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This is from the question list at the UC Davis Website epsilon delta exercise list.

In the exercise list we have:

Prove that

Which concludes with:

Thus, if , it follows that [PLAIN]https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/img13.gif. [Broken] This completes the proof.

My curiosity is here: This concludes with iff-then rhetoric. IFF , THEN [PLAIN]https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/img13.gif. [Broken] Which is true here as was proven. However, I don't see anything which explicitly demonstrates that for all real ε>0 with ε/3, there is an x in the domain of f(x), such that [PLAIN]https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/img20.gif. [Broken] This seems intuitively obvious, and likely easy to demonstrate in this case, however if the δ = (function of ε) is sufficiently complex, it seems as though verifying 0 < |x-10| < δ may require another epsilon-delta proof itself. That is to say, the proof here does show that IFF, THEN, however I don't see anything which demonstrates the IFF is actually true.

Have I missed something here? Thanks in advance, everyone!

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# Epsilon delta limits if/then language

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