Lesson 1.6: Delta-Epsilon: The Formal Definition of a Limit

Thus far we have studied limits without concerning ourselves with the formal definition of a limit. This intuitive understanding of limits is generally sufficient for the purposes of Calculus BC, but because the study of mathematics of based on rigor, it is important to at least understand that the definition of a limit can be formalized. This proper definition, known as the delta-epsilon definition, is not on the AP exam, but you will be tested on it in class.

Let’s take the function f(x)=2x+1 for x neq 2, f(x)=4 for x=2 and consider lim(x->2)f(x). Examining the graph of y=f(x) makes it intuitively clear that lim(x->2)f(x)=5. A limit’s formal definition is rooted in error tolerances, so we ask ourselves, “How close does x have to be to 2 so that f(x) is within 0.1 of 5?” The distance from x to 2 is represented as |x-2| and the distance from f(x) to 5 is |f(x)-5|, so we want to find a number delta (the Greek letter delta) such that |f(x)-5| is less than 0.1 if |x-2| is less than delta but x neq 2. Equivalently, we can look for delta such that x neq 2 if 0 is less than |x-2| is less than delta.

Graph of y=f(x)=2x+1 for x neq 2, f(x)=4 for x=2
The graph of y=f(x)=2x+1 for x neq 2, f(x)=4 for x=2

We can perform a final rewrite of our challenge as follows: find a delta such that 4.9 is less than f(x) is less than 5.1 if 0 is less than |x-2| is less than delta. Therefore, we graph y=f(x), y=4.9, and y=5.1. We find that f(x)=4.9 at x=1.95 and f(x)=5.1 at x=2.05. This gives us 1.95 is less than x is less than 2.05, or |x-2| is less than 0.05, so delta=0.05.

Graph of y=f(x)=2x+1 for x neq 2, f(x)=4 for x=2 and y=4.9 and y=5.1
The graph of y=f(x)=2x+1 for x neq 2, f(x)=4 for x=2
along with y=4.9 and y=5.1

Now let’s try to get closer: find delta such that |f(x)-5| is less than 0.01 if 0 is less than |x-2| is less than delta. We graph y=f(x), y=4.99, and y=5.01, finding that f(x)=4.99 at x=1.995 and f(x)=5.01 at x=2.005. Therefore 1.995 is less than x is less than 2.005, or |x-2| is less than 0.005, so delta=0.005.

Graph of y=f(x)=2x+1 for x neq 2, f(x)=4 for x=2 and y=4.99 and y=5.01
The graph of y=f(x)=2x+1 for x neq 2, f(x)=4 for x=2
along with y=4.99 and y=5.01

Let’s assign epsilon, the Greek letter epsilon, to the error tolerance (for example, we have been working with |f(x)-5| is less than epsilon). Incidentally, epsilon is used generally to denote any small number; the eccentric mathematician Paul Erdős referred to children as epsilons. If we continue this process with ever-decreasing values of epsilon, we will continue the pattern that has been established thus far: delta=(1/2)*epsilon for this function f. We define a limit as follows: if we state lim(x->a)f(x)=L, that means that for every epsilon>0 there exists a delta>0 such that |f(x)-L| is less than epsilon whenever 0 is less than |x-a| is less than delta.

Note that we have not yet proven that lim(x->2)f(x)=5 because we have not proven that delta=(1/2)*epsilon, we have only observed that from a pattern. In the next lesson, we will work on proving that limit.

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