MHB Trouble with Delta Epsilon Problem

[tfeuerbach]

I understand and can do part (a) of the problem but when it gets to part (b) I'm lost, how do I find/explain why there is no value of delta > 0 that satisfies the the problem.
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[Evgeny.Makarov]

how do I find/explain why there is no value of delta > 0 that satisfies the the problem.

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.

 

[Andrei1]

Part (b) is not a sentence to prove because $$x$$ is free.

(b) Explain why there can be no value of $$\delta>0$$ such that for all $$x\in [0,\ 2]$$ if $$0<|x-1|<\delta$$, then $$|f(x)-4|<1$$.

Now this is a sentence to prove.