Apply the epsilon-delta definition to find the limit of a function
Quantifying Closeness
Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between two points [latex]a[/latex] and [latex]b[/latex] on a number line is given by [latex]|a-b|[/latex].
The statement [latex]|f(x)-L|<\varepsilon [/latex] may be interpreted as: The distance between [latex]f(x)[/latex] and [latex]L[/latex] is less than [latex]\varepsilon[/latex].
The statement [latex]0<|x-a|<\delta [/latex] may be interpreted as: [latex]x\ne a[/latex] and the distance between [latex]x[/latex] and [latex]a[/latex] is less than [latex]\delta[/latex].
It is also important to look at the following equivalences for absolute value:
The statement [latex]|f(x)-L|<\varepsilon [/latex] is equivalent to the statement [latex]L-\varepsilon <f(x)<L+\varepsilon[/latex].
The statement [latex]0<|x-a|<\delta [/latex] is equivalent to the statement [latex]a-\delta <x<a+\delta [/latex] and [latex]x\ne a[/latex].
With these clarifications, we can state the formal epsilon-delta definition of the limit.
Definition
Let [latex]f(x)[/latex] be defined for all [latex]x\ne a[/latex] over an open interval containing [latex]a[/latex]. Let [latex]L[/latex] be a real number. Then
[latex]\underset{x\to a}{\lim}f(x)=L[/latex]
if, for every [latex]\varepsilon >0[/latex], there exists a [latex]\delta >0[/latex] such that if [latex]0<|x-a|<\delta[/latex], then [latex]|f(x)-L|<\varepsilon[/latex].
This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every [latex]\varepsilon >0[/latex]), an existential quantifier (there exists a [latex]\delta >0[/latex]), and, last, a conditional statement (if [latex]0<|x-a|<\delta[/latex] then [latex]|f(x)-L|<\varepsilon[/latex]). Let’s take a look at the table below, which breaks down the definition and translates each part.
Translation of the Epsilon-Delta Definition of the Limit
Definition
Translation
1. For every [latex]\varepsilon >0[/latex],
1. For every positive distance [latex]\varepsilon[/latex] from [latex]L[/latex],
2. there exists a [latex]\delta >0[/latex],
2. There is a positive distance [latex]\delta[/latex] from [latex]a[/latex],
3. such that
3. such that
4. if [latex]0<|x-a|<\delta[/latex], then [latex]|f(x)-L|<\varepsilon[/latex].
4. if [latex]x[/latex] is closer than [latex]\delta [/latex] to [latex]a[/latex] and [latex]x\ne a[/latex], then [latex]f(x)[/latex] is closer than [latex]\varepsilon[/latex] to [latex]L[/latex].
Try It
We can get a better handle on this definition by looking at the definition geometrically. Figure 1 shows possible values of [latex]\delta[/latex] for various choices of [latex]\varepsilon >0[/latex] for a given function [latex]f(x)[/latex], a number [latex]a[/latex], and a limit [latex]L[/latex] at [latex]a[/latex]. Notice that as we choose smaller values of [latex]\varepsilon[/latex] (the distance between the function and the limit), we can always find a [latex]\delta[/latex] small enough so that if we have chosen an [latex]x[/latex] value within [latex]\delta[/latex] of [latex]a[/latex], then the value of [latex]f(x)[/latex] is within [latex]\varepsilon[/latex] of the limit [latex]L[/latex].
Figure 1. These graphs show possible values of [latex]\delta[/latex], given successively smaller choices of [latex]\varepsilon[/latex].
The example below shows how you can use this definition to prove a statement about the limit of a specific function at a specified value.
Example: Proving a Statement about the Limit of a Specific Function
Prove that [latex]\underset{x\to 1}{\lim}(2x+1)=3[/latex].
Show Solution
Let [latex]\varepsilon >0[/latex].
The first part of the definition begins “For every [latex]\varepsilon >0[/latex].” This means we must prove that whatever follows is true no matter what positive value of [latex]\varepsilon[/latex] is chosen. By stating “Let [latex]\varepsilon >0[/latex],” we signal our intent to do so.
The definition continues with “there exists a [latex]\delta >0[/latex].” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find [latex]\delta[/latex]. So, where exactly did [latex]\delta =\varepsilon/2[/latex] come from? There are two basic approaches to tracking down [latex]\delta[/latex]. One method is purely algebraic and the other is geometric.
We begin by tackling the problem from an algebraic point of view. Since ultimately we want [latex]|(2x+1)-3|<\varepsilon[/latex], we begin by manipulating this expression: [latex]|(2x+1)-3|<\varepsilon[/latex] is equivalent to [latex]|2x-2|<\varepsilon[/latex], which in turn is equivalent to [latex]|2||x-1|<\varepsilon[/latex]. Last, this is equivalent to [latex]|x-1|<\varepsilon/2[/latex]. Thus, it would seem that [latex]\delta =\varepsilon/2[/latex] is appropriate.
