An extensive explanation about the Epsilon-Delta definition of limits

One of the most important topics in elementary calculus is the \epsilon-\delta definition of limits.  The definition says that the \lim_{x \to a} f(x)= L if and only if, for all \epsilon > 0, there exists a \delta >0 such that if |x-a| < \delta , then |f(x) - L |<\epsilon. In this article, we are going to discuss what this definition means. Readers of this article must have knowledge about elementary calculus and the concept of limits.

Review of Limit Basics

Consider the function f(x) = 2x. We have learned from elementary calculus that \lim_{x \to 2} 2x = 4.  Aside from algebraic computation, this is evident from the color-coded graph and the table shown in Figure 1.  The yellow arrows in the graph and the values in the yellow cells in the table indicate that as the value of x approaches 2 from the left of the x-axis, the value of f(x) approaches 4 from below of the y-axis. On the other hand, the red arrows in the graph and the values in the red cells in the table indicate that as the value of x approaches 2 on from the right of the x-axis, the value of f(x) approaches 4 from above of the y-axis.

Figure 1 – The table and the graph showing the value of f(x) as x approaches 2 from both sides.

From the above discussion, it is noteworthy to mention three things:

  1. We can get f(x) as close to 4 as we please by choosing an x sufficiently close to 2.  For example, I can set x to 1.9999 \cdots99 (with 100 nines) to get an f(x) very close to 4, which is 3.9999 \cdots 98 (99 nines).
  2. No matter how small is the distance of x from 2, a distance less than it may still be chosen. For example, if we choose the point which is very close to 2, say a point with coordinate 1.9999 \cdots 99 with (100 nines), we can still choose a value closer than this to 2.  For instance, we can choose 1.9999 \cdots 99 with 101 nines. This can be repeated for every chosen distance.
  3. Although x can be very very close to 2, it does not necessarily mean that x equals 2.

Now we go back to the definition of limits. In a specific example, the limit definition states that the \lim_{x \to 2} 2x = 4 if (and only if) for all distance (denoted by the Greek letter \epsilon) from 4  along the y-axis (directly above or below 4) – no matter how small – we can always find a certain distance (denoted by \delta) from 2 along the x-axis (left or right of 2) such that if x is between 2 - \delta and 2 + \delta,  then f(x) would lie between 4 - \epsilon and 4 + \epsilon.

To give you a more concrete example, suppose we want the distance \epsilon from 4, which is our limit, to be 0.1 then the interval of our f(x) is (4 -0.1, 4+ 0.1) = (3.9,4.1). The definition of limit says that given a distance \epsilon = 0.1, we can find a distance \delta in the x-axis such that if x is between 2 - \delta and 2 + \delta, we are sure that f(x) is between 3.9 and 4.1. We do not know the value of \delta yet, but we will calculate it later.

Figure 2 – The epsilon-delta definition given epsilon = 0.1.

In Figure 2, x is between 2 - \delta and 2 + \delta or 2 - \delta < x < 2 + \delta. Subtracting 2 from all terms of the inequality, we have - \delta < x - 2 < \delta. If you recall the definition of absolute value, this is precisely the same as |x - 2| < \delta. The comparison among the notations is in Table 1.

Using the notations in the table, we can conclude that the following statements are equivalent:

  1. Words: Given \epsilon = 0.1, we can find a \delta such that if x is between 2 - \delta, 2 + \delta, then f(x) is between 3.9 and 4.1.
  2. Set Notation: Given \epsilon = 0.1, we can find a \delta such that if x \in (2 -\delta, 2 + \delta), then f(x) \in (3.9,4.1).
  3. Relational Operator: Given \epsilon = 0.1, we can find a \delta such that if 2 - \delta < x < 2 + \delta, then 3.9< f(x) < 4 .1.
  4. Absolute Value: Given \epsilon = 0.1, we can find a \delta such that if |x-2| < \delta, then |f(x) - 4| < 0.1.

We have discussed that we can get f(x) as close to 4 as we please

 by choosing an x sufficiently close to 2.  This is equivalent to choosing an extremely small \epsilon, no matter how small, as long as \epsilon>0. Our next task is to find the \delta that corresponds to that \epsilon.

Applying this definition to our example, we can say the \lim_{x \to 2} 2x = 4 if and only if, given \epsilon > 0 (any small distance above and below 4), we can find a \delta > 0 (any distance from x to the left and right of 2) such that if |x - 2| < \delta, then |f(x) - 4| < \epsilon.

