Why monotone bounded function may not have limit and monotone bounded sequence must have limit. This paper analyzes the essential difference between the limit of function and sequence

Why monotone bounded function may not have limit and monotone bounded sequence must have limit. This paper analyzes the essential difference between the limit of function and sequence

"Monotone bounded sequence must have limit" is one of the basic theorems of calculus. The limit of sequence is relatively simple, which refers to the limit when n →∞ (actually n → + ∞), so we just need to find the limit of a certain sequence (not to say how N changes), we all understand it
The limit of a function is more complex. If we only want to find the limit of a certain function, others will not understand it. We must also indicate how the independent variable (such as x) changes
Considering the change trend of the independent variable, there are x → x0 (x0 is a real number, how many kinds are there?) and X →∞; if subdivided, there are x tending to x0 from the left, x0 from the right, positive infinity and negative infinity
Don't forget that there are preconditions for us to study the limit of function
In order to study the limit of X → x0, it is necessary to define the function in a centreless neighborhood of x0; in order to study the limit of X →∞, it is necessary to have a positive number X. when | x | > x, it is necessary to define the function
The existence and nonexistence of the limit of this function can be discussed only when the precondition is satisfied
You only give the monotone bounded function, you don't know what the domain of function is, and you don't know how the independent variable changes. In this case, it's meaningless to talk about the limit of function

A 120cm long wire is divided into two parts, each part is bent into a square. What is the sum of their areas? What's the area and minimum of them?

Let the iron wire be divided into two parts xcm and (120-x) cm, and the sum of the areas is y. The equation shows that y = (x4) 2 + (120 − x4) 2 = 18 (X-60) 2 + 450 ∥ the sum of their areas is 18 (X-60) 2 + 450, and the minimum value is 450cm2

Given that a (1,1), B (4,. 5) and the moving point P (x, y) move on the line AB, what is the maximum value of 2x-y?
Given that a (1,1), B (4,. 5) and the moving point P (x, y) move on the line AB, what is the maximum value of 2x-y

A:
Point a (1,1), point B (4,5)
AB slope k = (5-1) / (4-1) = 4 / 3
AB line Y-1 = (4 / 3) (x-1) = 4x / 3-4 / 3
So: y = 4x / 3-1 / 3
The point P (x, y) is on the line ab
So: 2x-y = 2x - (4x / 3-1 / 3) = 2x / 3 + 1 / 3
Because: 1

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 °. D is the midpoint of BC, de ⊥ AB is proved at point E: EB = 3EA

It is proved that: ∵ AB = AC, ∵ BAC = 120 ∵ B = ∵ C = 30 ∵ D is the midpoint of BC ∵ ad ⊥ BC and ad bisects ∵ BAC, ∵ bad = 60 ∵ ADB = 90 ∵ ad = 12ab and ∵ de ⊥ ab ∵ DEA = 90 ∵ ade = ∵ DEA -