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Math 220-1 Single-variable differential calculus: Kursobjekt

Aaron Greicius

Section 4 Limits: formal definition

As mentioned before, what makes Definition 2.6 less than rigorous is the use of the vague phrases “arbitrarily close” and “sufficiently close”. The “epsilon-delta” formulation given in Definition 4.1 is a mathematically precise way of of capturing these notions.

Definition 4.1. Limit (formal).

Suppose \(f\) is a function defined everywhere on an open interval containing the point \(a\in \R\text{,}\) except possibly at \(a\) itself. We say that the limit of \(f\) as \(x\) approaches \(a\) exists if there is a value \(L\) satisfying the following property: for all \(\epsilon > 0\text{,}\) there exists a \(\delta > 0\) such that if \(0< \abs{x-a}< \delta\text{,}\) then \(\abs{f(x)-L}< \epsilon\text{.}\) Using logical shorthand:
\begin{equation} \text{for all } \epsilon > 0, \text{ there is } \delta> 0, \text{ s.t., } 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon\text{.}\tag{4.1} \end{equation}

Remark 4.2. Anatomy of a definition.

Definition 4.1 can come off as forbiddingly technical. This is due to its combination of mathematical and logical details. We endeavor now to unpack and explicate some of these details. We will start with the mathematical nuts and bolts in the interior of the statement and work our way outward through the logical layers.
  • Absolute value.
    The absolute value expressions \(\abs{f(x)-L}\) and \(\abs{x-a}\) are measures of how close \(f(x)\) is to \(L\) and how close \(x\) is to \(a\text{.}\) To say \(\abs{f(x)-L}< \epsilon\) is to say \(f(x)\) is within a distance \(\epsilon\) of \(L\text{;}\) similarly \(\abs{x-a}< \delta\) says \(x\) is within a distance \(\delta\) of \(a\text{.}\)
  • Conditional.
    The conditional (or if-then) statement
    \begin{equation} 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon\tag{4.2} \end{equation}
    asserts that if \(x\) is within a distance \(\delta\) of \(a\) (but not equal to \(a\)), then \(f(x)\) is within a distance \(\epsilon\) of \(L\text{.}\)
  • Existential quantifier.
    How does adding the existential quantifier “there exists a \(\delta\)” change the meaning? Considering \(\epsilon\) to be fixed for the moment, the statement
    \begin{equation} \text{ there is } \delta> 0, \text{ s.t., } 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon\tag{4.3} \end{equation}
    asserts that for all \(x\) sufficiently close (but not equal) to \(a\text{,}\) \(f(x)\) is within a distance \(\epsilon\) of \(L\text{.}\) Indeed, it provides a sort of “safety distance” \(\delta\) and says that as long as \(x\) is within this safety distance of \(a\) (but not equal to \(a\)), then \(f(x)\) is within a distance \(\epsilon\) of \(L\text{.}\)
  • Universal quantifier.
    Lastly, consider what effect the universal quantifier “for all \(\epsilon\)” has. The full statement
    \begin{equation*} \text{for all } \epsilon > 0, \text{ there is } \delta > 0, \text{ s.t., } 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon \end{equation*}
    asserts that for any positive \(\epsilon\text{,}\) the value \(f(x)\) is within a distance \(\epsilon\) for all \(x\) sufficiently close to \(a\text{.}\) In particular, choosing \(\epsilon\) to be as small as you like, we have \(f(x)\) within that small distance \(\epsilon\) of \(L\) for all \(x\) sufficiently close to \(a\text{.}\) In other words, we can make \(f(x)\) arbitrarily close to \(L\) for all \(x\) sufficiently close (but not equal) to \(a\text{.}\)

Remark 4.3. Challenge and response.

It is useful to think of the process of verifying Definition 4.1 to establish a limit claim \(\lim_{x\to a}f(x)=L\) as a challenge and response type of procedure:
  • for each positive \(\epsilon\text{,}\) we are challenged to show that for \(x\) sufficiently close to \(a\text{,}\) \(\abs{f(x)-L}< \epsilon\text{;}\)
  • to meet this \(\epsilon\)-challenge we respond by providing a \(\delta\) for which \(0< \abs{x-a}< \delta\) implies \(\abs{f(x)-L}< \epsilon\text{.}\)
Note that the \(\delta\) we provide in response to a particular \(\epsilon\)-challenge will typically depend on \(\epsilon\text{:}\) and indeed, we typically see that the smaller the \(\epsilon\text{,}\) the smaller the \(\delta\text{.}\) This is illustrated in the examples below.
Our first example illustrating (4.4) treats a function of the form \(f(x)=ax+b\text{.}\) Such functions are called affine functions.