We may also find [latex]\delta [/latex] through geometric methods. Figure 2 demonstrates how this is done.
Figure 2. This graph shows how we find [latex]\delta[/latex] geometrically.
Assume [latex]0<|x-1|<\delta[/latex]. When [latex]\delta[/latex] has been chosen, our goal is to show that if [latex]0<|x-1|<\delta[/latex], then [latex]|(2x+1)-3|<\varepsilon[/latex]. To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.”
Thus,
[latex]\begin{array}{lllll}|(2x+1)-3| & =|2x-2| & & & \\ & =|2(x-1)| \\ & =|2||x-1| & & & \text{property of absolute values:} \, |ab|=|a||b| \\ & =2|x-1| & & & \\ & <2 \cdot \delta & & & \text{here’s where we use the assumption that} \, 0<|x-1|<\delta \\ & =2 \cdot \frac{\varepsilon}{2}=\varepsilon & & & \text{here’s where we use our choice of} \, \delta =\varepsilon/2 \end{array}[/latex]
Analysis
In this part of the proof, we started with [latex]|(2x+1)-3|[/latex] and used our assumption [latex]0<|x-1|<\delta[/latex] in a key part of the chain of inequalities to get [latex]|(2x+1)-3|[/latex] to be less than [latex]\varepsilon[/latex]. We could just as easily have manipulated the assumed inequality [latex]0<|x-1|<\delta[/latex] to arrive at [latex]|(2x+1)-3|<\varepsilon[/latex] as follows:
Watch the following video to see the worked solution to Example: Proving a Statement about the Limit of a Specific Function.
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For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
The following Problem-Solving Strategy summarizes the type of proof we worked out above.
Problem-Solving Strategy: Proving That [latex]\underset{x\to a}{\lim}f(x)=L[/latex] for a Specific Function [latex]f(x)[/latex]
Let’s begin the proof with the following statement: Let [latex]\varepsilon >0[/latex].
Next, we need to obtain a value for [latex]\delta[/latex]. After we have obtained this value, we make the following statement, filling in the blank with our choice of [latex]\delta[/latex]: Choose [latex]\delta =[/latex] _______.
The next statement in the proof should be (filling in our given value for [latex]a[/latex]):
Assume [latex]0<|x-a|<\delta[/latex].
Next, based on this assumption, we need to show that [latex]|f(x)-L|<\varepsilon[/latex], where [latex]f(x)[/latex] and [latex]L[/latex] are our function [latex]f(x)[/latex] and our limit [latex]L[/latex]. At some point, we need to use [latex]0<|x-a|<\delta[/latex].
We conclude our proof with the statement: Therefore, [latex]\underset{x\to a}{\lim}f(x)=L[/latex].
Example: Proving a Statement about a Limit
Complete the proof that [latex]\underset{x\to -1}{\lim}(4x+1)=-3[/latex] by filling in the blanks.
In the examples above, the proofs were fairly straightforward, since the functions with which we were working were linear. In the example below, we see how to modify the proof to accommodate a nonlinear function.
Example: Proving a Statement about the Limit of a Specific Function (Geometric Approach)
Prove that [latex]\underset{x\to 2}{\lim}x^2=4[/latex].
Solution
Let [latex]\varepsilon >0[/latex]. The first part of the definition begins “For every [latex]\varepsilon >0[/latex],” so we must prove that whatever follows is true no matter what positive value of [latex]\varepsilon[/latex] is chosen. By stating “Let [latex]\varepsilon >0[/latex],” we signal our intent to do so.
Without loss of generality, assume [latex]\varepsilon \le 4[/latex]. Two questions present themselves: Why do we want [latex]\varepsilon \le 4[/latex] and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for [latex]\delta[/latex], we will discover that [latex]\delta[/latex] involves the quantity [latex]\sqrt{4-\varepsilon}[/latex]. Consequently, we need [latex]\varepsilon \le 4[/latex]. In answer to the second question: If we can find [latex]\delta >0[/latex] that “works” for [latex]\varepsilon \le 4[/latex], then it will “work” for any [latex]\varepsilon >4[/latex] as well. Keep in mind that, although it is always okay to put an upper bound on [latex]\varepsilon[/latex], it is never okay to put a lower bound (other than zero) on [latex]\varepsilon[/latex].
Choose [latex]\delta =\text{min}\{2-\sqrt{4-\varepsilon},\sqrt{4+\varepsilon}-2\}[/latex]. Figure 3 shows how we made this choice of [latex]\delta[/latex].
Figure 3. This graph shows how we find [latex]\delta[/latex] geometrically for a given [latex]\varepsilon[/latex] for the proof in this example.
