- #1
adelin
- 32
- 0
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Homework Help Overview
The discussion revolves around understanding a proof related to limits in calculus, specifically focusing on the definitions and assumptions involved in limit proofs. The subject area is calculus, with an emphasis on the concept of limits and the epsilon-delta definition.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants are encouraged to explain their understanding of the proof and identify specific areas of confusion. There is a focus on the assumptions made, particularly regarding the values of delta and their implications for the proof.
Discussion Status
The discussion is ongoing, with participants sharing their interpretations and questioning the assumptions made in the proof. Some guidance has been offered regarding the implications of certain assumptions, but there is no explicit consensus on the understanding of the proof as participants explore different aspects of it.
Contextual Notes
Participants note that the problem involves a quadratic function, which is considered qualitatively different from previous linear examples. There is also mention of specific constraints regarding the values of delta and epsilon, which are central to the discussion.
- #2
- 22,170
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You have a previous thread asking the same thing. Was that thread helpful? Did you understand everything there?
If so, can you start by explaining what you think they're doing? And can you explain what you don't get??
If so, can you start by explaining what you think they're doing? And can you explain what you don't get??
- #3
adelin
- 32
- 0
micromass said:
This is another proof
- #4
- 22,170
- 3,333
adelin said:
It is very similar. So please, tell us what you think first.
- #5
adelin
- 32
- 0
micromass said:
in the next step they arrive to the same conclusion. If δ <1 then δ<10. ( I may be wrong)
The next step is what become problematic for me to understand.
- #6
Mark44
Mentor
- 38,107
- 10,669
This problem is qualitatively different from the other one. In the earlier problem, the function was linear. Here the function is a quadratic.
In the second line, which is what I believe you're asking about, they make the assumption that ##\delta < 1##. Then if ##|x - 4 | < \delta < 1##, they can say that x will be between 3 and 5. Note that I'm ignoring the part where it says 0 < |x - 4|. All this does is eliminate the possibility of x being equal to 4.
Since 3 < x < 5, the largest that |x + 5| can be is 10. From this, they can write
## |x + 5||x - 4| < 10|x - 4|##
If we take ##\delta = \epsilon/10##, then when ##|x - 4| < \delta##, it will follow that
##|x + 5||x - 4| < 10|x - 4| < 10 * \delta < 10 * \epsilon/10 = \epsilon ##
In the second line, which is what I believe you're asking about, they make the assumption that ##\delta < 1##. Then if ##|x - 4 | < \delta < 1##, they can say that x will be between 3 and 5. Note that I'm ignoring the part where it says 0 < |x - 4|. All this does is eliminate the possibility of x being equal to 4.
Since 3 < x < 5, the largest that |x + 5| can be is 10. From this, they can write
## |x + 5||x - 4| < 10|x - 4|##
If we take ##\delta = \epsilon/10##, then when ##|x - 4| < \delta##, it will follow that
##|x + 5||x - 4| < 10|x - 4| < 10 * \delta < 10 * \epsilon/10 = \epsilon ##
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