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

1. Jan 31, 2012

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!

Last edited: Jan 31, 2012
2. Jan 31, 2012

### micromass

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

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

$$|x-3|<\delta$$

Thus

$$|x-3|<\delta\leq 1~\text{and}~|x-3|<\varepsilon/7$$

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

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