# Definition of a Limit epsilon delta

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?

• "Why does the other way around not work as well?" What other way? – Simply Beautiful Art Sep 23 '17 at 19:19
• 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 – user420309 Sep 23 '17 at 19:26
• You mean to ask why it is not reversed so that:$$|f(x)-L|<\epsilon\implies0<|x-a|<\delta$$? – Simply Beautiful Art Sep 23 '17 at 19:29
• yes that is what I wanted to ask – user420309 Sep 23 '17 at 19:30
• 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 $k>\delta$. – Simply Beautiful Art Sep 23 '17 at 19:30

hint:

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

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