<s>[INST] Spivak Calculus, Ch. 5 Limits: Finding the smallest $b$ in $|x-3|Consider the function $f(x)=x^2$.
In the text of chapter 5 of Spivak's Calculus, he goes through the following argument to show that $f(x)$ approaches the limit $9$ near $x=3$.
We want that
$$|x^2-9|=|x-3||x+3|<\epsilon$$
Assume $|x-3|<1$.
$$\implies 2<x<4$$
$$\implies 5<x+3<7$$
$$\implies |x+3|<7$$
Therefore
$$|x^2-9|=|x-3||x+3|<7|x-3|<\epsilon$$
$$\implies |x-3|<\frac{\epsilon}{7}$$
provided $|x-3|<1$ as we assumed initially, ie $|x-3|<min(1,\frac{\epsilon}{7})$
I have a few observations about this argument:

*

*$|x+3|=7$ only happens if $|x-3|=1$, the largest distance from $3$ that we allow, by assumption.

*Assuming this "worst case scenario", $|x-3|$ has to be relatively smaller so that the product $|x-3||x+3|$ is smaller than $\epsilon$.

*In fact, depending on $\epsilon$, $|x-3|$ may have to be smaller than $1$. However, $\epsilon$ may also be so large that $|x-3|$ has to be smaller than a number larger than 1.

*All we know is that if $|x-3|<1$ then $|x+3|$ is definitely smaller than 7, and so to achieve a product smaller than $\epsilon$, we need $|x-3|<\frac{\epsilon}{7}$.

*Assume that $\epsilon<7$. Then, $|x-3|<\frac{\epsilon}{7}$<1. But then we can be more precise about how large $|x+3|$ is:

$$-\frac{\epsilon}{7}<x-3<\frac{\epsilon}{7}$$
$$6-\frac{\epsilon}{7}<x+3<6+\frac{\epsilon}{7}$$
$$|x+3|<6+\frac{\epsilon}{7}<7$$
$$|x^2-9|<(6+\frac{\epsilon}{7})|x-3|<\epsilon$$
$$|x-3|<\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
The interval allowed for $x$ near $3$ is now larger. But that means we can go back and calculate what this means for $|x+3|$ again:
$$-\frac{\epsilon}{6+\frac{\epsilon}{7}}<x-3<\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
$$6-\frac{\epsilon}{6+\frac{\epsilon}{7}}<x+3<6+\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
Note that $\frac{\epsilon}{6+\frac{\epsilon}{7}}>\frac{\epsilon}{7}$, so the interval for $|x+3|$ is now larger than before. And if we go back and recalculate the interval for $|x-3|$ it will now be smaller again.
My questions are

*

*Are my calculations correct? Ie, is it ok to do this thing where I narrow down the intervals one at a time?

*How do I figure out what the smallest interval is for $x-3$ such that we have $|x^2-9|=|x-3||x+3|<\epsilon$?

 [/INST] Let $x = 3+h$ then the problem is to estimate $h$, now you need $$-\varepsilon< (3+h)^2 - 9<\varepsilon$$
which is same as saying $ 9-\varepsilon < (3+h)^2 < 9+\varepsilon$, Now we need to check cases if $\varepsilon < 9$ and othercases, using that you will be able to get the interval for $h$
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