The Epsilon-Delta Definition of a Limit

Learning Outcomes

Describe the epsilon-delta definition of a limit
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 $a$ and $b$ on a number line is given by $|a-b|$ .

The statement $|f(x)-L|<\varepsilon$ may be interpreted as: The distance between $f(x)$ and $L$ is less than $\varepsilon$ .
The statement $0<|x-a|<\delta$ may be interpreted as: $x\ne a$ and the distance between $x$ and $a$ is less than $\delta$ .

It is also important to look at the following equivalences for absolute value:

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

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

Definition

Let $f(x)$ be defined for all $x\ne a$ over an open interval containing $a$ . Let $L$ be a real number. Then

lim x \to a f (x) = L

$\underset{x\to a}{\lim}f(x)=L$

if, for every $\varepsilon >0$ , there exists a $\delta >0$ such that if $0<|x-a|<\delta$ , then $|f(x)-L|<\varepsilon$ .

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 (for every $\varepsilon >0$ ), an (there exists a $\delta >0$ ), and, last, a (if $0<|x-a|<\delta$ then $|f(x)-L|<\varepsilon$ ). 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 $\varepsilon >0$ ,	1. For every positive distance $\varepsilon$ from $L$ ,
2. there exists a $\delta >0$ ,	2. There is a positive distance $\delta$ from $a$ ,
3. such that	3. such that
4. if $0<\|x-a\|<\delta$ , then $\|f(x)-L\|<\varepsilon$ .	4. if $x$ is closer than $\delta$ to $a$ and $x\ne a$ , then $f(x)$ is closer than $\varepsilon$ to $L$ .

Try It

We can get a better handle on this definition by looking at the definition geometrically. Figure 1 shows possible values of $\delta$ for various choices of $\varepsilon >0$ for a given function $f(x)$ , a number $a$ , and a limit $L$ at $a$ . Notice that as we choose smaller values of $\varepsilon$ (the distance between the function and the limit), we can always find a $\delta$ small enough so that if we have chosen an $x$ value within $\delta$ of $a$ , then the value of $f(x)$ is within $\varepsilon$ of the limit $L$ .

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 $\delta$ , given successively smaller choices of $\varepsilon$ .

Interactive

Visit the following applet to experiment with finding values of $\delta$ for selected values of $\varepsilon$ .

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 $\underset{x\to 1}{\lim}(2x+1)=3$ .

Show Solution

Let $\varepsilon >0$ .

The first part of the definition begins “For every $\varepsilon >0$ .” This means we must prove that whatever follows is true no matter what positive value of $\varepsilon$ is chosen. By stating “Let $\varepsilon >0$ ,” we signal our intent to do so.

Choose $\delta =\frac{\varepsilon}{2}$ .

The definition continues with “there exists a $\delta >0$ .” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find $\delta$ . So, where exactly did $\delta =\varepsilon/2$ come from? There are two basic approaches to tracking down $\delta$ . 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 $|(2x+1)-3|<\varepsilon$ , we begin by manipulating this expression: $|(2x+1)-3|<\varepsilon$ is equivalent to $|2x-2|<\varepsilon$ , which in turn is equivalent to $|2||x-1|<\varepsilon$ . Last, this is equivalent to $|x-1|<\varepsilon/2$ . Thus, it would seem that $\delta =\varepsilon/2$ is appropriate.

We may also find $\delta$ 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 $\delta$ geometrically.

Assume $0<|x-1|<\delta$ . When $\delta$ has been chosen, our goal is to show that if $0<|x-1|<\delta$ , then $|(2x+1)-3|<\varepsilon$ . To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.”

Thus,

$\begin{array}{lllll}|(2x+1)-3| & =|2x-2| & & & \\ & =|2(x-1)| \\ & =|2||x-1| & & & \text{property of absolute values:} \, |ab|=|a||b| \\ & =2|x-1| & & & \\ & <2 \cdot \delta & & & \text{here’s where we use the assumption that} \, 0<|x-1|<\delta \\ & =2 \cdot \frac{\varepsilon}{2}=\varepsilon & & & \text{here’s where we use our choice of} \, \delta =\varepsilon/2 \end{array}$

Analysis

In this part of the proof, we started with $|(2x+1)-3|$ and used our assumption $0<|x-1|<\delta$ in a key part of the chain of inequalities to get $|(2x+1)-3|$ to be less than $\varepsilon$ . We could just as easily have manipulated the assumed inequality $0<|x-1|<\delta$ to arrive at $|(2x+1)-3|<\varepsilon$ as follows:

[latex]\begin{array}{ll} 0<|x-1|<\delta & \implies |x-1|<\delta \\ & \implies -\delta

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

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

Let $\varepsilon >0$ .

