It is known that the function f (x) is a subtractive function of the definition field on (0, positive infinity), and satisfies f (XY) = f (x) + F (y), f (1 / 3) = 1 if f (x) - 2F (2-x) < 2 Find the value range of X

It is known that the function f (x) is a subtractive function of the definition field on (0, positive infinity), and satisfies f (XY) = f (x) + F (y), f (1 / 3) = 1 if f (x) - 2F (2-x) < 2 Find the value range of X

2＝1＋1＝f(1/3)+f(1/3)＝f[(1/3)(1/3)]＝f(1/9)
So f (x) - 2F (2-x)

Given the function f (x) = - 2 ^ X / (2 ^ x + 1). (1) prove that the function f (x) is a subtractive function on (negative infinity, positive infinity) by using the definition field (2) If x ∈ [1,2], find the value range of function f (x); (3) If G (x) = A / 2 + F (x), and G (x) ≥ 0 is constant when x ∈ [1,2], find the value range of real number a.

F (x) = - 2 ^ X / (2 ^ x + 1) = - 1-2 ^ x + 1) / (2 ^ x + 1) = - 1 + 1 / (2 ^ x + 1) since 2 ^ x + 1 is an increasing function, 1 / (2 ^ x + 1) is a decreasing function, so f (x) = - 2 ^ X / (2 ^ x + 1) = - 1 + 1 / (2 ^ x + 1) is a decreasing function. When x = 1, f (1) = - 2 / 3x = 2, f (1) = - 4 / 5, so x ∈ [1,2], find the value range [- 4 / 5, - 2 / 3] g of function f (x)

lim x-sinx/x+sinx x→0

(x → 0) LIM (x-sinx) / (x + SiNx). Robida's law
=(x→0)lim (1-cosx)/(1+cosx)
=0/2
=0

How to prove LIM (x → 0) SiNx / x = 1?

This theorem is an important theorem in advanced mathematics or calculus, which has been proved in books
Repeat for reference

LIM (x tends to 0) SiNx / x = 1, then LIM (x tends to 0) x / SiNx =? How?

The same is 1 because they are equivalent infinitesimals

LIM (SiNx / x) ^ (1 / x ^ 2) x tends to 0 Find the limit with lobida's law,

Original limit = LIM (x tends to 0) e ^ [ln (SiNx / x) * 1 / x ^ 2] obviously, when x tends to 0, SiNx / X tends to 1, then ln (SiNx / x) = ln (1 + SiNx / X - 1) is equivalent to SiNx / X - 1, so ln (SiNx / x) * 1 / x ^ 2 is equivalent to (SiNx / X - 1) / x ^ 2 = (SiNx - x) / x ^ 3. Use lobitar's law