Epsilon-Delta Proof Homework: Understanding the Thought Process

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Homework Statement



For every ε> 0, there is a δ> 0 such that 1- δ< x <1 + implies 2- ε <7-5x <2 + ε


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The Attempt at a Solution



My understanding of epsilon-delta proofs is very minimal at this point. Was hoping someone would be able to explain your thought process when attempting these kinds of proofs. Thank you!
 
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hi dainty77! :smile:
dainty77 said:
Was hoping someone would be able to explain your thought process when attempting these kinds of proofs.

my thought process would be to attempt the simplest possible solution first …

ie try δ = kε first (for a constant k),

then δ = kε2, then δ = k√ε and so on :wink:
 
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I would start with 2- \epsilon&lt; 7- 5x and see that -5- \epsilon&lt; -5x so that x&lt; 1+ \epsilon/5 and x- 1&lt; \epsilon/5. Then turn to 7- 5x&lt; 2+ \epsilon, so that -5x&lt; -5+ \epsilon and x&gt; 1- \epsilon/5, x- 1&gt; -\epsilon/5. Now do you see what \delta must be?
 
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tiny-tim said:
hi dainty77! :smile:


my thought process would be to attempt the simplest possible solution first …

ie try δ = kε first (for a constant k),

then δ = kε2, then δ = k√ε and so on :wink:

I see where you are gettiing at! Let me work on it some more. Thank you for your help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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