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title: Some epsilon-delta proofs
---

Update (May 2022): Like a lot of things in math, the main idea here now seems obvious but it was not at all obvious to me when I was first learning it. I'd probably write the page/proof very differently now, but I'm just leaving this up as a piece of history.

From Salas's *Calculus*, 10th edition, page 104: Chapter 2 review exercise 45.
Below, the important thing to keep in mind is that we want to use the "piecewise function idea": that if a function can be thought of as a piecewise function, we first want to restrict it to where it is essentially nonpiecewise, and then show that the limit exists there.

*Proof*. We want to show $\lim_{x\to-4} |2x+5| = 3$.
If $|x+4|<1$, then $-1<x+4<1$ so $-2<2x+8<2$ so $-5<2x+5<-1$, which means $2x+5$ is negative.
But then $|2x+5| = -(2x+5) = -2x -5$.
So we want to show that for all $x$ satisfying $0<|x+4|<\delta$, that $\big||2x+5|-3\big|<\epsilon$ holds.
But using what we know, $\big| |2x+5|-3 \big| <\epsilon$ is the same thing as $|-2x-5-3| = |-2x-8| = 2|x+4|<\epsilon$ as long as $|x+4|<1$.
So let $\delta = \min\{1, \epsilon/2\}$.
Then $|x+4|<\epsilon/2$ so
$$\begin{align*}
-\epsilon/2 < x&+4 < \epsilon/2 \\
-\epsilon < 2x&+8 < \epsilon \\
-3 -\epsilon < 2x &+ 5 < -3 +\epsilon \ \ \ \ \ (\star) \\
\end{align*}$$
But since $|x+4|<1$, we know
$$\begin{align*}
-1 < x&+4 < 1\\
-2 < 2x &+ 8 < 2 \\
-5 < 25 &+ 5 < -1
\end{align*}$$
So $2x+5 < 0$ i.e. $|2x+5| = -2x -5$.
From $(\star)$ we have then
$$\begin{align}
3 + \epsilon > -2x - 5 > 3 - \epsilon \\
3 + \epsilon > |2x +5| > 3-\epsilon\\
\epsilon > |2x+5|-3 > \epsilon \\
\big| |2x+5| -3\big| < \epsilon \ \ \ \ \ \square
\end{align}$$