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Why is it that the limit exists if: for all numbers epsilon (in the epsilon range close to L) => that there’s a delta (in the delta range close to a) and if this number Epsilon exists and it turns out that there is a delta for this epsilon then this implies that there is a limit? Why does the other way around not work as well? (it seems like from this definition that the y axis (epsilon range) is greater than the x-axis delta range according to the definition.

Could someone clarify this for me?

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    $\begingroup$ "Why does the other way around not work as well?" What other way? $\endgroup$ – Simply Beautiful Art Sep 23 '17 at 19:19
  • $\begingroup$ 0<|x-a|<\delta implies that |f(x)-L|<\epsilon Why is it not reversed so that if for every epsilon we can find a delta then there is a limit $\endgroup$ – user420309 Sep 23 '17 at 19:26
  • $\begingroup$ You mean to ask why it is not reversed so that:$$|f(x)-L|<\epsilon\implies0<|x-a|<\delta$$? $\endgroup$ – Simply Beautiful Art Sep 23 '17 at 19:29
  • $\begingroup$ yes that is what I wanted to ask $\endgroup$ – user420309 Sep 23 '17 at 19:30
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    $\begingroup$ Consider $f(x)=\sin(x)$. Then $|f(x)-0|<\epsilon$ does not imply $0<|x-0|<\delta$, since we could have $x=k\pi$, where $k$ is an integer $k>\delta$. $\endgroup$ – Simply Beautiful Art Sep 23 '17 at 19:30
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hint: enter image description here

This gif show the performance of $$\epsilon,\delta$$ . Hope it help you

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  • $\begingroup$ thanks, could you elaborate any further? $\endgroup$ – user420309 Sep 23 '17 at 19:23

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