = 0, where an = f ^ n (0). It is easy to know that A0 = 1, A1 = - 1, we can prove a (3n) = (3n)

Find the n-order derivative of the function y = x / (1-x ^ 2)

y=x/(1-x^2)
=1/2[1/(1-x)-1/(1+x)]
y=1/(1-x)
y'=1/(1-x)^2
y''=2/(1-x)^3
y^(n)=n!/(1-x)^(n+1)
y=1/(1+x)
y'=-1/(1+x)^2
y''=2/(1+x)^3
y^(n)=(-1)^n*n!/(1+x)^(n+1)
therefore
y=x/(1-x^2)
The n-th derivative of is;
y^n=1/2[n!/(1-x)^(n+1)-(-1)^n*n!/(1+x)^(n+1)]
=n!/2[1/(1-x)^(n+1)-(-1)^n/(1+x)^(n+1)]

Let f (x) have a second order continuous derivative in a certain field of point a, and
[f(a+h)+f(a-h)-2f(a)]/(h^2)
The limit value at x → 0
The answer is f (a), which is the second derivative of F (a)

Let f (x) have a second order continuous derivative in a certain field of point a, so: LIM (H → 0) {[f (a + H) + F (A-H) - 2F (a)] / h ^ 2}. It is an indeterminate form of (0 / 0). We can use lobita's rule I, and pay attention to the composition function

If f (x) has a first order continuous derivative in (- ∞, + ∞) and f (0) = 0, then when a =? G (x) = f (x) / x, X ≠ 0; a, x = 0 is continuous in (- ∞, + ∞)

If f (x) has a first order continuous derivative in (- ∞, + ∞) and f (0) = 0, then:
F '(0) = [f (0 + DX) - f (0)] / DX, DX approaches 0
=f(dx)/dx
If G (x) = f (x) / X is continuous at x = 0, then when x approaches 0, it should be equal to a
X approaches 0, f (x) / x = f '(0)
So a = f '(0)

Derivative exercise f (x) = (x-a) × g (x), G (x) has continuous second derivative at x = a, find f '(a), f ′
What is the result of finding F "(a)?

f′（a）=lim {f(x)-f(a)}/(x-a)=lim {(x-a)g(x)-0}/(x-a)=lim g(x)=g(a)

Let f (x), G (x) be continuous on [a, b], have second derivative in (a, b), f (a) = f (b) = 0, f (c) > 0, C belong to (a, b), then s belong to (a, B)
Let the second derivative of F (s) be 0

The function f (x) is continuous in the interval [a, b], and has second derivative in the interval (a, b)
From Lagrange mean value theorem f '(c) - f' (a) = (C-A) f '' (ξ) > 0, so f '(ξ) > 0; ξ ∈ (a, c)
Similarly, f '(b) - f' (c) = (B-C) f '' (η)

E ^ (x + y) + xy = 1, find the value of n-order derivative of F (x) at x = 0

This is a function derivative problem determined by the equation
e^(x+y)+xy=1
When x = 0, y = 0
The two sides of the equation are derived from X
e^(x+y)(1+y')+y+xy'=0 (1)
Substituting x = 0, y = 0 to get 1 + y '(0) = 0, the solution is y' (0) = - 1
(1) We continue to derive x on both sides of X, and we get
e^(x+y)(1+y')^2+e^(x+y)y"+2y'+xy"=0
Substitute x = 0, y = 0, y '(0) = - 1 to get y "(0) = 2
With the above process, we can judge that all the items with (1 + y ') factor are 0, and the items with X and y are 0 at last
Therefore, the n-th derivative of the original equation for X is obtained
(1+y')g(x)+e^(x+y)y^(n)+(n-1)y^(n-1)+xh(x)=0
So we get: y ^ (n) (0) = - (n-1) y ^ (n-1) (0) (2)
Repeated use of (2) resulted in
y^(n)(0)=(-1)^(n-1)(n-1)!y'(0)
=(-1)^n×(n-1)!