Prove the limit is true using the epsilon,delta definition of a limit.

In summary: You can make the same error if you are not careful before in your proof. In fact, I'm not sure it's the same error because I'm not sure I understand what you are saying. Can you explain please?My apologies, I misunderstood your notation. Yes, you are correct that |x - 1|^2 < |x - 1| if δ < 1. However, in your proof, you wrote "assuming |x - 1| < 1" without specifying that it was also assuming δ < 1. That was my confusion.
  • #1
WK95
139
1

Homework Statement


Prove the following:
lim x2 - x = 0
x→1

Homework Equations



If 0<|x-a|<δ then |f(x)-L|<ε

The Attempt at a Solution


Part I - Set up
0<|x-a|<δ |f(x)-L|<ε
If 0<|x-1|<δ then |(x^2 - x) - 0|<ε
x|x-1|<ε
x|x-1|<ε
|x-1|<ε/x
δ=ε/x

Part II - Proof
Given ε < 0, choose δ=ε/x
If 0<|x-1|<δ then
|(x^2 - x) - 0| = |x^2 - x| = x|x - 1| < xδ = x(ε/x) = ε
 
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  • #2
No! Delta may not be dependent on x. Also your definition is incorrect, it is for every ε>0 there is a ... etc.

Try this:

|x||x-1| <δ|x| = δ |x-1+1|<δ^2+δ < δ if δ < 1,..., etc.
 
  • #3
I'm afraid you can't set ##δ=ε/x##. It needs to be a particular number not an arbitrary value. Try setting up ##|x||x-1|## and then apply triangle inequality to find the upper bound ##f(x)## can take. It can be assured that |x-1| will at least be smaller than 1.
 
  • #4
dirk_mec1 said:
No! Delta may not be dependent on x. Also your definition is incorrect, it is for every ε>0 there is a ... etc.

Try this:

|x||x-1| <δ|x| = δ |x-1+1|<δ^2+δ < δ if δ < 1,..., etc.

I'm not quite sure why you added 1 to |x - 1|
|x||x-1| < δ|x| = δ|x-1+1| < δ^2 + δ < δ
|x^2 - x| < δ|x-1+1| < δ^2 + δ < δ <== I'm not getting this part
 
  • #5
Seydlitz said:
I'm afraid you can't set ##δ=ε/x##. It needs to be a particular number not an arbitrary value. Try setting up ##|x||x-1|## and then apply triangle inequality to find the upper bound ##f(x)## can take. It can be assured that |x-1| will at least be smaller than 1.

How can it be assured that |x - 1| will be smaller than 1? In a formal proof, how would I show this?

I found the Triangle Inequality to be |A + B| ≤ |A| + |B|

Set up
0<|x - a|<δ |f(x) - L|<ε
If 0<|x - 1|<δ then |(x^2 - x) - 0|<ε
x|x-1|<ε
 
Last edited:
  • #6
So you have this :

##|x^2 - x| = |x||x-1| < δ|x|##

Now this is quite a common trick and you should use it often. It allows you to use the triangle inequality on your second factor :

##|x| = |x-1+1| ≤ |x-1| + 1 < δ + 1##

Therefore :

##|x^2 - x| = |x||x-1| < δ|x| < δ(δ+1)##

Now at this point, it's very useful to place a bound on ##δ##, let's say ##δ ≤ 1## to make life simple. So we get :

##δ(δ+1) ≤ 2δ##

Now you can find ##δ## in terms of ##ε##.

What is your final value(s) for ##δ##?
 
  • #7
Here is what I have now.
##|(x^2 - x) - 0| = |x - 1| |x|##
##= |x - 1| |(x - 1) + 1| ≤ |x - 1| (|x - 1| + |1|)##
##= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|##
assuming
##|x - 1| < 1##
Given
##ε > 0, δ = min{1, ε/2}##
Then
##0 < |x - 1| < δ##
##|(x^2 - x) - 0| < 2|x - 1| < 2(ε/2) = ε##
 
Last edited:
  • #8
It looks fine to me. ##δ=min(1,\frac{ε}{2})##
 
  • #9
WK95 said:
Here is what I have now.
##|(x^2 - x) - 0| = |x - 1| |x|##
##= |x - 1| |(x - 1) + 1| ≤ |x - 1| (|x - 1| + |1|)##
##= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|##
assuming
##|x - 1| < 1##
Given
##ε > 0, δ = min{1, ε/2}##
Then
##0 < |x - 1| < δ##
##|(x^2 - x) - 0| < 2|x - 1| < 2(ε/2) = ε##

I'm not so sure I agree with the third line :

##= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|##

How were you able to deduce ##|x-1|^2 < |x-1|##. This is certainly not true in general. Your value for ##δ## is indeed correct though.
 
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  • #10
Yes it is true if delta<1.
 
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FAQ: Prove the limit is true using the epsilon,delta definition of a limit.

1. What is the epsilon-delta definition of a limit?

The epsilon-delta definition of a limit is a mathematical concept that is used to formally define the concept of a limit in calculus. It states that for a function f(x), the limit of the function as x approaches a certain value c is equal to L if for every positive number ε, there exists a positive number δ such that if the distance between x and c is less than δ, then the distance between f(x) and L is less than ε.

2. Why is the epsilon-delta definition of a limit important?

The epsilon-delta definition of a limit is important because it provides a rigorous and precise way to define the concept of a limit. It helps to avoid any ambiguity or confusion when determining whether a limit exists and what its value is.

3. How do you use the epsilon-delta definition of a limit to prove a limit is true?

To prove a limit is true using the epsilon-delta definition, you must first choose a value for ε and then find a corresponding value for δ that satisfies the definition. You must then show that for any x value within a distance of δ from the limit point, the distance between f(x) and the limit is less than ε.

4. Can you provide an example of using the epsilon-delta definition of a limit to prove a limit is true?

Yes, for example, let's say we want to prove that the limit of f(x) = 3x + 2 as x approaches 2 is equal to 8. We choose ε = 0.5, and then we need to find a δ such that if the distance between x and 2 is less than δ, then the distance between f(x) and 8 is less than 0.5. We can find that δ = 0.1 satisfies this condition, as if the distance between x and 2 is less than 0.1, then the distance between f(x) and 8 is less than 0.5.

5. What are some common mistakes when using the epsilon-delta definition of a limit?

Some common mistakes when using the epsilon-delta definition of a limit include choosing the wrong value for ε, not finding a corresponding value for δ, and not properly showing that the distance between f(x) and the limit is less than ε for all x values within a distance of δ from the limit point. It is important to carefully follow the steps of the definition and make sure all conditions are satisfied to accurately prove a limit is true.

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