Example 4.5. Epsilon-delta: affine function.

Let \(f(x)=-4x+7\text{.}\) Verify that \(\lim\limits_{x\to 1}f(x)=3\) using the epsilon-delta definition of the limit.
Solution.
We follow Procedure 4.4.

Step 1.

Treating \(\epsilon\) as a fixed arbitrary constant we attempt to solve the inequality \(\abs{f(x)-3}< \epsilon\text{:}\)
\begin{align*} \abs{f(x)-3}< \epsilon\amp \iff \abs{-4x+7-3}< \epsilon \\ \amp \iff \abs{-4x+4}< \epsilon\\ \amp \iff \abs{-4(x-1)}< \epsilon\\ \amp \iff \abs{-4}\abs{x-1}< \epsilon \amp \text{(abs. val. prop.)}\\ \amp \iff 4\abs{x-1}< \epsilon\\ \amp \iff \abs{x-1}< \frac{\epsilon}{4} \text{(abs. val. prop.)}\\ \amp \iff -\frac{\epsilon}{4}< x-1< \frac{\epsilon}{4}\\ \amp \iff 1-\frac{\epsilon}{4}< x < 1+\frac{\epsilon}{4}\text{.} \end{align*}

Step 2.

By step 1, we see that the solution \(S\) to our starting inequality is the set of \(x\) satisfying \(1-\frac{\epsilon}{4}< x < 1+\frac{\epsilon}{4}\text{.}\) In other words, \(S=(1-\epsilon/4, 1+\epsilon/4)\text{.}\) Thus \(S\) itself is of the form \((1-\delta, 1+\delta)\text{,}\) where \(\delta=\frac{\epsilon}{4}\text{.}\)

Step 3.

Thus given any \(\epsilon > 0\text{,}\) setting \(\delta=\frac{\epsilon}{4}\text{,}\) we have \(\abs{f(x)-3}< \epsilon\) for all \(x\) satisfying \(\abs{x-1}< \delta=\epsilon/4\text{.}\) We conclude that \(\lim\limits_{x\to 1}f(x)=3\text{.}\)

Interactive example 4.1. Visualizing epsilon-delta proofs.

The Geogebra interactive below provides a means of visualizing the challenge-response nature of the epsilon-delta proof of a limit claim \(\lim\limits_{x\to a}f(x)=L\) in terms of the graph of \(f\text{.}\) (The window below is a bit narrow. Go to the Geogebra page
 1 
www.geogebra.org/m/degz8kgk
of the interactive for a larger rendition.) We elucidate how to parse this graphical representation.
  • The specific \(\epsilon\)-challenge is indicated by a horizontal band centered about \(y=L\text{.}\)
  • The \(\delta\)-response is indicated by a vertical band centered about \(x=a\text{.}\)
  • The game is, given a specific \(\epsilon\)-challenge, find an appropriate \(\delta\) such that when \(x\) is within \(\delta\) of \(a\text{,}\) the values \(f(x)\) are within \(\epsilon\) of \(L\text{.}\) Visually, this is accomplished when all points \((x,f(x))\) on the segment of the graph of \(f\) lying in the vertical band about \(x=a\) also lie within the horizontal band about \(y=L\text{.}\)
Figure 4.6. Visualizing epsilon-delta proofs. Made with Geogebra
 2 
www.geogebra.org/m/degz8kgk

Remark 4.7. Finding \(\delta\) such that \((a-\delta, a+\delta)\subseteq S\).