Choose $\delta =\varepsilon/2$ .

Assume $0<|x-1|<\delta$ .

Thus,

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

Therefore, $\underset{x\to 1}{\lim}(2x+1)=3$ .

Watch the following video to see the worked solution to Example: Proving a Statement about the Limit of a Specific Function.

Closed Captioning and Transcript Information for Video

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

Problem-Solving Strategy: Proving That $\underset{x\to a}{\lim}f(x)=L$ for a Specific Function $f(x)$

Let’s begin the proof with the following statement: Let $\varepsilon >0$ .
Next, we need to obtain a value for $\delta$ . After we have obtained this value, we make the following statement, filling in the blank with our choice of $\delta$ : Choose $\delta =$ _______.
The next statement in the proof should be (filling in our given value for $a$ ):
Assume $0<|x-a|<\delta$ .
Next, based on this assumption, we need to show that $|f(x)-L|<\varepsilon$ , where $f(x)$ and $L$ are our function $f(x)$ and our limit $L$ . At some point, we need to use $0<|x-a|<\delta$ .
We conclude our proof with the statement: Therefore, $\underset{x\to a}{\lim}f(x)=L$ .

Example: Proving a Statement about a Limit

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

Let _____.

Choose $\delta =$ ________.

Assume $0<|x-\text{___}|<\delta$ .

Thus, $|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \varepsilon$ .

Therefore, $\underset{x \to -1}{\lim}(4x+1)=-3$ .

Show Solution

We begin by filling in the blanks where the choices are specified by the definition. Thus, we have

Let $\varepsilon >0$ .

Choose $\delta =$ _______. (Leave this one blank for now — we’ll choose $\delta$ later)

Assume $0<|x-(-1)|<\delta$ (or equivalently, $0<|x+1|<\delta$ ).

Thus, $|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta = \text{_______} = \varepsilon$ .

Focusing on the final line of the proof, we see that we should choose $\delta =\frac{\varepsilon}{4}$ .

We now complete the final write-up of the proof:

Let $\varepsilon >0$ .

Choose $\delta =\frac{\varepsilon}{4}$ .

Assume $0<|x-(-1)|<\delta$ (or equivalently, $0<|x+1|<\delta$ ).

Thus, $|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta =4(\varepsilon/4)=\varepsilon$ .

Try It

Complete the proof that $\underset{x\to 2}{\lim}(3x-2)=4$ by filling in the blanks.

Let _____.

Choose $\delta =$ ________.

Assume $0<|x-\text{___}|<\delta$ .

Thus, $|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \varepsilon$ .

Therefore, $\underset{x\to 2}{\lim}(3x-2)=4$ .

Hint

Show Solution

Let $\varepsilon >0$ ; choose $\delta =\frac{\varepsilon}{3}$ ; assume $0<|x-2|<\delta$ .

Thus, $|(3x-2)-4|=|3x-6|=|3| \cdot |x-2|<3 \cdot \delta =3 \cdot (\varepsilon/3)=\varepsilon$ .

Therefore, $\underset{x\to 2}{\lim}3x-2=4$ .

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 $\underset{x\to 2}{\lim}x^2=4$ .

Solution

Let $\varepsilon >0$ . The first part of the definition begins “For every $\varepsilon >0$ ,” so we must prove that whatever follows is true no matter what positive value of $\varepsilon$ is chosen. By stating “Let $\varepsilon >0$ ,” we signal our intent to do so.
Without loss of generality, assume $\varepsilon \le 4$ . Two questions present themselves: Why do we want $\varepsilon \le 4$ and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for $\delta$ , we will discover that $\delta$ involves the quantity $\sqrt{4-\varepsilon}$ . Consequently, we need $\varepsilon \le 4$ . In answer to the second question: If we can find $\delta >0$ that “works” for $\varepsilon \le 4$ , then it will “work” for any $\varepsilon >4$ as well. Keep in mind that, although it is always okay to put an upper bound on $\varepsilon$ , it is never okay to put a lower bound (other than zero) on $\varepsilon$ .
Choose $\delta =\text{min}\{2-\sqrt{4-\varepsilon},\sqrt{4+\varepsilon}-2\}$ . Figure 3 shows how we made this choice of $\delta$ .

