Article contents:

1. Introduction.
2. Definition statement.
3. Graph of definition statement components.
4. Definition statement, broken into pieces.
5. Example of what it looks like when the definition statement conditions aren’t met.
6. Applying the definition generally, with a detailed example for a simple linear equation.

The epsilon-delta definition of a limit is the rigorous definition of a limit. It can be applied as a proof to determine if a value is the true limit of a function or series. It’s not used to find the limit of a function; there are other techniques to do that.

The definition is abstract and took me a while to unpack. Don’t be discouraged if it takes you a while too. Interestingly, it was formulated well after Newton and Leibniz had put together their generalized systems of Calculus. In fact, it took another 150 years! So, why is this formalization necessary? Prior to this definition, limits were regarded as values that we “approach” or “get closer and closer to” until we’re comfortable with the level of precision. This approach worked well enough to solve the practical problems that Calculus was created to solve, but it’s arbitrary. How close is close enough? The epsilon-delta definition is a clever way to prove that a value is the limit.

Here’s the definition:

The components of the definition are much easier to understand with a graph. For this example, we’ll use the linear function \(y=2x+2\), and show an \(\varepsilon = 4\) and a \(\delta = 1\).

Figure 1. A graph of a linear function showing the limit \(L\) of the function \(y=2x+2\) as \(x\) approaches \(a\), and example epsilon and delta values.

A quick note: an important concept to emphasize is that we are evaluating a limit, not finding the value that a function takes at a particular point. For the function above, it is clear that both the limit as \(x\to 6\) is \(14\), and also the value that the function takes when \(x=6\) is \(14\). You should understand that the limit as \(x\) approaches a value is different than finding the value that the function takes at that point. For example, consider a function that is undefined at \(x=2\). The limit of this function as \(x\to 2\) could be a finite number, but the value of the function at \(x=2\) is undefined.

Now that you can visualize the components of the definition, let’s translate each piece of the definition statement into plainer language.

Definition Statement	Meaning
The function \(f(x)\) need not be defined at \(x=a\), but it must be defined for all other \(x\) in some interval which contains \(a\).	The function needs to be defined through the segment of the function we’re interested in (“some interval”). However, the function can be undefined at the the x-value (\(a\)) that corresponds to the limit that we’re testing. This allows the case where we have a limit that is undefined, such as when it goes to infinity, or if there’s a discontinuity in the function.
For every \(\varepsilon > 0\),…	This statement informs us that we will need to show that for every \(\varepsilon > 0\) that exists, we must prove that the following holds true:
…there exists a \(\delta > 0\)…	An interval of input (\(\delta\)) to the function exists whereby…
such that for all values of \(x\) in the domain of \(f\) with \(\left| x – a \right| < \delta \)…	…all \(x\)-values that fall within the interval defined by \(\delta\), excluding \(x=a\)…
…we have \(\left| f(x) – L \right| < \varepsilon\)	…will yield function output that fall within the output interval defined by \(\varepsilon\).

Compare the conditions outlined in this definition with Figure 1 above. The conditions laid out in the definition are met, at least for \(\varepsilon = 4\). We can easily define an interval of \(x\)-values around \(x=a\) (as described by the variable \(\delta\)) that will in turn correspond to function output that falls within the \(\varepsilon\) interval.

But what does it look like when the conditions of this definition are not met?

To demonstrate this visually, take the same function \(y=2x+2\). However, let’s incorrectly assert that the limit \(L\) of the function as \(x\to 6=10\). We know this isn’t correct, but let’s pretend like it is for a moment. The figure below shows how things play out with this new (and incorrect) limit and the same \(\varepsilon\) and \(\delta=1\) values:

Figure 2. A graph of a linear function showing an (incorrect) limit \(L\) of the function \(y=2x+2\) as \(x\) approaches \(a\), and example epsilon and delta values. For the specified The specified \(\varepsilon\) and the chosen \(\delta\) value, the criteria of the limit definition are no longer met.

In Figure 2 above, part of the function defined by \(\delta=1\) is now outside the part of the function defined by \(\varepsilon = 4\), and so the definition criteria are not met. There is no \(\delta\) value that you can choose that will work; the segment of the function defined by \(\delta\) to the right of \(x = 6\) will always fall outside the segment defined by \(\varepsilon\). If you choose larger \(\delta\) values, the \(\delta\) segment will start to extend outside the \(\varepsilon\) segment on the left side too!

Applying the epsilon-delta definition generally using algebra.