In the final step of applying Procedure 4.4 to a limit statment \(\lim\limits_{x\to a}f(x)=L\text{,}\) we usually have found an open interval \((c,d)\) containing \(a\text{,}\) and then must find a \(\delta\) such that
\begin{equation*} (a-\delta, a+\delta)\subseteq (c,d)\text{.} \end{equation*}
The following approach is useful in this regard. Observe that
\begin{align*} c \amp =a-(a-c)\\ d \amp =a+(d-a) \end{align*}
and thus \((c,d)=(a-\delta_1, a+\delta_2) \text{,}\) where
\begin{align*} \delta_1 \amp = a-c\\ \delta_2 \amp = d-a\text{,} \end{align*}
are the distances from \(a\) to the left and right endpoints of the interval.
Visualizing distances to endpoints of interval
Figure 4.8. Visualizing distances to endpoints of interval
Setting \(\delta=\min\{\delta_1,\delta_2\}\) as the minimum of these distances, we now have
\begin{equation*} (a-\delta, a+\delta)\subseteq (a-\delta_1,a+\delta_2)=(c,d)\text{,} \end{equation*}
as desired.
Depending on the example, we can sometimes easily determine which of \(\delta_1\) and \(\delta_2\) is the minimum, giving us an explicit expression for \(\delta\text{.}\) When things are too complicated, however, it is perfectly fine to set \(\delta=\min\{\delta_1,\delta_2\}\text{.}\)

Example 4.9. Epsilon-delta: radical function.

Let \(f(x)=\sqrt{x-2}\text{.}\) Verify that \(\lim\limits_{x\to 3}f(x)=1\) using the epsilon-delta definition of the limit.
Solution.
Following (4.4) we treat \(\epsilon\) as an arbitrary positive constant and attempt to solve the inequality
\begin{equation*} \abs{f(x)-1}< \epsilon\text{.} \end{equation*}
We have
\begin{align*} \abs{f(x)-1}< \epsilon\amp \iff -\epsilon < f(x)-1 < \epsilon\\ \amp \iff -\epsilon < \sqrt{x-2}-1 < \epsilon\\ \amp \iff 1-\epsilon < \sqrt{x-2}< \epsilon+1\text{.} \end{align*}
We are tempted now to square all terms in the above inequality to get rid of the radical, but recall that in order for the rule
\begin{equation*} a< b \iff a^2< b^2 \end{equation*}
to apply, we need \(a\) and \(b\) to both be nonnegative. (Example: \(-3< -1\text{,}\) but \(9\not\lt 1\text{.}\)) To deal with this, we will treat two separate cases: \(\epsilon\leq 1\) and \(\epsilon> 1\text{.}\)

Case: \(\epsilon \leq 1\).

When \(\epsilon \leq 1\text{,}\) all terms in the last inequality above are positive, and we may continue on to conclude
\begin{align*} \abs{f(x)-1}< \epsilon\amp \iff 1-\epsilon < \sqrt{x-2}< \epsilon+1\\ \amp \iff (1-\epsilon)^2 < x-2 < (\epsilon +1)^2\\ \amp \iff 2+(1-\epsilon)^2 < x < 2+(\epsilon +1)^2\\ \amp \iff 3-\epsilon(2-\epsilon)< x < 3+\epsilon(2 +\epsilon)\text{.} \end{align*}
Thus the set of solutions to our original inequality, assuming \(\epsilon\geq 1\text{,}\) is
\begin{equation*} S=(3-\epsilon(2-\epsilon), 3+\epsilon(2 +\epsilon))=(3-\delta_1, 3+\delta_2)\text{,} \end{equation*}
where
\begin{align*} \delta_1 \amp = \epsilon(2-\epsilon)\\ \delta_2 \amp = \epsilon(2 +\epsilon)\text{.} \end{align*}
Note that since \(0< \epsilon\leq 1\text{,}\) both \(\delta_1\) and \(\delta_2\) are positive. It is clear that \(3\in S\text{.}\) Setting \(\delta\) to be the minimum of \(\delta_1, \delta_2\text{,}\) guarantees that \((3-\delta, 3+\delta)\subseteq S\text{.}\) We conclude that for any positive \(\epsilon\leq 1\text{,}\) setting \(\delta=\min\{(\delta_1, \delta_2\}\text{,}\) we have \(\abs{f(x)-1}< \epsilon\) for all \(x\) satisfying \(\abs{x-3}< \delta\text{.}\)
Note: in fact, it is not difficult to see in this case that
\begin{equation*} \delta_1=\epsilon(2-\epsilon)\leq \delta_2=\epsilon(2+\epsilon) \end{equation*}
and so we could have more explicitly set \(\delta=\epsilon(2-\epsilon)\) in this case. Nonetheless, declaring \(\delta=\min\{\delta_1, \delta_2\}\) works just as well here, as well as in other examples.

Case: \(\epsilon> 1\).

