The Precise Definition of a Limit: Learn It 1

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The Precise Definition of a Limit: Learn It 1

Use the epsilon-delta method to determine the limit of a function
Demonstrate how limit rules work using the epsilon-delta approach

By now you have progressed from the very informal definition of a limit in the introduction of this module to the intuitive understanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function means and how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precise mathematical language.

Epsilon-Delta Definition of the Limit

Before stating the formal definition of a limit, we must introduce a few preliminary ideas. 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] means 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] means [latex]x\ne a[/latex] and the distance between [latex]x[/latex] and [latex]a[/latex] is less than [latex]\delta[/latex].

It’s important to recognize these equivalences for absolute value:

The statement [latex]|f(x)-L|<\varepsilon[/latex] is equivalent to the statement [latex]L-\varepsilon
The statement [latex]0<|x-a|<\delta[/latex] is equivalent to the statement [latex]a-\delta

With these clarifications, we can state the formal epsilon-delta definition of the limit.

epsilon-delta definition of the limit

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.

Translation of the Epsilon-Delta Definition of the Limit
Definition	Meaning
For every [latex]\varepsilon >0[/latex],	For every positive distance [latex]\varepsilon[/latex] from [latex]L[/latex],
there exists a [latex]\delta >0[/latex],	There is a positive distance [latex]\delta[/latex] from [latex]a[/latex],
such that	such that
if [latex]0<\|x-a\|<\delta[/latex], then [latex]\|f(x)-L\|<\varepsilon[/latex].	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].

By breaking down the definition into these parts, we can better understand and apply the formal epsilon-delta definition of a limit.

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].

There are three graphs side by side showing possible values of delta, given successively smaller choices of epsilon. Each graph has a decreasing, concave down curve in quadrant one. Each graph has the point (a, L) marked on the curve, where L is the limit of the function at the point where x=a. On either side of L on the y axis, a distance epsilon is marked off - namely, a line is drawn through the function at y = L + epsilon and L – epsilon. As smaller values of epsilon are chosen going from graph one to graph three, smaller values of delta to the left and right of point a can be found so that if we have chosen an x value within delta of a, then the value of f(x) is within epsilon of the limit L. — Figure 1. These graphs show possible values of [latex]\delta[/latex], given successively smaller choices of [latex]\varepsilon[/latex].

Visit the following applet to experiment with finding values of [latex]\delta[/latex] for selected values 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.

Prove that [latex]\underset{x\to 1}{\lim}(2x+1)=3[/latex].

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.

Choose [latex]\delta =\frac{\varepsilon}{2}[/latex].

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.

This graph shows how to find delta geometrically. The function 2x + 1 is drawn in red from x=0 to 2. A straight line is drawn at y=3 in green, which intersects the function at (1,3). Two blues lines are drawn at 3 + epsilon and 3 – epsilon, which are graphed here between 5 and 6 and between 0 and 1, respectively. Finally, two pink lines are drawn down from the points of intersection of the function and the blue lines – the taller between 1 and 2, and the shorter between 0 and 1. Since the blue lines and the function intersect, we can solve for x. For the shorter, corresponding to the line y = 3 – epsilon, we have 3 – epsilon = 2x + 1, which simplifies to x = 1 – epsilon / 2. For the taller, corresponding to the line y = 3 + epsilon, we have 3 + epsilon = 2x + 1, which simplifies to x = 1 + epsilon / 2. Delta is the smaller of the two distances between 1 and where the pink lines intersect with the x axis. We have delta is the min of 1 + epsilon / 2 -1 and 1 – (1 – epsilon / 2), which is the min of epsilon / 2 and epsilon / 2, which is simply epsilon / 2. — 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:

[latex]\begin{array}{ll} 0<|x-1|< \delta & \implies |x-1|< \delta \\ & \implies -\delta < x-1< \delta \\ & \implies -\frac{\varepsilon}{2} < x-1 < \frac{\varepsilon}{2} \\ & \implies -\varepsilon < 2x-2 < \varepsilon \\ & \implies |2x-2| < \varepsilon \\ & \implies |(2x+1)-3| < \varepsilon \end{array}[/latex]

Therefore, [latex]\underset{x\to 1}{\lim}(2x+1)=3[/latex]. (Having completed the proof, we state what we have accomplished.)

After removing all the remarks, here is a final version of the proof:

Let [latex]\varepsilon >0[/latex].

Choose [latex]\delta =\varepsilon/2[/latex].

Assume [latex]0<|x-1|<\delta[/latex].

Thus,

[latex]\begin{array}{ll} |(2x+1)-3|& =|2x-2| \\ & =|2(x-1)| \\ & =|2||x-1| \\ & =2|x-1| \\ & <2 \cdot \delta \\ & =2 \cdot \frac{\varepsilon}{2} \\ & =\varepsilon \end{array}[/latex]

Therefore, [latex]\underset{x\to 1}{\lim}(2x+1)=3[/latex].

Watch the following video to see the worked solution to this example.

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The following Problem-Solving Strategy summarizes the type of proof we worked out above.

How to: Prove 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].

Complete the proof that [latex]\underset{x\to -1}{\lim}(4x+1)=-3[/latex] by filling in the blanks.

Let _____.

Choose [latex]\delta =[/latex] ________.

Assume [latex]0<|x-\text{___}|<\delta[/latex].

Thus, [latex]|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \varepsilon[/latex].

Therefore, [latex]\underset{x \to -1}{\lim}(4x+1)=-3[/latex].

In the example above, the proof was fairly straightforward, since the function with which we were working with was linear. In the example below, we see how to modify the proof to accommodate a nonlinear function.

Prove that [latex]\underset{x\to 2}{\lim}x^2=4[/latex].

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,

[latex]-(2-\sqrt{4-\varepsilon})\le -\delta < x-2 < \delta \le \sqrt{4+\varepsilon}-2[/latex].

We simplify the expression on the left:

[latex]-2+\sqrt{4-\varepsilon} < x-2 < \sqrt{4+\varepsilon}-2[/latex].

Then, we add 2 to all parts of the inequality:

[latex]\sqrt{4-\varepsilon} < x < \sqrt{4+\varepsilon}[/latex].

We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:

[latex]4-\varepsilon < x^2 < 4+\varepsilon[/latex].

We subtract 4 from all parts of the inequality:

[latex]-\varepsilon < x^2-4 < \varepsilon[/latex].

Last,

[latex]|x^2-4| < \varepsilon[/latex].
Therefore,
[latex]\underset{x\to 2}{\lim}x^2=4[/latex].

The geometric approach to proving limits works well for some functions and offers valuable insight into the formal definition of limits. However, an algebraic approach can also be useful. It often provides additional insight and can be simpler. Algebraic methods are the primary tools for proving statements about limits. The example below demonstrates a purely algebraic approach to limit proofs.

Prove that [latex]\underset{x\to -1}{\lim}(x^2-2x+3)=6[/latex].

Let’s use our outline from the How To:

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

[latex]|(x^2-2x+3)-6|=|x+1| \cdot |x-3|<\frac{\varepsilon}{5} \cdot 5=\varepsilon[/latex]

Therefore,

[latex]\underset{x\to -1}{\lim}(x^2-2x+3)=6[/latex]

Watch the following video to see the worked solution to this example.

Closed Captioning and Transcript Information for Video
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 can view the transcript for this segmented clip of “2.5 Precise Definition of a Limit” here (opens in a new window).

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.