Proving Limit Laws: Epsilon/Delta Definition

It is just a matter of taste.In summary, the proof shows that by setting the limit of f(x) as L and the limit of g(x) as M, we can establish that the limit of the combined function f(x)+g(x) is equal to the sum of the individual limits, L and M respectively. This is achieved by setting an epsilon value for both functions and finding a corresponding delta value that ensures the combined function stays within the epsilon range. The significance of choosing epsilon/2 is that it simplifies the calculations and makes the proof more concise.
  • #1
ciubba
65
2

Homework Statement


Prove that [tex]\lim_{x->a}[f(x)+g(x)]=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)][/tex]

Homework Equations


Epsilon/delta definition

The Attempt at a Solution



The book says:

Let the limit of f(x)=L and the limit of g(x)=M. Then,

[tex]\mid f(x)-L \mid<\frac{\epsilon}{2}[/tex] whenever [tex]0<\mid x-a \mid<\delta_{1}[/tex]
and
[tex]\mid g(x)-M \mid<\frac{\epsilon}{2}[/tex] whenever [tex]0<\mid x-a \mid<\delta_{2}[/tex]

Let [tex] \delta=min \big\{ \delta_{1} ,\delta_{2} \big\}[/tex] and suppose [tex]0<\mid x-a \mid<\delta[/tex]. Because [tex]\delta\leq\delta_{1}[/tex], it follows that [tex]0<|x-a|\delta_{1}[/tex] and [tex]|f(x)-L<\frac{\epsilon}{2}[/tex].

Similarly, [tex]\delta\leq\delta_{2}[/tex], it follows that [tex]0<|x-a|\delta_{2}[/tex] and [tex]|g(x)-L<\frac{\epsilon}{2}[/tex]. Therefore,

[tex]|[f(x)+g(x)]-(L+M)|<\epsilon[/tex], which implies that [tex]]\lim_{x->a}[f(x)+g(x)]=L+M=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)][/tex]

My biggest question is from where they obtained a value of epsilon/2 from. I feel that I cannot even begin to understand the proof until I understand that point.

Edit: The latex didn't come out as clearly as I'd hoped, so here is a picture: http://s9.postimg.org/7gfz3owul/Proof.gif
 
Last edited:
Physics news on Phys.org
  • #2
Read the first sentence after SOLUTION in the picture: "Assume that ε > 0 is given " You can require that both functions deviate from their limit less than half of the given ε.
 
  • #3
ehild said:
Read the first sentence after SOLUTION in the picture: "Assume that ε > 0 is given " You can require that both functions deviate from their limit less than half of the given ε.

Yes, I saw that, but I also could have defined it as < ε/3 or 2piε/4. What is the significance of making it ε/2? I looked at a proof for the product limit laws and saw a similar technique employed.
 
  • #4
You have to find δ so as the deviation of the combined function f(x)+g(x) from its limit is less than ε. The simplest is to choose equal ε1 and ε2 for the functions f and g, so as ε1+ε2 = ε. But you have the freedom to choose ε/3 for one function and 2/3 ε for the other one, or anything else if they are positive and their sum is ε.
 
  • #5
ciubba said:
Yes, I saw that, but I also could have defined it as < ε/3 or 2piε/4. What is the significance of making it ε/2? I looked at a proof for the product limit laws and saw a similar technique employed.

It really does not matter. For any (small) ##\epsilon' > 0## we can find ##\delta > 0## such that ##|f(x)-L| < 3 \epsilon'## and ##|g(x) -M |< 2 \pi \epsilon'## for ##|x - a| < \delta##. Then we have
[tex] |f(x) + g(x) - (L+M)| < (3 + 2 \pi) \epsilon'[/tex]
for ##|x-a| < \delta##. Now just put ##\epsilon = (3 + 2 \pi) \epsilon'##. Given any (small) ##\epsilon > 0## we can thus find ##\delta > 0## such that ##|f(x) + g(x) - (L+M)| < \epsilon## if ##|x-a| < \delta##. We do that by putting ##\epsilon' = \epsilon/(3 + 2 \pi)## and using the ##\delta## that goes along with ##\epsilon'##. It is just a lot neater to put ##\epsilon' = \epsilon/2##.
 

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

The epsilon/delta definition of a limit is a mathematical concept used to rigorously prove the behavior of a function as it approaches a specific point or value. It is based on the idea that for any arbitrary distance (epsilon) from the limit, there exists a corresponding distance (delta) from the input such that if the input is within delta of the limit, the output will be within epsilon of the limit.

2. Why is the epsilon/delta definition important?

The epsilon/delta definition is important because it provides a rigorous and precise way to prove the behavior of a function near a specific point. It allows us to make general statements about the behavior of a function without needing to know the exact value of the function at that point.

3. How do you prove a limit using the epsilon/delta definition?

In order to prove a limit using the epsilon/delta definition, you must first choose an arbitrary epsilon (distance from the limit) and then find a corresponding delta (distance from the input) such that if the input is within delta of the limit, the output will be within epsilon of the limit. This can be done by manipulating the equation and using algebraic techniques to find a suitable delta value.

4. What are some common mistakes when proving limit laws using the epsilon/delta definition?

One common mistake is not choosing an arbitrary epsilon value, but instead using a specific value. Another mistake is not manipulating the equation properly to find a suitable delta value. It is also important to remember that the delta value must be positive and cannot depend on the input value.

5. Can the epsilon/delta definition be used to prove all limit laws?

Yes, the epsilon/delta definition can be used to prove all limit laws. However, for some limit laws, such as the Squeeze Theorem, a different approach may be more efficient. It is important to carefully consider which method is most appropriate for each specific limit law.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
400
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
146
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
792
  • Calculus and Beyond Homework Help
Replies
2
Views
792
  • Calculus and Beyond Homework Help
Replies
5
Views
811
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Back
Top