4.2. Functional Limits 115
116 Chapter 4. Functional Limits and Continuity
c−δ c c+δ L−
L L+
V(L)
Vδ(c)
Figure 4.4: Definition of Functional Limit.
ǫ-neighborhood Vǫ(L) centered at L, there is a point in the sequence—call it aN—after which all of the terms an fall in Vǫ(L). Eachǫ-neighborhood repre- sents a particular challenge, and each N is the respective response. For func- tional limit statements such as limx→cf(x) =L, the challenges are still made in the form of an arbitraryǫ-neighborhood around L, but the response this time is a δ-neighborhood centered atc.
Definition 4.2.1 (Functional Limit). Letf :A →R, and let c be a limit point of the domain A. We say that limx→cf(x) = L provided that, for all ǫ >0, there exists a δ >0 such that whenever 0<|x−c|< δ (and x∈A) it follows that |f(x)−L|< ǫ.
This is often referred to as the “ǫ–δversion” of the definition for functional limits. Recall that the statement
|f(x)−L|< ǫis equivalent to f(x)∈Vǫ(L).
Likewise, the statement
|x−c|< δ is satisfied if and only ifx∈Vδ(c).
The additional restriction 0<|x−c|is just an economical way of sayingx=c.
Recasting Definition 4.2.1 in terms of neighborhoods—just as we did for the definition of convergence of a sequence in Section 2.2—amounts to little more than a change of notation, but it does help emphasize the geometrical nature of what is happening (Fig.4.4).
Definition 4.2.1B (Functional Limit: Topological Version). Letc be a limit point of the domain of f : A → R. We say limx→cf(x) = L provided
4.2. Functional Limits 117 that, for everyǫ-neighborhoodVǫ(L) of L, there exists aδ-neighborhoodVδ(c) aroundcwith the property that for all x∈Vδ(c) different fromc (withx∈A) it follows that f(x)∈Vǫ(L).
The parenthetical reminder “(x∈ A)” present in both versions of the def- inition is included to ensure that x is an allowable input for the function in question. When no confusion is likely, we may omit this reminder with the understanding that the appearance off(x) carries with it the implicit assump- tion thatxis in the domain off. On a related note, there is no reason to discuss functional limits at isolated points of the domain. Thus, functional limits will only be considered asxtends toward a limit point of the function’s domain.
Example 4.2.2. (i) To familiarize ourselves with Definition4.2.1, let’s prove that if f(x) = 3x+ 1, then
xlim→2f(x) = 7.
Let ǫ > 0. Definition 4.2.1 requires that we produce a δ > 0 so that 0<|x−2|< δ leads to the conclusion|f(x)−7|< ǫ. Notice that
|f(x)−7|=|(3x+ 1)−7|=|3x−6|= 3|x−2|.
Thus, if we choose δ = ǫ/3, then 0 < |x−2| < δ implies |f(x)−7| <
3 (ǫ/3) =ǫ.
(ii) Let’s show
xlim→2g(x) = 4,
where g(x) =x2. Given an arbitraryǫ >0, our goal this time is to make
|g(x)−4| < ǫ by restricting |x−2| to be smaller than some carefully chosenδ. As in the previous problem, a little algebra reveals
|g(x)−4|=|x2−4|=|x+ 2||x−2|.
We can make |x−2|as small as we like, but we need an upper bound on
|x+2|in order to know how small to chooseδ. The presence of the variable xcauses some initial confusion, but keep in mind that we are discussing the limit as xapproaches 2. If we agree that ourδ-neighborhood around c= 2 must have radius no bigger thanδ= 1, then we get the upper bound
|x+ 2| ≤ |3 + 2|= 5 for allx∈Vδ(c).
Now, chooseδ= min{1, ǫ/5}. If 0<|x−2|< δ, then it follows that
|x2−4|=|x+ 2||x−2|<(5)ǫ 5 =ǫ, and the limit is proved.
118 Chapter 4. Functional Limits and Continuity
Sequential Criterion for Functional Limits
We worked very hard in Chapter 2 to derive an impressive list of proper- ties enjoyed by sequential limits. In particular, the Algebraic Limit Theorem (Theorem2.3.3) and the Order Limit Theorem (Theorem2.3.4) proved invalu- able in a large number of the arguments that followed. Not surprisingly, we are going to need analogous statements for functional limits. Although it is not difficult to generate independent proofs for these statements, all of them will follow quite naturally from their sequential analogs once we derive the sequen- tial criterion for functional limits motivated in the opening discussion of this chapter.
Theorem 4.2.3 (Sequential Criterion for Functional Limits). Given a function f :A→R and a limit pointc of A, the following two statements are equivalent:
(i) lim
x→cf(x) =L.
(ii) For all sequences(xn)⊆Asatisfyingxn=c and(xn)→c, it follows that f(xn)→L.
