No problem, glad I could help!

In summary, when using the delta-epsilon definition of limits, it can be proven that if lim f(x) = l and lim f(x) = m, then l=m by considering the smallest delta that satisfies both conditions. This is because a smaller delta can always be chosen to satisfy the limit as it still satisfies the definition.
  • #1
lionel_hutz
3
0

Homework Statement


Using the delta-epsilon definition of limits, prove that of lim f(x) = l and lim f(x) =m, then l=m

Homework Equations


Delta-epsilon definiition of the limit of f(x), as x approaches a:
For all e>0, there is a d s.t if for all x, |x-a|<d, then |f(x) -l|<e

The Attempt at a Solution


Suppose not. Then
(1): For all e>0, there is a d1 s.t. |f(x) - l|<e, provided |x-a|<d1 for all x
(2): For all e>0, there is a d2 s.t. |f(x) - m|<e, provided |x-a|<d2 for all x

I'm getting stuck at finding the delta that works for both of them. I see that the next step is to pick d=min(d1, d2), but I'm confused as to why this is the case

For instance, if d1=1, and d2=2, I don't see how d=1 would satisfy (2) above
 
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  • #2
Given some ##\delta## that works in an epsilon-delta proof, you can always take a smaller delta, ##\delta'##, instead (since ##|x-a|<\delta' < \delta \rightarrow |x-a| < \delta##). So if ##\delta_1## works for (1) and ##\delta_2## works for (2), then ##min(\delta_1,\delta_2)## works for both.
 
  • #3
LastOneStanding said:
Given some ##\delta## that works in an epsilon-delta proof, you can always take a smaller delta, ##\delta'##, instead (since ##|x-a|<\delta' < \delta \rightarrow |x-a| < \delta##). So if ##\delta_1## works for (1) and ##\delta_2## works for (2), then ##min(\delta_1,\delta_2)## works for both.

Alright, I see. I was thinking max and confusing myself, lol. Thanks!
 
Last edited:

1. What is the definition of uniqueness of limits?

The uniqueness of limits is a concept in mathematics that states that a limit of a function is unique, meaning that there can only be one value that the function approaches as the input variable gets closer and closer to a certain value.

2. Why is uniqueness of limits important?

Uniqueness of limits is important because it allows us to accurately describe the behavior of a function near a certain point. It also helps us to prove the existence of limits and determine when a limit does not exist.

3. How is uniqueness of limits proven?

The uniqueness of limits can be proven using the epsilon-delta definition of a limit. This involves showing that for any given epsilon (a small positive number), there exists a corresponding delta (a small positive number) such that the distance between the function's output and the limit is less than epsilon whenever the distance between the input and the limit is less than delta.

4. Can uniqueness of limits be applied to all functions?

No, uniqueness of limits only applies to functions that are continuous at a certain point. This means that the function's output approaches the same value as the input approaches a particular value. Functions that have discontinuities or jumps cannot have unique limits.

5. What role does continuity play in the uniqueness of limits?

Continuity is essential in the uniqueness of limits because it ensures that the function's output does not have any sudden changes or breaks near the point of interest. A continuous function is one that can be drawn without lifting the pen from the paper, and this allows us to accurately determine the behavior of the function near a particular point.

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