Now assume \(\epsilon\geq 1\text{.}\) We need to find a \(\delta\) such that \(0< \abs{x-3}< \delta\) implies \(\abs{f(x)-1}< \epsilon\text{.}\) From the previous case (\(\epsilon\leq 1\)), we know there is a \(\delta\) such that \(0< \abs{x-3}< \delta\) implies \(\abs{f(x)-1}< 1\text{.}\) But since \(\epsilon > 1\text{,}\) we have
\begin{align*} 0< \abs{x-3}< \delta \amp\implies \abs{f(x)-1}< 1 \\ \amp \implies \abs{f(x)-1}< \epsilon\text{,} \end{align*}
as desired.
Our two cases, taken together, show that for all \(\epsilon > 0\text{,}\) there exists a \(\delta> 0\) such that
\begin{equation*} 0 <\abs{x-3}< \delta\implies \abs{f(x)-1}< \epsilon\text{.} \end{equation*}
Thus \(\lim\limits_{x\to 3}f(x)=1\text{.}\)
As the last example illustrates, as our function \(f\) becomes more complicated, solving the relevant inequality (4.4) can be a delicate and difficult affair. Mindful of this fact, instead of asking you to give a full epsilon-delta proof of a given limit claim, we will sometimes ask you to give a partial proof: namely we will give you one specific \(\epsilon\) as a challenge, and ask you to find a \(\delta\) that satisfies (4.1) for this particular \(\epsilon\text{.}\)

Example 4.10. Finding \(\delta\) for specific \(\epsilon\text{:}\) quadratic function.

Let \(f(x)=1-3x^2\text{.}\) It is a fact that \(\lim\limits_{x\to 1}f(x)=-2\text{.}\) Verify the epsilon-delta definition for this limit statement for the specific epsilon \(\epsilon=\frac{1}{2}\text{.}\)
Solution.
We wish to find \(\delta\) satisfying
\begin{equation*} 0< \abs{x-1}< \delta \implies \abs{f(x)-(-2)}< \frac{1}{2}\text{.} \end{equation*}
Proceeding as in Procedure 4.4, we first solve the inequality \(\abs{f(x)-(-2)}< \frac{1}{2}\text{:}\)
\begin{align*} \abs{f(x)-(-2)}< \frac{1}{2} \amp \iff \abs{(1-3x^2)+2}< \frac{1}{2} \\ \amp \iff \abs{3-3x^2}< \frac{1}{2}\\ \amp \iff \abs{(-3)(x^2-1)}< \frac{1}{2}\\ \amp \iff 3\abs{x^2-1}< \frac{1}{2}\\ \amp \iff \abs{x^2-1}< \frac{1}{6}\\ \amp\iff -\frac{1}{6}< x^2-1< \frac{1}{6}\\ \amp \iff\frac{5}{6}< x^2< \frac{7}{6}\\ \amp \iff\sqrt{\frac{5}{6}}< \abs{x}< \sqrt{\frac{7}{6}} \amp \text{(rad. ineq. prop.)}\\ \amp \iff -\sqrt{\frac{7}{6}}< x < -\sqrt{\frac{5}{6}} \text{ or } \sqrt{\frac{5}{6}}< x < \sqrt{\frac{7}{6}}\text{.} \end{align*}
Notice that in this case, the set of solutions \(S\) to the inequality is a union of two intervals:
\begin{equation*} S=\left(-\sqrt{\frac{7}{6}},-\sqrt{\frac{5}{6}}\right)\cup \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)\text{.} \end{equation*}
Notice further that our limit point \(1\) is an element of the second interval, and that we have
\begin{equation*} \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)=(1-\delta_1,1+\delta_2) \text{,} \end{equation*}
where
\begin{align*} \delta_1 \amp = 1-\sqrt{\frac{5}{6}} \\ \delta_2 \amp = \sqrt{\frac{7}{6}}-1\text{.} \end{align*}
Thus, setting \(\delta=\min\{\delta_1,\delta_2\}\text{,}\) we have
\begin{equation*} (1-\delta, 1+\delta)\subseteq \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)\subseteq S\text{.} \end{equation*}
It follows that
\begin{equation*} 0< \abs{x-1}< \delta \implies \abs{f(x)-(-2)}< \frac{1}{2}\text{,} \end{equation*}
as desired.
We gather here some inequality properties that came to light in the examples above. These are essential for solving inequalities, and thus for performing epsilon-delta proofs.
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