I just started Calculus and am making my way through limits. They make sense to me and the $\epsilon$-$\delta$ Definition of a Limit is also clear (I think).
However, I'm having trouble groking this example:
I get that I want to get the left side of the $\epsilon$ inequality in the format of the $\delta$ inequality.
So I have:
$$
0 < |x-2| < \delta
$$
and I have:
$$
\\ |x^2 – 4| < \epsilon
\\ |x-2||x+2| < \epsilon
$$
Now we have to do something about the $|x+2|$ term on the left side. This is where I'm having trouble following along.
"For all x in the interval $(1,3)$, $x + 2 < 5$ and thus $|x + 2| < 5$." I understand what they're saying, but where does the $(1,3)$ interval come from? I see that it's $c\pm1$, but that seems arbitrary. Based on this arbitrary interval we get the inequality $|x + 2| < 5$. If I had picked an interval for $c\pm2$ of $(0,4)$ than I'd get an inequality of $|x+2| < 6$. It looks like based on this interval, they continued the problem like this:
$$
\\|x+2| < 5
\\|x-2|*5<\epsilon
\\|x-2|<\frac{\epsilon}{5} = \delta \therefore
\\|x-2||x+2|<(\frac{\epsilon}{5})(5) = \epsilon
$$
But if I had used a different interval, say $(0,4)$, then I'd be using $6$ everywhere instead of $5$, right?
"So, letting $\delta$ be the minimum of $\epsilon / 5$ and $1$": I don't get what they're saying here at all.
Can $\delta$ ultimately be put into terms of $\epsilon$ via a single function, or do are we choosing a function just so that it satisfies the limit definition?
What am I missing?
Thanks for any help.
Best Answer
You're right; the choice of making the radius $1$ to get an open interval of $(1,3)$ was indeed arbitrary. As you pointed out, we could have picked a radius of $2$ instead and then changed $5$ to $6$ instead.
By letting $\delta = \min\{\epsilon/5, 1\}$, we gain two pieces of information. Namely, if $|x - 2| < \delta$, then we know that $|x - 2| < \delta \leq \epsilon/5$ and $|x - 2| < \delta \leq 1$. The last inequality implies that $x \in (1,3)$ so that $|x + 2| < 5$, which was important for bounding our absolute value term. This strategy of defining $\delta$ as the minimum of two terms is a powerful technique that allows us to gain extra facts in our proof.