Rational set: irrational set: positive real set: real set: As follows:-11, root 5,3, root 9/11,0,2/3, root 196,-π,0.4, root 2/3

Rational set: irrational set: positive real set: real set: As follows:-11, root 5,3, root 9/11,0,2/3, root 196,-π,0.4, root 2/3

Rational number:-11,3,0,2/3, root 196(14),0.4
Irrational number: root number 5, root number 9/11,-π, root number 2/3
Positive real numbers: all but 0,-11,-π are
Real: All are
Real numbers include rational numbers and irrational numbers, where irrational numbers are infinite non-recurring decimals and rational numbers include integers and fractions

Rational:-11,3,0,2/3, root 196(14),0.4
Irrational number: root number 5, root number 9/11,-π, root number 2/3
Positive real numbers: all but 0,-11,-π are
Real number: all are
Real numbers include rational numbers and irrational numbers, where irrational numbers are infinite non-recurring decimals and rational numbers include integers and fractions

Classified rational number of real number {real number {irrational number}

Real: Rational: Integer: positive,0, negative
Scores: positive, negative
Irrational number: positive irrational number, negative irrational number

ON THE UNDERSTANDING PROBLEM OF LOCAL NORMALITY OF FUNCTION LIMIT Taking A/2 in the definition proof only means that a number is found in the domain to make f (x)>0, that is, there is f (x)>0 in the domain. It can not be proved that f (x) is always greater than 0 in the domain. If it's not Heng greater or Heng less, how can we keep it? And when A =0, there's no guarantee? ON THE UNDERSTANDING PROBLEM OF LOCAL NORMALITY OF FUNCTION LIMIT A/2 in the definition proof only means that a number is found in the domain to make f (x)>0, that is, there is f (x)>0 in the domain. If it's not bigger and smaller, how can it be guaranteed? And when A =0, there's no guarantee? ON THE UNDERSTANDING PROBLEM OF LOCAL SIGNAL PROTECTION OF FUNCTIONAL LIMIT Taking A/2 in the definition proof only means that a number is found in the domain to make f (x)>0, that is, there is f (x)>0 in the domain. If it's not Heng greater or Heng less, how can it be guaranteed? And when A =0, there's no guarantee?

It's not like that.
See proof of insurability first:
First there is the function f (x) at x→x0(note: x0 can be a concrete number or infinite), there is a limit A >0(A0, there is δ>0, so that |x-x0|

It's not like that.
First see the proof of warranty:
First, when the function f (x) is x→x0(note: x0 can be a concrete number or infinite), there is a limit A >0(A0, there is δ>0, so that |x-x0|

Understanding of Limit Warranty I don't know what's going on, Understanding of Limit Protection He really didn't know what was going on,

Did you get it right?
Guaranteed, that is:
If x→a, f (x)→A,
If A >0
Then in some neighborhood N (a) of a, in this neighborhood f (x)>0,
This neighborhood can be very small, but it must exist
It can also be understood that you can find a point x1 near a so that f (x1)>0
When A <0, follow the above description.

Understanding and Understanding of Limit Algorithm It needs to be recognized and understood in words.

0

Limit Warranty Understanding? Why is it so important? Given the function limit A >0, any ε>0, there exists δ>0, so that |x-x0|<δ, has |f (x)-A |<ε Because ε is arbitrary,ε0=A/2 is taken Then δ>0, so that |x-x0|<δ, has |f (x)-A|<ε0=A/2 I.e. f (x)> A/2>0 What is the final decision for this f (x)> A/2? Must it be f (x)> A/2? Why do I only get 2A/3> f (x)>-A/2 when I go absolute?

I can only say that you are wrong
-A/2A/2

I can only say you're wrong.
-A/2A/2