On the preceding screen we used the Pancake Story to introduce $\epsilon$ (the Greek letter ‘epsilon’), which refers to the ‘error,’ or tolerance, in the output of a process, and $\delta$ (the lower-case Green letter ‘delta’), which refers to the difference in input values. We then began working toward the formal limit definition by referencing these quantities in our working conceptual definition:

\[\text{The limit of the function $f(x)$ as $x$ approaches $a$, written} \] \[\lim_{x \to a} f(x) = L \] \[\text{is a number $L$ (if one exists)}\] \[\text{such that $f(x)$ is $\underbrace{\text{as close to}}_{\Large{\text{within } \pm \epsilon} \text{ of}}$ $L$ as we want whenever $x$ is $\underbrace{\text{sufficiently close to}}_{\Large{\text{within } \pm \delta \text{ of}}}$ $a.$ }\]On this screen we will extend these ideas to develop the formal definition of the limit.

But if you’d like to learn the formal definition and see how closely it adheres to our working conceptual definition, along with how we actually *prove* that a limit exists, please continue on here:

Let’s use our refined definition of “limit,” with $\epsilon$ and $\delta$ in mind, to revisit the limit as $x \to 2$ of the function we examined at the start of this section, $f(x) = \dfrac{x^2 \,- \,4}{x \,-\, 2}.$

As we saw when we first examined this function, we can get as close to the output value of $f(x) = 4$ as we’d like by being sufficiently close to the input value $x=2,$ information that we capture with the limit statement

\[\lim_{\color{blue}{x \to 2}} \overbrace{ \dfrac{x^2\, -\, 4}{x \,-\, 2}}^{f(x)} = \color{brown}{4}\]
We think about this mathematical statement in words as

“We can get as close to $f(x)$ = 4 as we’d like by being sufficiently close to *x* = 2.”

Using the language and symbols we introduced on the preceding screen, we can say now that for this function,

- focusing on the
*output,*“as close to” in the phrase “as close to $f(x) = 4$ as we’d like” refers to the error, or tolerance, which we’re denoting by $\epsilon,$ while - focusing on the
*input,*“sufficiently close” in the phrase “sufficiently close to*x*= 2″ refers to $\delta.$

That is, if someone says to you,

First, I will give you a value for $\epsilon.$ (Perhaps, for instance, you are told $\epsilon = 0.5.$)

Then you need to tell me how close ($\delta$) your input value needs to be to $x = 2$ such that for

\[2- \delta \lt x \lt 2+ \delta \] the function’s output value will be within $\pm \epsilon$ of 4:

\[ 4 \,-\, \epsilon \lt f(x) \lt 4 + \epsilon\]

Show/Hide connection to this statement from the Pancake Story

This “first … then” flow here should remind you of your imaginary job at the Pancake House: Your boss gave you an allowable $\epsilon,$ or tolerance value, for the day (say, $\epsilon = 0.5$ inches). You then had to give him a value of $\delta$ for how “off” you could be in how much pancake batter you pour (say, $\delta = 1$ tablespoon) in order to be within that day’s tolerance for the pancake’s size.

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Notice that the requirement on the last line is an *open* interval: the function’s output must lie in the range $(4 \,-\,\epsilon \lt 4 + \epsilon),$ and the endpoints are *not* included. We’ll discuss why we exclude the endpoints later.

In the following activity, we’ll use the same linked-graphs we did in the Pancake Story to investigate this limit, and use that to build the formal definition of the limit.

The following linked-graphs illustrate the meaning of epsilon ($\epsilon$) and delta ($\delta$), just as we did on the preceding screen for the Pancake Story:

- The upper graph focuses on the function’s output near $x=2,$ and shows the maximum and minimum allowable output values of the function, $4 \pm \epsilon$, for a particular value of $\epsilon$ (initially $\epsilon =1).$
- The lower graph focuses on the proposed input range, as determined by the slider for $\delta$ beneath that graph, so we can easily see which values of $\delta$ meet our criterion and which do not.