The Definition of a Limit of a Function

Now, notice that 4 is the limit of the function as x approaches 2. If we let the limit of a function be equal to L and a be the fixed value that x approaches, then we can say that \lim_{x \to a} f(x) = L if and only if, for any \epsilon>0 (any small distance above and below L), we can find a \delta>0 (any small distance from to the left and to the right of a) such that if |x - a| < \delta then, |f(x) - L| < \epsilon. And that is precisely, the definition of limits that we have stated in the first paragraph of this article.

Figure 3 – The epsilon-delta definition given any epsilon.

In mathematics, the phrase “for any” is the same as “for all” and is denoted by the symbol \forall. In addition, the phrase “we can find” is also the same as “there exists” and is denoted by the symbol \exists. So, rephrasing the definition above, we have \lim_{x \to a} f(x) = L if and only if, \forall \epsilon > 0, \exists \delta >0, such that if |x - a| < \delta then, |f(x) - L| < \epsilon. A much shorter version of this definition is the phrase \lim_{x \to a} f(x) = L \iff, \forall \epsilon > 0, \exists \delta >0, such that |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon. The symbol \iff stands for if and only if and the symbol \Rightarrow is similiar to if-then. If P and Q are statements, the statement P \Rightarrow Q is the same as the statement of the form “If P then Q“.

Finding a specific delta

We said that given any positive \epsilon, we can find a specific \delta, no matter how small our \epsilon is. So let us try our first specific value \epsilon = 0.1.

From the definition, we have \lim_{x \to 2} 2x = 4 if and only if, given \epsilon> 0 (any small distance above and below 4), \exists \delta >0 such that if |x -2| < \delta then, |f(x) - 4| < \epsilon.

Now |f(x) - 4| = |2x - 4| < \epsilon. This implies that
|2||x-2| < \epsilon which implies that 2|x - 2| <0.1. Simplifying, we have |x- 2| < 0.05. This means that our x should be between 1.95 and 2.05 to be sure that our f(x) is between 3.9 and 4.1. This is shown in Figure 4.

Figure 4 – The table showing some of the values of epsilon and delta satisfying the definition of limit of 2x as x approaches 2.

Now, let \epsilon = 0.05. This means that our interval is (3.95,4.05). Now |f(x) - 4| = |2x - 4| < 0.05. Thus, |2||x -2| < \epsilon which implies that 2|x - 2| <0.05. Solving, we have |x - 2| < 0.025. This means that our x should be between 1.975 and 2.025 to be sure that our f(x) is between 3.95 and 4.05. There are only two examples above, but the definition tells us that we can choose any \epsilon > 0 so let us generalize our statement by doing so.

Now |f(x) - 4| = |2x - 4| < \epsilon. This results to |2||x -2| < \epsilon which implies that 2|x - 2| <\epsilon. Solving, we have |x - 2| < \epsilon/2. From the condition above, |x - 2| < \delta so we can let \delta = \epsilon/2.

This means given any \epsilon, we just let our \delta equal to \epsilon/2 and we are sure that if x is between 2-\delta/2 and 2+\delta to be sure that our f(x) is between 4 - \epsilon and 4 - \epsilon.

In the next calculus post, we are going to discuss the strategies on how to get  \delta given an arbitrary \epsilon value, so keep posted.

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23 thoughts on “An extensive explanation about the Epsilon-Delta definition of limits

  1. Why cant we use the more intuitive way of defining limits by saying as x moves closer ro a fx moves closer to L, and the lesser we make the distance from x to a subsequent distance from fx to L should become lesser and lesser than before. Is this not enough?
    Thanks
    Ayesha

  2. Ayesha,

    Thank you for your suggestion. I agree with you. A more intuitive way of introducing limits is very important; however, symbols are required and the parallelism between conceptual understanding and the ability to use symbols is needed to solve problems in calculus. For example, elementary calculus discusses the \epsilon - \delta definition of limits and exams usually require students to get the value of \delta in a given limit problem.

    Even though a student understands the concept of limits, he will still need the understanding of the symbols, procedures used and strategies in solving a limit problem by definition. That is why, it is also important to introduce mathematical notation. 🙂

    Best,
    Guillermo

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  5. An exposition of d topic rarely found in texts around and you wonder what such texts mean when statements are made using such notations. Great work! guy.

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