Now that you have the intuition behind the definition, the real power lies in the ability to apply it using algebra. In the above examples, we limited ourselves to a single \(\varepsilon\) value. However, the definition statement states that we must be able to find a \(\delta\) for any \(\varepsilon>0\). This is often phrased as if it’s a game between two people; one person supplies an \(\varepsilon>0\), the other proves that a \(\delta>0\) exists. But we don’t want to play this game forever. We need to show generally that a \(\delta>0\) exists for any \(\varepsilon>0\), and we do this by putting \(\delta>0\) in terms of \(\varepsilon\). By doing this, we’ll have a formula that will tell us the appropriate \(\delta\) no matter what \(\varepsilon\) is given.

There is a step-by-step process to find \(\delta\) in terms of \(\varepsilon\). The following example will show you the process for a linear equation. Once you understand how to do this for a simple linear equation, you’ll have a sense of how this works and can practice with more complicated equations, which may have their own peculiarities, but idea is the same.

Example 1. For the equation \(f(x)=2x+2\), prove that the limit as \(x\to 6=14\).

1	\(L=14\) \(a=6\) \(f(x) = 2x+2\)	Write out the pieces of given information. We have a function, a limit that we’re testing, and an x-value (“\(a\)”) associated with the limit.
2	\(\left| f(x) – L \right| < \varepsilon\)	We need to make a general expression for what we ultimately want: an interval of function output within (i.e., smaller than) the interval defined by \(\varepsilon\). This expression is given in the last part of the definition statement. We can define an interval of output generally as the difference between the function and the limit that we’re testing. The absolute value of this difference ensures that we’re not working with a negative distance. Finally, this distance is set to be less than \(\varepsilon\) because we must ensure that it is within the epsilon interval.
3	\(\left| (2x+2) – 14 \right| < \varepsilon\) \(\left| 2x-12 \right| < \varepsilon\)	Plug in the specific function and limit that we’ve been given and simplify. We now have an expression for our specific situation that represents the interval of function output that is smaller than \(\varepsilon\), no matter what \(\varepsilon\) is given to us.
4	\( \left| x – a \right| < \delta \)	Next, we must construct a similar generalized expression for the interval of input within (i.e., smaller than) \(\delta\). Recall that we need to demonstrate that this input will yield output that falls within \(\varepsilon\). The expression is constructed like the expression shown in step 3 for \(\varepsilon\), by taking the absolute value of a difference and then using an inequality. Only this time, the difference is between \(x\) and \(a\).
5	\( \left| x – 6 \right| < \delta \)	Plug in the specific \(a\) that we’ve been given.
…	…	At this point, we have two different expressions; one for \(\varepsilon\) from step 3 and one for \(\delta\) from step 5. Now, we need to put \(\delta\) in terms of \(\varepsilon\). You’ll see that both have \(x\) in their expressions. You can manipulate the expression for \(\varepsilon\) until you have the form \(\left| x-a \right|\) in terms of \(\varepsilon\), which will then link these two equations. Sometimes this form will arrive easily without much algebraic manipulation, other times it may take more work.
6	\(\left| 2x-12 \right| < \varepsilon\) \(2\left| x-6 \right| < \varepsilon\) \(\left| x-6 \right| < \frac{\varepsilon}{2} \)	Manipulate the expression for \(\varepsilon\) to get \(\left| x-a \right|\) on one side of the inequality. Taking the equation in step 3, factor out \(2\) from the terms inside the absolute value brackets, then divide both sides of the equation by \(2\). We now have an equation in the form \(\left| x-a \right|<\varepsilon\).
7	Choose \(\delta= \frac{\varepsilon}{2} \)	According to equation 5, \( \left| x-6 \right| < \delta \). According to equation 6, \(\left| x-6 \right| < \frac{\varepsilon}{2} \). Because \(\frac{\varepsilon}{2} \) is smaller than \(\varepsilon\), we can logically say that if \(\delta = \frac{\varepsilon}{2} \), we will satisfy the conditions of the definition; in other words, no matter what \(\varepsilon>0\) is supplied to us, we can just divide it by 2, and this number gives us an appropriate \(\delta\) that will in turn yield an interval of function output that falls within the interval defined by \(\varepsilon\).
…	…	At this point, most of the work has been done. However, we must demonstrate that our chosen \(\delta\) will work. This involves working in the other direction. We’ll start with the expression \( \left| x – a \right| < \delta \) from the definition statement (and from step 4 here) and plug in our chosen \(\frac{\varepsilon}{2} \) value in place of \(\delta\), then work backwards and see if we algebraically manipulate this expression to arrive back at what we ultimately want: \(\left| f(x) – L \right| < \varepsilon\).
8	\( \left| x – 6 \right| < \frac{\varepsilon}{2} \) \( 2\left| x – 6 \right| < \varepsilon\) \( \left| 2x – 12 \right| < \varepsilon\)	Plug in our chosen delta value into the expression given in (5). Multiply each side by 2, then distribute the 2 into the absolute value expression. You’ll note that we’ve arrived back at the expression shown in step (3). Done!