**PART I** (like Day 1 of the Pancake Story): $\epsilon = 1$

We arbitrarily set $\delta = 1.8$ below to start you off. You can see that with this value of $\delta,$ some input values in the range $2- \delta \lt x \lt 2+ \delta$ have output values $f(x)$ that fall *outside* the allowable range $4\, -\, \epsilon \lt f(x) \lt 4 + \epsilon$ for $\epsilon = 1.$ For instance, let’s consider just the single value $x = 0.5,$ which is certainly within the range $2- 1.8 \lt x \lt 2+ 1.8$. The function’s corresponding output value $f(0.5) = 2.5$ is *less* than the allowable smallest value $4\, -\, \epsilon = 3$. The fact that even just this single value of *x* in the input-range $2 \pm \delta$ for $\delta = 1.8$ does *not* meet the requirements for $\epsilon = 1$ leads us to the conclusion that this value of $\delta$ is not acceptable. Of course there are many other values of *x* in this input range that also do not work, as well as many that do. The point is that *all* values of *x* in the entire range *must* work; otherwise the $\delta$ is unacceptable.

Use the slider to decrease the value of $\delta$ in order to determine which values *do* satisfy the condition for $\epsilon = 1.$ This process mimics what you did in the Pancake Story, when you found the range of batter you needed to use in order to produce pancakes sized within the permitted tolerance of ±1 inch.

Further instructions will appear below once you have completed this task.

Focus on Function Output:

The function’s output must fall within $4 \pm \epsilon.$ Currently $\epsilon$ = 1.

The function’s output must fall within $4 \pm \epsilon.$ Currently $\epsilon$ = 1.

Focus on Function Input:

How small must $\delta$ be so the output is within the desired range?

How small must $\delta$ be so the output is within the desired range?

Use the slider to change the value of $\delta$:

Currently $\delta$ = 1.8.

Currently $\delta$ = 1.8.

For $\epsilon$ = 1 and for this value of $\delta,$ some output values fall outside $4 \\pm \\epsilon.$ ✗

[Further instructions will appear here once you find values of $\delta$ that work for $\epsilon = 1.$]

**PART 2**: Other values of $\epsilon$

Now that you’ve determined values of $\delta$ that satisfy $\epsilon = 1,$ please change to the other values of $\epsilon$ using the buttons below and then verify that there are values of $\delta$ that satisfy the new condition.

You you have verified that the condition can be met for each value of $\epsilon.$ Nice! Now, when you’re sure you understand what the graph is showing in each case, please continue.

In the preceding Activity you verified (the probably obvious result) that, for each of the given $\epsilon$ values, you can find at least one value of $\delta$ such that for *any* input value *x* that lies in the interval

\[2- \delta \lt x \lt 2+ \delta \]
the function’s output value $f(x)$ will be within $\pm \epsilon$ of 4:

\[ 4 \,-\, \epsilon \lt f(x) \lt 4 + \epsilon\]
That’s what we *mean* when we write $\displaystyle{\lim_{x \to 2}f(x) = 4.}$

You can probably imagine doing similar verifications for other values of $\epsilon$ we asked you try, for $\epsilon = 0.000000000001,$ or $\epsilon = 10^{-21},$ or *any* value (greater than zero) of $\epsilon$ we chose.

The preceding paragraph illustrates a very math-y statement:

The limit **exists** *if* such a $\delta$ exists for *any* value of $\epsilon \gt 0$ that we choose.

That is, whatever arbitrary value of $\epsilon$ we choose, be it large or small – even smaller than you can conceive – in order for the limit to exist there *must* be at least a single value of $\delta$ that satisfies the requirement that all output values lie within the range $L \pm \epsilon,$ where *L* is the limit. Considering our function above, regardless of the value of $\epsilon$ someone specifies, we can find (or imagine finding, anyway) a value of $\delta$ such that all values of *x* that lie within the range of $\pm \delta$ of $x=2$ produce output values that lie within the range of $L \pm \epsilon,$ where $L = 4.$

With our understanding of $\epsilon$ and $\delta$ in place, we can now state the formal definition of limits:

Let

Then a number

\[\text{if} \quad a\, -\, \delta \lt x \lt a + \delta, \quad \text{ then } \quad L\, – \,\epsilon \lt f(x) \lt L + \epsilon. \] We write this limit as $\displaystyle{\lim_{x \to a}f(x) = L. }$

While the formal definition may look abstract, you can see how it exactly matches our conclusion directly above for why $\displaystyle{\lim_{x \to 2}\dfrac{x^2\, -\, 4}{x \,-\, 2} = 4.}$ We have $a=2,$ and the function is not defined for that input value, as allowed by the phrase in the first sentence, “except possibly at *a* itself.” The rest of the definition matches the reasoning you explored in Activity 1 above: for any $\epsilon$ you’re given, you can find an appropriate number $\delta$ that meets the requirements.

