Limit proof (using delta-epsilon definition)

In summary, the epsilon-delta definition is used to prove that the limit of the product of two functions, f and g, at a limit point c is equal to the product of their individual limits, L and M. This is done by finding a delta value that ensures that the absolute value of the difference between f(x)g(x) and LM is less than any given epsilon value. By using the triangle inequality and the individual limits of f and g, we can manipulate the expression and find a delta value that satisfies the definition.
  • #1
dancergirlie
200
0

Homework Statement


Let f, g : A [tex]\rightarrow[/tex] R be functions and let c [tex]\in[/tex] R be a limit point of A. Assume that lim(x-> c): f(x) = L
and lim(x->c): g(x) = M. Use the epsilon-delta definition to prove:
lim(x->c) f(x)g(x)=LM


Homework Equations





The Attempt at a Solution


alright in order to prove this, for epsilon>0, I need to find a delta>0 so that for 0<|x-c|<delta,
|f(x)g(x)-LM|<epsilon.

This is what I tried so far:

|f(x)g(x)-LM|= |f(x)g(x)-Lg(x)+Lg(x)-LM|= |(f(x)-L)(g(x))+L(g(x)-M)|
By the triangle inequality, that is less than or equal to:

|(f(x)-L)(g(x)|+|L(g(x)-M)|= |f(x)-L||g(x)| + |L||g(x)-M|

Well I know that since lim(x->c) f(x)=L, that means for epsilon>0 there exists a delta(1) so that for 0<|x-c|<delta(1)
|f(x)-L|<epsilon

and also since lim(x->c) g(x)=M, that means for epsilon>0 there exists a delta(2) so that for 0<|x-c|<delta(2)
|g(x)-M|<epsilon

For |f(x)-L||g(x)| + |L||g(x)-M|, I am trying to make each part at most equal to epsilon/2.

This is where I get stuck, I don't know what to do with |f(x)-L||g(x)|, because g(x) isn't a constant... any help would be great!
 
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  • #2
dancergirlie said:

Homework Statement


Let f, g : A [tex]\rightarrow[/tex] R be functions and let c [tex]\in[/tex] R be a limit point of A. Assume that lim(x-> c): f(x) = L
and lim(x->c): g(x) = M. Use the epsilon-delta definition to prove:
lim(x->c) f(x)g(x)=LM


Homework Equations





The Attempt at a Solution


alright in order to prove this, for epsilon>0, I need to find a delta>0 so that for 0<|x-c|<delta,
|f(x)g(x)-LM|<epsilon.

This is what I tried so far:

|f(x)g(x)-LM|= |f(x)g(x)-Lg(x)+Lg(x)-LM|= |(f(x)-L)(g(x))+L(g(x)-M)|
By the triangle inequality, that is less than or equal to:

|(f(x)-L)(g(x)|+|L(g(x)-M)|= |f(x)-L||g(x)| + |L||g(x)-M|

Well I know that since lim(x->c) f(x)=L, that means for epsilon>0 there exists a delta(1) so that for 0<|x-c|<delta(1)
|f(x)-L|<epsilon

and also since lim(x->c) g(x)=M, that means for epsilon>0 there exists a delta(2) so that for 0<|x-c|<delta(2)
|g(x)-M|<epsilon

For |f(x)-L||g(x)| + |L||g(x)-M|, I am trying to make each part at most equal to epsilon/2.

This is where I get stuck, I don't know what to do with |f(x)-L||g(x)|, because g(x) isn't a constant... any help would be great!
You may want to define epsilon1 and epsilon2 applied to the two separate limits. I think you'll have to again use the closeness of g(x) to M in the part where you are stuck so that epsilon1 (for f) is also a function of epsilon2 (for g). (Or maybe the other way around?)

Beyond that you can try further alternative manipulation and application of the triangle inequality to see if you can isolate out that troublesome case.

Hmmm...just trying some expressions...
(f-L)(g-M) = fg -Lg -fM + MG... (f-L)(g+M)/2 + (f+L)(g-M)/2 = fg -LM ...? (f-L)(g-M + 2M)/2 < ?

Something there might be useful or hint at something useful... or maybe not. I haven't gone through the proof myself yet.
 
  • #3
Let epsilon=1. Then there must exist a corresponding delta such that |g(x)-M|<1. How does this help?
 

1. What is a limit proof using delta-epsilon definition?

A limit proof using delta-epsilon definition is a mathematical method used to rigorously prove the limit of a function as its input approaches a certain value. It involves using two variables, delta (δ) and epsilon (ε), to show that for any given epsilon, there exists a corresponding delta such that the difference between the function's output and the limit is less than epsilon when the input is within a certain range.

2. Why is the delta-epsilon definition important in calculus?

The delta-epsilon definition is important in calculus because it provides a rigorous and precise way to prove the limit of a function. It allows us to make accurate predictions about the behavior of a function near a specific input, which is crucial in many real-world applications.

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

To use the delta-epsilon definition to prove a limit, you must first determine the value of epsilon (ε) that you want to prove the limit for. Then, you must find a corresponding delta (δ) value that satisfies the definition for that epsilon. This involves manipulating the function and the given values to find an appropriate delta value. Once you have found a suitable delta, you can prove that the limit exists by showing that the difference between the function's output and the limit is less than epsilon for all inputs within a certain range of the given value.

4. What is the role of the delta-epsilon definition in the concept of continuity?

The delta-epsilon definition is closely related to the concept of continuity in calculus. It states that a function is continuous at a given point if the limit of the function at that point exists and is equal to the function's value at that point. This means that the delta-epsilon definition is used to prove continuity by showing that the function's output remains close to the limit as the input approaches the given value.

5. Can you provide an example of a limit proof using delta-epsilon definition?

Yes, for example, to prove that the limit of the function f(x) = x2 as x approaches 3 is 9, we can use the delta-epsilon definition. Let ε > 0 be given. Then, we need to find a δ > 0 such that if 0 < |x - 3| < δ, then |x2 - 9| < ε. We can choose δ = ε/6, so that if 0 < |x - 3| < ε/6, then |x2 - 9| = |x - 3||x + 3| < ε/6 * (|x - 3| + 6) < ε/6 * (δ + 6) = ε/6 * (ε/6 + 6) = ε. Therefore, we have proven that the limit of f(x) as x approaches 3 is 9, as desired.

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