النهايات - التعريف - الجزء الثاني | Part 2: Definition de la Limite
Summary
TLDRThe script explains the concept of limits in mathematics through an interactive game between two players. Player 1 selects an interval around the number 1 on the y-axis. Player 2 then restricts the domain around the number 2 on the x-axis to keep the function within player 1's interval. As player 1 narrows their interval, pushing the function outside the bounds, player 2 responds by tightening the domain to pull the function back between the lines. This back and forth models how for any epsilon interval around a limit 'L', you can find a delta interval around a point 'a' where function values remain trapped between the lines. This models the formal definition of a limit.
Takeaways
- 😀 To find the limit of a function as x approaches a value, substitute values increasingly close to that value and see what happens to the function
- 👉 Draw horizontal lines on a graph to represent an interval around the expected limit value
- 🌟 The game involves narrowing the interval on the y-axis while also narrowing the domain around the limit point
- 📏 Use mathematical notation like |y - L| < ε to represent getting closer to the limit L within some tolerance ε
- 🔢 Delta (δ) represents the narrowing domain, while epsilon (ε) represents the range tolerance
- 🤝 The formal definition of a limit involves quantifiers: for all ε > 0, there exists δ > 0, such that...
- ⚖️ The choice of epsilon comes first, and delta is chosen accordingly after
- 🎯 L represents the limit value, while a represents the number we take x towards
- ✏️ The process works for finding limits of any function at any point
- 📈 Understanding this concept is key for studying calculus and analysis
Q & A
What is the purpose of the 'game' introduced in the transcript?
-The purpose of the game is to demonstrate and arrive at the formal definition of a limit, by having one player choose intervals on the y-axis, while the other player chooses intervals on the x-axis to keep the function confined.
Why does the first player start by choosing an interval centered around the number 1?
-Because 1 is the assumed limit that the function approaches as x approaches 2. So the first player centers their interval around this assumed limit.
What do the variables epsilon and delta represent?
-Epsilon represents the radius or distance that defines the interval chosen by the first player on the y-axis. Delta represents the radius or distance that defines the interval chosen by the second player on the x-axis.
Why must epsilon and delta be greater than 0?
-Epsilon and delta must be positive to ensure that actual intervals with non-zero widths are being selected on each axis.
What is the strategy of each player in the game?
-The first player tries to make their interval smaller and smaller to force the function outside the bounds. The second player responds by making their interval smaller to bring the function back inside the bounds.
When does the game end?
-The game theoretically never ends, as each player can always make their interval smaller, continuing the process indefinitely.
What does it mean when the game continues indefinitely?
-It demonstrates that as x gets arbitrarily close to 2, the function gets arbitrarily close to 1, formalizing the limit definition.
Why are general symbols like L and a used?
-To generalize the ideas and make the formal definition apply to finding limits at any number for any function, not just at x=2 for this particular function.
What is represented by |f(x) - L| < ε?
-This states that the distance between the function value and the assumed limit L is less than the predefined proximity ε.
Why must the first player choose ε first?
-Because the second player must choose δ based on the ε chosen by the first player in order to properly relate the x-axis interval to the y-axis interval.
Outlines
😊 Defining limits through an interactive game
This paragraph explains the concept of limits by framing it as a game between two players. Player 1 chooses an interval around the y-value that the function is approaching. Player 2 then chooses a corresponding interval around the x-value that makes the function approach that y-value. By iterating this process and narrowing the intervals, it is shown mathematically how the function can get arbitrarily close to a certain y-value as x approaches a certain point, thus defining the limit.
📝 Formalizing the limit definition mathematically
This paragraph formalizes the interactive game into mathematical notation for the formal definition of a limit. It introduces the epsilon-delta definition, with epsilon representing the y-axis interval chosen by player 1, and delta representing the x-axis interval chosen by player 2. It shows how, for any epsilon, player 2 can choose a delta to make the function stay within an epsilon distance of the limit.
Mindmap
Keywords
💡limit
💡epsilon
💡delta
💡absolute value
💡domain
💡interval
💡approaches
💡arbitrarily small
💡formal definition
💡confine
Highlights
To answer the question, I must give x values close to 2 and see what happens.
In this game, you will be player 1, who plays first. You must choose an interval on the y-axis with the number 1 in its center.
The goal of this game is to make the function never exit the horizontal lines, i.e., never exit your interval, not even a little.
As long as I can move, no matter how small you make your interval and push the function out of the horizontal lines, I can choose a smaller interval and confine the function within your lines again.
Thus, the interval you choose will get closer and closer to the number 1 from both sides, as long as I get closer and closer to the number 2 from both sides.
This means that the function's limit, as long as x approaches the number 2 from both sides, will equal 1.
The absolute value of y - 1 is the distance between a random point y and the number 1, and the distance can never be negative.
The absolute value of the function minus the number 1 is smaller than epsilon is true for all x close to and different from the number 2.
The first player's move was to choose Epsilon definitely positive. And the second player's move was to choose delta definitely positive.
Ultimately, this is valid for any function and any number.
It means that for every choice you make for a definitely positive epsilon, I can choose a definitely positive delta through it.
My choice of delta must be based on your initial choice of epsilon.
To answer the question, I must give x values close to 2 and see what happens.