Looking at a graph like the one above, it’s probably obvious to *see* (or imagine, at least) that no matter how small we choose to make $\epsilon,$ we can find such a value for $\delta$ to satisfy the requirement that the output values fall within $4 \pm \epsilon$. *Proving* that this is the case, by contrast, can be a real challenge depending on the particular function. A proof is in the box below. (You don’t have to look at the proof in order to continue; we include it for completeness and in case you’re interested.)

Hide/Show Proof and Discussion of this limit

Note than in most Calculus 1 courses you are

First, since

\[\dfrac{x^2 \, -\,4}{x\,-\,2} = \dfrac{(x+2)\cancel{(x\,-\,2)}}{\cancel{(x\,-\,2)}}\]
we can rewrite our function as

\[

f(x) = \begin{cases}

x+2 & \text{for }x \ne 2 \\[8px]
\text{undefined} & \text{for } x =2

\end{cases}\]
We’ve saw this equivalence earlier when we first graphed the function: $f(x)$ is a line with a hole at $x=2.$

Rewriting *f* as a linear function (undefined at one point) makes the proof much easier than dealing with a fraction.

Now, let’s consider the definition of the limit as applied to this function at $a=2$ and with $L = 4$: the requirement is that for every number $\epsilon \gt 0,$ there is a number $\delta \gt 0$ such that

\[\text{if } \overbrace{ \overbrace{2}^a \,-\, \delta \;\lt\; x \;\lt\; \overbrace{2}^a + \delta}^{\text{the input is within $\pm \delta$ of 2}} \quad \text{then} \quad \overbrace{\overbrace{4}^L \,-\, \epsilon \lt \overbrace{x+2}^{f(x)} \lt \overbrace{4}^L + \epsilon}^{\text{the output is within $\pm \epsilon$ of 4}}.\]
Let’s start by focusing on the second inequality, which tells us that we need the function’s *output* to lie in the range $4 \pm \epsilon$ (“as close to 4 as we want”):

\[4 \, – \, \epsilon \;\lt\; x+2 \;\lt \;4 + \epsilon \quad \triangleleft\]
We want to manipulate this requirement-statement (for reasons you’ll see) to isolate *x* in the middle, so let’s subtract 2 from each component of the inequality and then simplify:

\begin{align*}

(4 -\epsilon) -2 &\;\lt\; (x +2) – 2 \lt (4 + \epsilon) -2 \\[8px]
2 \,-\,\epsilon &\;\lt\; x \;\lt\; 2 + \epsilon \quad \blacktriangleleft

\end{align*}

Ah, that simple manipulation gave us a big insight: *if* we choose the input value of *x* to be within $\pm \epsilon$ of 2, then we are *guaranteed* that the output will lie within $\pm \epsilon$ of 4 (the initial equality $\triangleleft$). That guarantee comes from the tiny bit of algebraic manipulation we did.

That insight is exactly what the definition of the limit requires: Compare that last line $(\blacktriangleleft)$ to the “if … then” statement in the definition of the limit:

\[\text{if } 2 \,-\, \delta \lt x \lt 2 + \delta, \quad \text{then (we meet the output requirements and be within $\pm \epsilon$ of 4)} \]
We manipulated the line marked $\triangleleft$ to turn it into $\blacktriangleleft,$ with the *x *isolated in the middle of the inequality, precisely because we can now compare $\blacktriangleleft$ to the “if…” part of the definition statement and draw the necessary conclusion: a value of $\delta$ that “works” to meet the requirement that the function’s output be $4 \pm \epsilon$ is in this case $\delta = \epsilon.$ The fact that we have shown that there is a $\delta$ that meets the requirement for *any* $\epsilon$ we might choose completes the proof: the limit exists, and equals 4.$\;\cmark$

