Response to homework: 20math401retHW12.pdf
KBe20math401srcDeltaEpsilonReview1.png \(|x-2||x+4| < \epsilon\) so near \(x=2\), \(|x-2|\) is smol, so the primary term contributing to the value of the total function is \(|x+4|\). Using the above condition, \(|x-2| < 1 \Rightarrow 1 < x < 3 \Rightarrow 5 < x+4 < 7\)
So, \(|x+4|\) is at most \(7\), we could try substituting it in and getting \(7|x-2| < \epsilon\). Also do the other side: \(5|x-2| < \epsilon\)
You also need to do this: KBe20math401srcDeltaEpsilonProof2.png for the actual proof.
