Change the order of delta-epsilon definition?

[FycxzX]

We all know that the normal delta-epsilon definition is to make sure we can find a delta to satisfy the relationship of whatever epsilon that we chose. So can we do it the other way, find a epsilon to satisfy the relationship of whatever delta that we chose, that is:

For any delta, we can always find an epsilon such that:

|f(x) - L| < epsilon implies |x - a| < delta ?

 

[HallsofIvy]

"Do it the other way" to do what? To prove that a limit is a specified number, we have to "given \(\displaystyle \epsilon\)> 0[/tex], find \(\displaystyle \delta> 0\) such that ... " Doing it "the other way" wouldn't prove anything.

 

[Ishuda]

You can under certain conditions. For simplicity let's only talk of continuous functions [limits exist 'everywhere'] for now. If we can find a function f^--1 such that x = f^--1(f(x)), then the order can be changed around.

As a simple example, let
y = x + 10
then y has a limiting value for all x. But we also have
x = y - 10
and x has a limiting value for all y. The function having an inverse is a sufficient condition, I'm not sure right now whether it is a necessary condition.

 

[FycxzX]

"Do it the other way" to do what? To prove that a limit is a specified number, we have to "given \(\displaystyle \epsilon\)> 0[/tex], find \(\displaystyle \delta> 0\) such that ... " Doing it "the other way" wouldn't prove anything.

Yes, but what I want to know is that in which situation is the "do it in another way" would not work?

 

[FycxzX]

You can under certain conditions. For simplicity let's only talk of continuous functions [limits exist 'everywhere'] for now. If we can find a function f^-1 such that x = f^-1(f(x)), then the order can be changed around.

As a simple example, let
y = x + 10
then y has a limiting value for all x. But we also have
x = y - 10
and x has a limiting value for all y. The function having an inverse is a sufficient condition, I'm not sure right now whether it is a necessary condition.

So, for example, the function x = 5 (a vertical line) can not be done in the other way?

 

[Ishuda]

So, for example, the function x = 5 (a vertical line) can not be done in the other way?

In this situation, I'm not sure what you mean because by the definition of function in Calculus we mean a rule where for a given value in the Domain (x) of the function there is a unique functional value (y=f(x)). So x = 5 is not a function. NOTE: Just being pedantic here.

However, turning the problem into what I think you meant, we can approach it as
(1) y = f(x) = (x-5) / a
and thus
(2) x = f^--1(y) = 5 + ay
At a = 0, (1) f is undefined so there is no epsilon - delta argument and (2) x = f^--1(y) = 5 and is perfectly ok [who said x=5 wasn't a function]

All of that to just say, yes the function x = 5 (a vertical line) can not be done in the other way. WOW!