You actually discovered the relationship between $\delta$ and $\epsilon$ for this function in the Activity above: when we set $\epsilon = 1,$ you found that $\delta = 1$ (or smaller) produces “all output values fall within $4 \pm \epsilon.$” And when we set $\epsilon = 0.2,$ once you set $\delta = 0.2$ “all output values fall within $4 \pm \epsilon.$” (In case you’re wondering: the result that $\delta = \epsilon$ for this function is because it’s a linear function with slope equal to 1.) While the Desmos activity wasn’t a proof, the numerical results mirror the proof: you could always find (or imagine finding) a value for $\delta$ that “works” by simply setting $\delta \le \epsilon.$ As we noted above, the Desmos activity does not constitute a proof; by contrast, the one line of algebraic manipulation and accompanying discussion does.

Let’s tie this all together and first write out our conceptual definition of the limit with this more precise result now in mind:

\[\text{The limit of the function $f(x) =\dfrac{x^2 \, -\,4}{x\,-\,2}$ as $x$ approaches $\overbrace{2}^a$ is the number $\overbrace{4}^L$, written} \]
\[\lim_{x \to 2}f(x) = 4,\]
\[\text{because we can make $f(x)$ $\underbrace{\text{as close to}}_{\Large{\text{within } \pm

\epsilon} \text{ of}}$ $4$ as we want whenever $x$ is $\underbrace{\text{sufficiently close to}}_{\Large{\text{within }

\pm \delta = \epsilon \text{ of}}}$ $2.$ }\]

You can imagine that for some other function, you might find that that $\delta = \dfrac{4}{5}\epsilon,$ or $\delta = \sqrt{\epsilon},$ or whatever. Such a proof would similarly show that the limit exists and equals

That said, while our proof consists of essentially one line of algebra, you can imagine that for other functions the proofs are significantly more complicated and often require a clever insight of some form or another. If you’re interested in learning how to complete such proofs and developing your own ‘cleverness’ further, consider taking an Upper Division Mathematics course in “Analysis,” which covers the theoretical underpinnings of Calculus and is focused primarily on such proofs. Good stuff!

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For now, let’s keep our focus on epsilon, delta . . . and on the limit-value *L* itself. For our function above, it was pretty easy to see that the “height of the hole” in the graph is at *y* = 4, and so $\displaystyle{\lim_{x \to 2}\dfrac{x^2-4}{x-2} = 4 }.$ On the next screen, we’ll examine a function where we can’t identify the precise *y*-value of the hole in the graph as easily as we could here.

- Our conceptual definition of a limit is:

\[\text{The limit of the function $f(x)$ as $x$ approaches $a$, written} \] \[\lim_{x \to a} f(x) = L \] \[\text{is a number $L$ (if one exists)}\] \[\text{such that $f(x)$ is $\underbrace{\text{as close to}}_{\Large{\text{within } \pm

\epsilon} \text{ of}}$ $L$ as we want whenever $x$ is $\underbrace{\text{sufficiently close to}}_{\Large{\text{within }

\pm \delta \text{ of}}}$ $a.$ }\] - We have formalized this idea into the formal
**limit definition**:

Let*f*be a function defined at each point on an open interval containing*a*, except possibly at*a*itself.

Then a number*L*is the**limit of***f***at***a*if for every number $\epsilon \gt 0,$ there is a number $\delta \gt 0$ such that

\[\text{if} \quad a \,-\, \delta \lt x \lt a + \delta, \quad \text{ then } \quad L\, -\, \epsilon \lt f(x) \lt L + \epsilon. \] We write this limit as $\displaystyle{\lim_{x \to a}f(x) = L. }$

This screen got rather math-y. If there are pieces that aren’t clear to you, please ask on the Forum and we’ll do our best to clarify — and modify what’s above to help future students. 🙂

Of course we’ll keep exploring these ideas on upcoming screens too, so please don’t stop here. Often ideas that are initially quite abstract become familiar through use, and we’ll be using these ideas a lot. So let’s keep going!

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