Help Understanding this simple proof? [lim x->3 x^2 = 9]

[nickadams]

Edit: Sorry i seem to have lost my attachment! I will upload it again tomorrow when i get to a scanner...

Disclaimer: I have no experience with proofs so go easy on me
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I don't understand what they are doing on example number 4 of the attached page. It makes sense until they say, "it is reasonable to assume that x is within a distance 1 from 3.." Why not a distance .1, or .5, or .001? How were they able to come up with the arbitrary (to me) number "1" and say that is the distance x is from 3? And how did they know this would work for the purposes of their proof?

Also I don't the part under showing that this works where they seem to use both conditions |x-3|<1 and |x-3|<ε/7 even though they said earlier they would only use the smaller of the two restrictions... Because if they use both restrictions, isn't that restricting the value of ε to be <7, because if ε was greater than 7 then both restrictions could not be satisfied?

Lastly, I have trouble understanding how all their steps even end up proving anything? I think the goal was to prove that a δ>0 exists such that if 0<|x-3|<δ is true then it will guarantee the truth of |x^2 -9|<ε for all ε>0. But I am extremely confused about how their steps prove that...











Any clarification will be greatly appreciated!

nickadams

 

[micromass]

Disclaimer: I have no experience with proofs so go easy on me
____________________________________________________________________I don't understand what they are doing on example number 4 of the attached page. It makes sense until they say, "it is reasonable to assume that x is within a distance 1 from 3.." Why not a distance .1, or .5, or .001? How were they able to come up with the arbitrary (to me) number "1" and say that is the distance x is from 3? And how did they know this would work for the purposes of their proof?

.1, .5 and .001 would have worked as well. They chose 1 because 1 is easy to work with, I guess.

Also I don't the part under showing that this works where they seem to use both conditions |x-3|<1 and |x-3|<ε/7 even though they said earlier they would only use the smaller of the two restrictions... Because if they use both restrictions, isn't that restricting the value of ε to be <7, because if ε was greater than 7 then both restrictions could not be satisfied?

We take \delta the smallest of 1 and \varepsilon/7. So we have both
\delta \leq 1 and \delta \leq \varepsilon/7.

By assumption, we know that

|x-3|&lt;\delta

Thus

|x-3|&lt;\delta\leq 1~\text{and}~|x-3|&lt;\varepsilon/7

So we can use both |x-3|<1 and |x-3|&lt;\varepsilon/7.

Lastly, I have trouble understanding how all their steps even end up proving anything? I think the goal was to prove that a δ>0 exists such that if 0<|x-3|<δ is true then it will guarantee the truth of |x^2 -9|<ε for all ε>0. But I am extremely confused about how their steps prove that...

Given a certain \varepsilon&gt;0, we choose \delta=\min\{1,\varepsilon/3\}. We must prove that if |x-3|&lt;\delta, then |x^2-9|&lt;\varepsilon.
Step (2) indeed assumes that |x-3|&lt;\delta and deduces |x^2-9|&lt;\varepsilon.