Let $$f(x)$$ be a function defined on an open interval containing the number $$A$$, except possible at $$A$$ itself. We say that $\lim_{x \to A} f(x) = L$ if for every $$\varepsilon > 0$$, there exists some $$\delta > 0$$ so that $| f(x) - L | < \varepsilon$ whenever $0 < | x-A | < \delta$ If no such $$\delta$$ exists, the limit at $$A$$ does not exist.

Using the Applet

Adjust the sliders to see how epsilon and delta relate to one another in the definition of a limit. After choosing a value of epsilon, try to find a value of delta so that the "delta tube" representing the interval $$[A-\delta,A+\delta]$$ is blue. If the value of delta is too large for your chosen epsilon, the delta tube will turn red. This happens when there's a value of $$x$$ between $$A-\delta$$ and $$A+\delta$$ where $$f(x)$$ is outside of the interval $$[L-\varepsilon,L+\varepsilon]$$, in green. The "problem sections" of the graph will turn red as well. Notice that $$A$$ is important here as well - the same values of epsilon and delta may not work for all values of $$A$$.

For any $$A \neq 3$$, you should be able to find a delta small enough to satisfy the definition no matter how small epsilon is. For very small values of epsilon, the value of delta will also need to be quite small. This applet only allows values of epsilon of at least 0.01 (smaller values would not show up on the graph), but keep in mind that by definition, epsilon can be as close to zero as you want.

Note that at $$A=3$$, the limit does not exist. If you choose a sufficiently large epsilon here, you may find a delta that works. However, the definition requires that such a delta exists for any value of epsilon. For instance, when $$A=3$$ and $$\varepsilon = 0.5$$, no such delta exists.

You might also notice that there is another "hole" in the graph at $$x=1$$. Despite the fact that the function is not defined here, the limit of the function still exists as $$x$$ approaches 1. Both one sided limits exist, and are the same; the definition does not require the function to be defined at 1.

To keep experimenting, you can try disabling the automatic calculation of the limit for each value of $$A$$. This can also reinforce the definition. If you choose the wrong limit, you may not be able to find a delta that satisfies the definition for your chosen epsilon.