Limx tends to infinity [(2 ^ x + 3 ^ x) / (2 ^ (x + 1) + 3 ^ (x + 1))],

Limx tends to infinity [(2 ^ x + 3 ^ x) / (2 ^ (x + 1) + 3 ^ (x + 1))],

X tends to infinity Lim [(2 ^ x + 3 ^ x) / (2 ^ (x + 1) + 3 ^ (x + 1))]
Divide the numerator and denominator by 3 ^ (x + 1)
Limit = Lim [(1 / 3) * (2 / 3) ^ x + (1 / 3)] / [(2 / 3) ^ (x + 1) + 1]
= (0 + 1/3)/(0 +1)
=1/3

Limx tends to infinity (2 * x-1) * e ^ (1 / x) - 2 * x

If x approaches infinity, easy to get 1 / x approaches 0, then e ^ (1 / x) approaches 1,
Then (2 * x-1) * e ^ (1 / x) - 2 * x can be abbreviated as (2 * x-1) - 2 * x, then the answer is - 1

Limx tends to infinity ((2 + e ^ (1 / x)) / (1 + e ^ (4 / x) + SiNx / | x |)

What is the derivative of INX ^ (- x)

[InX^(-x)]'=[-xlnx]'=-lnx-1

If the derivative function f (x) defined on R satisfies that the derivative of function f (x) defined on R is
F '(x), if (x-1) f' (x) ≥ 0, then (d)
A.f(0)+f(2) ＜2f(1) B.f(0)+f(2) ≤2f(1) C.f(0)+f(2) ＞2f(1) D.f(0)+f(2) ≥2f(1)
In the solution, the function is reduced on (negative infinity, 1) and monotonically increased on (1, positive infinity), so f (0) ≥ f (1), f (2) ≥ f (1), so how does f (0) + F (2) ≥ 2F (1) equal sign come from?

It's very simple. If the derivative on the domain of definition is always zero, then it also satisfies (x-1) f '(x) ≥ 0, so it gets the equal sign. Remember, monotone subtraction is not strictly monotone subtraction. The former only needs to be less than or equal to, while the latter is more demanding, and the requirement must be less than

It is known that for any real number x, if f (- x) = - f (x), G (- x) = - G (- x), and x > 0, the derivative of F (x) is greater than 0, and the derivative of G (x) is greater than 0, then X

F (x) is an odd function, so f '(x) is an even function
x> 0, f '(x) > 0, f' (x) about Y-axis symmetry
So x0, G '(x) > 0
So x

Let f (x) = x − ax − 1, M = {x | f (x) ＜ 0}, P = {x | f ′ (x) ＞ 0}, if M ⊆ P, then the value range of real number a is______ ．

Because f (x) < 0 is equivalent to (x-1) (x-a) < 0 and f ′ (x) = a − 1 (x − 1) 2, f ′ (x) > 0 is equivalent to a − 1 (x − 1) 2 > 0, f ′ (x) > 0 is equivalent to a − 1 (x − 1) 2 > 0. When a ＜ 1, set P has no solution and does not satisfy the problem. When a = 1, both sets are empty and conform to the problem

Let f (x) = (LNX + m) / E, G (x) = x + ㏑ x + e, and the minimum value of G = (x) is 1
(1) Find the value of M (2) find the monotone interval of F (x)!

F (x) = (LNX + m) / e ^ x = (LNX + m) e ^ (- x) f '(x) = [e ^ (- x)] / x-e ^ (- x) (LNX + m) g (x) = x + LNX + e ^ XF' (x) = x + LNX + 1 / X - (LNX + m) g '(x) = 1 + 1 / X - 1 / X & # 178; - 1 / x = 1-1 / X & # 178; let g' (x) = 0, the minimum value of x = 1 or x = - 1 (rounding off) g (x) is 1, so g (1) = 1, that is g (1) = 1 + 0

The function f (x), when x is not equal to 0, f (x) = SiNx / X; when x = 0, f (x) = 1, why is the n-th derivative of F (0) substituted into 0 after the derivation of F (x) = SiNx / X instead of the n-th derivative of F (0) = 1?

Because when x = 0, f (x) = 1 means that when x approaches 0, f (x) = 1
This is the meaning of taking the limit

1. 6 cubic decimeter 300 cubic centimeter = () cubic centimeter 2, 8 cubic meter 40 cubic decimeter = () cubic decimeter

6 cubic decimeter 300 cubic centimeter = (6300) cubic centimeter
8 cubic meters 40 cubic decimeters = (8040) cubic decimeters
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