We must show: If [latex]0<|x-2|<\delta[/latex], then [latex]|x^2-4|<\varepsilon[/latex], so we must begin by assuming
[latex]0<|x-2|<\delta[/latex].
We don’t really need [latex]0<|x-2|[/latex] (in other words, [latex]x\ne 2[/latex]) for this proof. Since [latex]0<|x-2|<\delta \implies |x-2|<\delta[/latex], it is okay to drop [latex]0<|x-2|[/latex].
So, [latex]|x-2|<\delta[/latex], which implies [latex]-\delta <x-2<\delta[/latex].
Recall that [latex]\delta =\text{min}\{2-\sqrt{4-\varepsilon},\sqrt{4+\varepsilon}-2\}[/latex]. Thus, [latex]\delta \le 2-\sqrt{4-\varepsilon}[/latex] and consequently [latex]-(2-\sqrt{4-\varepsilon})\le -\delta[/latex]. We also use [latex]\delta \le \sqrt{4+\varepsilon}-2[/latex] here. We might ask at this point: Why did we substitute [latex]2-\sqrt{4-\varepsilon}[/latex] for [latex]\delta [/latex] on the left-hand side of the inequality and [latex]\sqrt{4+\varepsilon}-2[/latex] on the right-hand side of the inequality? If we look at Figure 3, we see that [latex]2-\sqrt{4-\varepsilon}[/latex] corresponds to the distance on the left of 2 on the [latex]x[/latex]-axis and [latex]\sqrt{4+\varepsilon}-2[/latex] corresponds to the distance on the right. Thus,
The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For the example below, we take on a purely algebraic approach.
Example: Proving a Statement about the Limit of a Specific Function (Algebraic Approach)
Prove that [latex]\underset{x\to -1}{\lim}(x^2-2x+3)=6[/latex].
Show Solution
Let’s use our outline from the Problem-Solving Strategy:
Let [latex]\varepsilon >0[/latex].
Choose [latex]\delta =\text{min}\{1,\varepsilon/5\}[/latex]. This choice of [latex]\delta[/latex] may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: [latex]|(x^2-2x+3)-6|<\varepsilon[/latex]. This inequality is equivalent to [latex]|x+1|\cdot |x-3|<\varepsilon[/latex]. At this point, the temptation simply to choose [latex]\delta =\frac{\varepsilon}{x-3}[/latex] is very strong. Unfortunately, our choice of [latex]\delta[/latex] must depend on [latex]\varepsilon[/latex] only and no other variable. If we can replace [latex]|x-3|[/latex] by a numerical value, our problem can be resolved. This is the place where assuming [latex]\delta \le 1[/latex] comes into play. The choice of [latex]\delta \le 1[/latex] here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since [latex]\delta \le 1[/latex] and [latex]|x+1|<\delta \le 1[/latex], we are able to show that [latex]|x-3|<5[/latex]. Consequently, [latex]|x+1| \cdot |x-3|<|x+1| \cdot 5[/latex]. At this point we realize that we also need [latex]\delta \le \varepsilon/5[/latex]. Thus, we choose [latex]\delta =\text{min}\{1,\varepsilon/5\}[/latex].
Assume [latex]0<|x+1|<\delta[/latex]. Thus,
[latex]|x+1|<1[/latex] and [latex]|x+1|<\frac{\varepsilon}{5}[/latex]
Since [latex]|x+1|<1[/latex], we may conclude that [latex]-1<x+1<1[/latex]. Thus, by subtracting 4 from all parts of the inequality, we obtain [latex]-5<x-3<−1[/latex]. Consequently, [latex]|x-3|<5[/latex]. This gives us
Watch the following video to see the worked solution to Example: Proving a Statement about the Limit of a Specific Function (Algebraic Approach).
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For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits.
Try It
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CC licensed content, Original
2.5 Precise Definition of a Limit. Authored by: Ryan Melton. License: CC BY: Attribution
![This graph shows how to find delta geometrically for a given epsilon for the above proof. First, the function f(x) = x^2 is drawn from [-1, 3]. On the y axis, the proposed limit 4 is marked, and the line y=4 is drawn to intersect with the function at (2,4). For a given epsilon, point 4 + epsilon and 4 – epsilon are marked on the y axis above and below 4. Blue lines are drawn from these points to intersect with the function, where pink lines are drawn from the point of intersection to the x axis. These lines land on either side of x=2. Next, we solve for these x values, which have to be positive here. The first is x^2 = 4 – epsilon, which simplifies to x = sqrt(4-epsilon). The next is x^2 = 4 + epsilon, which simplifies to x = sqrt(4 + epsilon). Delta is the smaller of the two distances, so it is the min of (2 – sqrt(4 – epsilon) and sqrt(4 + epsilon) – 2).](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/2332/2018/01/11203538/CNX_Calc_Figure_02_05_003.jpg)
