# Some epsilon-delta proofs

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 \[\begin{align}
3 + \epsilon > -2x - 5 > 3 - \epsilon \\
3 + \epsilon > |2x +5| > 3-\epsilon\\
\epsilon > |2x+5|-3 > \epsilon \\
\big| |2x+5| -3\big| < \epsilon \ \ \ \ \ \square
\end{align}\]