The discussion centers on the Delta Epsilon problem, specifically part (b), which requires proving that no value of delta greater than 0 can satisfy the condition for all x in the interval [0, 2]. The key argument presented is that the set of x satisfying 0 < |x-1| < delta always includes elements less than 1, where the function f(x) yields values less than or equal to 2. Consequently, the distance from f(x) to 4 is always greater than 1, invalidating the condition for any delta > 0.
PREREQUISITESStudents studying calculus, particularly those tackling limits and continuity, as well as educators looking for clear explanations of Delta Epsilon proofs.
Because the set of $x$ satisfying $0<|x-1|<\delta$ (the deleted neighborhood of 1) always has elements for which $|f(x)-4|\ge2$: namely, those $x$ that lie between 0 and 1. Whichever neighborhood of 1 you choose, it will always have elements smaller than 1, and for those elements the value of $f$ is $\le 2$, so its distance to 4 is greater than 1.tfeuerbach said:
