# Explore the Definition of a Limit

Specify a function $$f(x)$$, a point $$c$$, and a value $$L$$ in order to explore the limit $$\displaystyle \lim_{x \to c} f(x) = L$$.

Function: $$f(x) =$$
Point: $$c =$$
Limit: $$L =$$

Adjust the values of $$\epsilon$$ and $$\delta$$ using the panel on the left and mark the pairs $$(\epsilon,\delta)$$ for which $$0 < |x-c| < \delta$$ implies $$|f(x)-L| < \epsilon$$, using the graphs in the panel on the right.

You may attempt to guess a curve that predicts $$\delta$$ as a function of $$\epsilon$$ and view that curve. Type eps to represent $$\epsilon$$ in your formula.
Curve: $$\delta = \phi(\epsilon) =$$

## Discussion

The definition of the property that $$f(x)$$ has a limit $$L$$ as $$x \to c$$ is the logical statement,

For any $$\epsilon > 0$$, there exists $$\delta > 0$$ so that $$0 < |x-c| < \delta$$ implies $$|f(x)-L| < \epsilon$$.

The inequality $$|f(x)-L| < \epsilon$$ means that the vertical distance between $$y=f(x)$$ and $$y=L$$ is less than $$\epsilon$$. So the right panel shows $$y=L$$ as a dashed horizontal line along with two thresholds $$y=L+\epsilon$$ and $$y=L-\epsilon$$. We want to find conditions on $$x$$ that guarantee $$y=f(x)$$ is between these thresholds.

The inequality $$0 < |x-c| < \delta$$ means that the distance between $$x$$ and $$c$$ on the numberline is less than $$\delta$$ but also that $$x \ne c$$. This creates a punctured interval, $$(c-\delta, c) \cup (c, c+\delta)$$. So the right panel also shows dashed vertical lines at $$x=c$$ and the left- and right-thresholds at $$x=c-\delta$$ and $$x=c+\delta$$.

When $$x$$ is between the vertical lines, we know $$0 < |x-c| < \delta$$. We want to choose $$\delta$$ so that this guarantees that $$y=f(x)$$ is between the horizontal lines, which would mean that $$|f(x)-L| < \epsilon$$. Because $$\delta$$ needs to be found depending on $$\epsilon$$, the control for $$\epsilon$$ on the left panel is the slider on the independent variable's axis while the control for $$\delta$$ is on the dependent variable's axis.

When you reduce the value of $$\epsilon$$, requiring that $$f(x)$$ be closer to the value $$L$$, the set of $$x$$-values typically gets smaller. This then requires that you make the value of $$\delta$$ smaller in response. As you marked the pairs $$(\epsilon, \delta)$$, you should have seen this played out.

Why is a region shaded when I mark a pair $$(\epsilon, \delta)$$?

Once you find a pair $$(\epsilon, \delta)$$ for which $$0 < |x-c| < \delta$$ implies $$|f(x)-L| < \epsilon$$, any smaller value of $$\delta$$ would still make the implication true since the punctured interval is only getting smaller. Similarly, if we make $$\epsilon$$ any larger, we would still know that the graph $$y=f(x)$$ will be between the thresholds. The shaded region shows all points $$(\epsilon,\delta)$$ for which we already know the implication is true based on our earlier observations.

What is the purpose of an $$(\epsilon, \delta)$$ proof?

Finding individual pairs $$(\epsilon, \delta)$$ illustrates what is required between $$\epsilon$$ and $$\delta$$ so that $$0 < |x-c| < \delta$$ implies $$|f(x)-L| < \epsilon$$. For the limit statement to be true, we must verify that for any $$\epsilon > 0$$ we can find $$\delta > 0$$ so that the implication is true. While checking individual pairs gives us a picture of what is happening, it always leaves a gap below the lowest point we have checked. In this gap, we don't have any proof on what will happen.

A proof provides an argument that fills the gap. Elementary proofs actually provide a formula for where $$\delta$$ needs to be placed when $$\epsilon$$ is known so that the implication will be true. You can visualize this formula as being a curve in the $$\epsilon$$-$$\delta$$ plane on which your marked pairs can always be chosen. While you verify your choice visually, the proof, which is not visual, is required to provide an algebraic (or other) justification for why you know the implication will be true.

What goes wrong when the limit is not valid?

One thing that could go wrong is that the function has a limit but your value $$L$$ is not the right limit. Try the following example: $$f(x) = (2^x-1)\div x$$ with $$x \to 0$$ and $$L=0.7$$. When $$\epsilon$$ and $$\delta$$ are big enough, everything looks good. But if you make $$\epsilon$$ very small, you will actually see that the curve does not go through the cross-hairs of $$x=c$$ and $$y=L$$. For small $$\epsilon$$, you will not be able to find any $$\delta$$ that work.

A second possibility is that the function does not have a limit. In this case, it does not matter what you try for $$L$$, the function will not stay close to your value. Try the following example: $$f(x) = \sin(1/x)$$ with $$x \to 0$$ and $$L=0$$. Because of the rapid oscillation, for values $$\epsilon < 1$$, the graph will continue to cross the thresholds regardless of how small you make $$\delta$$. (In fact, this example might be able to trick the marking test because of how it tries to decide if the function stays between the thresholds.)