Figure 3. This graph shows how we find $\delta$ geometrically for a given $\varepsilon$ for the proof in this example.
We must show: If $0<|x-2|<\delta$ , then $|x^2-4|<\varepsilon$ , so we must begin by assuming
$0<|x-2|<\delta$ .

We don’t really need $0<|x-2|$ (in other words, $x\ne 2$ ) for this proof. Since $0<|x-2|<\delta \implies |x-2|<\delta$ , it is okay to drop $0<|x-2|$ .
So, $|x-2|<\delta$ , which implies [latex]-\delta

Recall that $\delta =\text{min}\{2-\sqrt{4-\varepsilon},\sqrt{4+\varepsilon}-2\}$ . Thus, $\delta \le 2-\sqrt{4-\varepsilon}$ and consequently $-(2-\sqrt{4-\varepsilon})\le -\delta$ . We also use $\delta \le \sqrt{4+\varepsilon}-2$ here. We might ask at this point: Why did we substitute $2-\sqrt{4-\varepsilon}$ for $\delta$ on the left-hand side of the inequality and $\sqrt{4+\varepsilon}-2$ on the right-hand side of the inequality? If we look at Figure 3, we see that $2-\sqrt{4-\varepsilon}$ corresponds to the distance on the left of 2 on the $x$ -axis and $\sqrt{4+\varepsilon}-2$ corresponds to the distance on the right. Thus,

[latex]-(2-\sqrt{4-\varepsilon})\le -\delta
We simplify the expression on the left:

[latex]-2+\sqrt{4-\varepsilon}
Then, we add 2 to all parts of the inequality:

[latex]\sqrt{4-\varepsilon}
We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:

[latex]4-\varepsilon
We subtract 4 from all parts of the inequality:

[latex]-\varepsilon
Last,

$|x^2-4|<\varepsilon$ .
Therefore,
$\underset{x\to 2}{\lim}x^2=4$ .

Try It

Find $\delta$ corresponding to $\varepsilon >0$ for a proof that $\underset{x\to 9}{\lim}\sqrt{x}=3$ .

Hint

Show Solution

Choose $\delta =\text{min}\{9-(3-\varepsilon)^2,(3+\varepsilon)^2-9\}$ .

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 $\underset{x\to -1}{\lim}(x^2-2x+3)=6$ .

Show Solution

Let’s use our outline from the Problem-Solving Strategy:

Let $\varepsilon >0$ .
Choose $\delta =\text{min}\{1,\varepsilon/5\}$ . This choice of $\delta$ may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: $|(x^2-2x+3)-6|<\varepsilon$ . This inequality is equivalent to $|x+1|\cdot |x-3|<\varepsilon$ . At this point, the temptation simply to choose $\delta =\frac{\varepsilon}{x-3}$ is very strong. Unfortunately, our choice of $\delta$ must depend on $\varepsilon$ only and no other variable. If we can replace $|x-3|$ by a numerical value, our problem can be resolved. This is the place where assuming $\delta \le 1$ comes into play. The choice of $\delta \le 1$ 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 $\delta \le 1$ and $|x+1|<\delta \le 1$ , we are able to show that $|x-3|<5$ . Consequently, $|x+1| \cdot |x-3|<|x+1| \cdot 5$ . At this point we realize that we also need $\delta \le \varepsilon/5$ . Thus, we choose $\delta =\text{min}\{1,\varepsilon/5\}$ .
Assume $0<|x+1|<\delta$ . Thus,
$|x+1|<1$ and $|x+1|<\frac{\varepsilon}{5}$

Since $|x+1|<1$ , we may conclude that [latex]-1 $|(x^2-2x+3)-6|=|x+1| \cdot |x-3|<\frac{\varepsilon}{5} \cdot 5=\varepsilon$

Therefore,

$\underset{x\to -1}{\lim}(x^2-2x+3)=6$

Watch the following video to see the worked solution to Example: Proving a Statement about the Limit of a Specific Function (Algebraic Approach).

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.

Module 2: Limits