In this game, player 1 chooses an interval around 1, player 2 keeps function inside by restricting domain around 2.
The game continues with players narrowing intervals, showing limit exists as x approaches 2.
Formalized with epsilon/delta definitions, valid for any function/number, delta based on epsilon.
Transcripts
(The last question from the previous episode)
mmm,this was the last question I ended the last episode with.
To answer it, I must give x values close to 2 and see what happens.
These close values can be taken either from the right side of 2 or the left.
For example, from the right, if x equals 2.01, the image of this number will be 1.01.
And from the left, if x equals 1.99, the image of this number will be 0.99.
As long as x approaches 2, the function, in turn, approaches 1, but it will never equal the number 1.
To generalize this, to make it a general rule applicable to any other situation,
we must play a certain game.
In this game, you will be player 1, who plays first.
You must choose an interval on the y-axis with the number 1 in its center.
Then, draw two dashed lines starting from the edges of that interval, parallel to the x-axis,
as you see on the screen.
Notice that I wrote in front of the interval's boundary 1 plus 0.75 and 1 minus 0.75.
Meaning that the boundaries of the interval are the same distance, which is 0.75, from the number 1.
The goal of this game is to make the function never exit the horizontal lines,
i.e., never exit your interval, not even a little.
Here it will be my turn, player two.
I will restrict the domain of the function as I want.
The condition imposed on me in choosing is that any number I choose must have the number 2 in its center.
Then I draw two dashed lines on the sides of the interval, parallel to the y-axis,
as you see on the screen.
Here I was able to confine the function and keep it inside the interval you specified.
Notice that I wrote at the boundary of the domain 2 minus 0.5 and 2 plus 0.5.
Meaning that the boundaries are the same distance, which is 0.5, from 2.
It's your turn again, and here you realize that you can tighten the noose on me if you narrow your interval more.
So, you chose this interval, whose boundaries are 0.25 distance from the number 1.
When you narrowed your interval, the function exited the horizontal lines you drew.
But now it's my turn again, and I will shorten my interval more to keep the function confined within your horizontal lines,
as you see on the screen.
The question now is: When will we reach the end of this game?
And the answer is: We will never reach it.
As long as I can move, no matter how small you make your interval and push the function out of the horizontal lines,
I can choose a smaller interval and confine the function within your lines again.
Thus, the interval you choose will get closer and closer to the number 1 from both sides,
as long as I get closer and closer to the number 2 from both sides.
This means that the function's limit, as long as x approaches the number 2 from both sides, will equal 1.
Now, how can I write these things mathematically?
First, the interval you chose the first time, can write it like this:
And you started to narrow your interval to get closer to the number 1.
Meaning 0.75 became 0.25,
then 0.1,
then 0.0001,
and so on.
Therefore, instead of writing all these intervals,
I will give that variable a name, which is epsilon.
Of course, epsilon must be positive for sure, it cannot be 0.
So, my interval will become:
Since I am now working on the ordinal axis,
Since I am now working on the y-axis, the numbers I have on the y-axis are usually denoted by the symbol y.
And if you watched the interval episode, you would know that I can write this interval as:
y confined between 1 - epsilon and 1 + epsilon.
Therefore, if I subtract 1 from all sides, I will have y - 1
confined between negative epsilon and epsilon.
And this can be written as the absolute value of y - 1 definitely smaller than epsilon.
Why?
Because the absolute value y - 1 is the distance between a random point y and the number 1,
and the distance can never be negative.
Now in my turn, I always had to keep the function confined within your specific interval lines.
This means I was always restricting the function's definition set.
So, all the remaining values of the function must be definitely smaller than the distance epsilon from 1.
Therefore, when I played, you concluded:
The absolute value of the function minus the number 1 is smaller than epsilon is true for all x close to and different from the number 2.
And following the same steps we did regarding your interval, my interval will be like this
And delta is a number definitely positive, like epsilon, so don't worry.
And if we repeat the same previous steps, we will find that the absolute value of x - 2 is definitely smaller than delta.
And since I do not want x to equal the number 2, I can add that it is definitely greater than zero,
as you see on the screen.
In short, the first player's move was to choose Epsilon definitely positive.
And the second player's move was to choose delta definitely positive.
This is to ensure that the absolute value of the function minus one is smaller than epsilon, as long as x satisfies
the condition that the absolute value of x minus 2 is definitely greater than 0 and definitely smaller than delta.
This embodies the idea that the function is very close to the number 1 if x is very close to the number 2.
What remains is for you to choose a very small value for epsilon,
and I will choose delta according to the epsilon you selected.
Ultimately, this is valid for any function and any number.
Therefore, I will replace the number 1 with the letter L, which means Limit.
And I will replace the number 2 with the letter a, which symbolizes a specific number, meaning it's not unknown like x.
The final rule, with full confidence, will be as follows:
It means that for every choice you make for a definitely positive epsilon,
I can choose a definitely positive delta through it.
So that the absolute value of the function minus its limit is definitely smaller than epsilon, it holds true for every x that satisfies
the absolute value of x minus the number a is definitely greater than zero and definitely smaller than delta.
And this is the official definition of the limit.
In the end, it's very important that your move should be the first.
My choice of delta must be based on your initial choice of epsilon.
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