True/False regarding Delta Neighborhoods

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SUMMARY

The discussion centers on the true/false statement regarding delta neighborhoods in response to epsilon challenges in calculus. It is established that if a specific delta has been constructed to satisfy an epsilon condition, a smaller positive delta does not necessarily suffice. The submitter's argument incorrectly suggests that demonstrating a larger delta's sufficiency implies the same for a smaller delta, which contradicts the original question. Understanding the relationship between delta and epsilon is crucial for accurate limit analysis.

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Homework Statement


True/False: If a particular delta has been constructed as a suitable response to a particular epsilon challenge, then any smaller positive delta will also suffice.

Homework Equations

The Attempt at a Solution



The submitted solution is as follows:
mathhw1.png


However, when I read this solution, I note that 0 < delta_1 < delta_2.

The submitter goes on to start with delta_1 to then show that delta 2 holds. Didn't the submitter show that if a particular delta has been constructed as a suitable response to a particular epsilon challenge, then any LARGER positive delta will also suffice? This is not what the question asks.
 
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RJLiberator said:

Homework Statement


True/False: If a particular delta has been constructed as a suitable response to a particular epsilon challenge, then any smaller positive delta will also suffice.

Homework Equations

The Attempt at a Solution



The submitted solution is as follows:
View attachment 113966

However, when I read this solution, I note that 0 < delta_1 < delta_2.

The submitter goes on to start with delta_1 to then show that delta 2 holds. Didn't the submitter show that if a particular delta has been constructed as a suitable response to a particular epsilon challenge, then any LARGER positive delta will also suffice? This is not what the question asks.
No, it is not saying that a larger delta works. Here's what I think is going on. A challenge value of ##\epsilon > 0## has been given, which is answered by a value of ##\delta_2##. In the image, a smaller value of ##\delta_1## is then selected. Now, if ##|x - c | < \delta_2## it will also be true (almost trivially) that ##|x - c | < \delta_1##, which in turn implies that ##|f(x) - f(c)| < \epsilon##
 
RJLiberator said:

Homework Statement


True/False: If a particular delta has been constructed as a suitable response to a particular epsilon challenge, then any smaller positive delta will also suffice.

There are two aspects to limits: an understanding of what you are trying to do; and, the nitty-gritty manipulation of epsilons and deltas etc.

In this case, the understanding should be clear: finding a delta means you are "close enough" to a point and if you reduce the delta you are "even closer". While, increasing the delta means you are "further away".

You really shouldn't be having any trouble seeing this.
 

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