Proof. (⇒) Let’s first assume that limx→cf(x) =L. To prove (ii), we consider an arbitrary sequence (xn), which converges tocand satisfiesxn =c. Our goal is to show that the image sequence f(xn) converges to L. This is most easily seen using the topological formulation of the definition.
Let ǫ > 0. Because we are assuming (i), Definition 4.2.1B implies that there exists Vδ(c) with the property that allx∈Vδ(c) different fromc satisfy f(x)∈Vǫ(L). All we need to do then is argue that our particular sequence (xn) is eventually in Vδ(c). But we are assuming that (xn)→c. This implies that there exists a pointxN after whichxn ∈Vδ(c). It follows that n≥N implies f(xn)∈Vǫ(L), as desired.
(⇐) For this implication we give a contrapositive proof, which is essentially a proof by contradiction. Thus, we assume that statement (ii) is true, and carefully negate statement (i). To say that
xlim→cf(x)=L
means that there exists at least one particularǫ0>0 for which noδis a suitable response. In other words, no matter whatδ >0 we try, there will always be at least one point
x∈Vδ(c) with x=c for which f(x)∈/ Vǫ0(L).
Now considerδn= 1/n. From the preceding discussion, it follows that for each n∈N we may pick anxn ∈Vδn(c) with xn =c andf(xn)∈/ Vǫ0(L). But now notice that the result of this is a sequence (xn) → c with xn = c, where the image sequencef(xn) certainly doesnot converge toL.
Because this contradicts (ii), which we are assuming is true for this argument, we may conclude that (i) must also hold.
4.2. Functional Limits 119 Theorem 4.2.3has several useful corollaries. In addition to the previously advertised benefit of granting us some short proofs of statements about how functional limits interact with algebraic combinations of functions, we also get an economical way of establishing that certain limits do not exist.
Corollary 4.2.4(Algebraic Limit Theorem for Functional Limits). Let f andgbe functions defined on a domainA⊆R, and assumelimx→cf(x) =L andlimx→cg(x) =M for some limit point c ofA. Then,
(i) lim
x→ckf(x) =kL for allk∈R, (ii) lim
x→c[f(x) +g(x)] =L+M, (iii) lim
x→c[f(x)g(x)] =LM, and (iv) lim
x→cf(x)/g(x) =L/M, provided M = 0.
Proof. These follow from Theorem 4.2.3and the Algebraic Limit Theorem for sequences. The details are requested in Exercise4.2.1.
Corollary 4.2.5(Divergence Criterion for Functional Limits). Let f be a function defined on A, and let c be a limit point of A. If there exist two sequences(xn) and(yn)inA with xn=c andyn =c and
limxn= limyn=c but limf(xn)= limf(yn),
then we can conclude that the functional limit limx→cf(x)does not exist.
Example 4.2.6. Assuming the familiar properties of the sine function, let’s show that limx→0sin(1/x) does not exist (Fig.4.5).
If xn = 1/2nπ and yn = 1/(2nπ+π/2), then lim(xn) = lim(yn) = 0.
However, sin(1/xn) = 0 for alln∈N while sin(1/yn) = 1. Thus, lim sin(1/xn)= lim sin(1/yn),
so by Corollary4.2.5, limx→0sin(1/x) does not exist.
Figure 4.5: The function sin(1/x) near zero.
120 Chapter 4. Functional Limits and Continuity
Exercises
Exercise 4.2.1. (a) Supply the details for how Corollary4.2.4part (ii) follows from the Sequential Criterion for Functional Limits in Theorem4.2.3and the Algebraic Limit Theorem for sequences proved in Chapter2.
(b) Now, write another proof of Corollary4.2.4part (ii) directly from Defini- tion4.2.1without using the sequential criterion in Theorem4.2.3.
(c) Repeat (a) and (b) for Corollary4.2.4part (iii).
Exercise 4.2.2. For each stated limit, find the largest possibleδ-neighborhood that is a proper response to the givenǫchallenge.
(a) limx→3(5x−6) = 9, whereǫ= 1.
(b) limx→4√x= 2, whereǫ= 1.
(c) limx→π[[x]] = 3, whereǫ = 1. (The function [[x]] returns the greatest integer less than or equal tox.)
(d) limx→π[[x]] = 3, whereǫ=.01.
Exercise 4.2.3. Review the definition of Thomae’s function t(x) from Section4.1.
(a) Construct three different sequences (xn), (yn), and (zn), each of which converges to 1 without using the number 1 as a term in the sequence.
(b) Now, compute limt(xn), limt(yn), and limt(zn).
(c) Make an educated conjecture for limx→1t(x), and use Definition4.2.1B to verify the claim. (Givenǫ >0, consider the set of points{x∈R:t(x)≥ǫ}. Argue that all the points in this set are isolated.)
Exercise 4.2.4. Consider the reasonable but erroneous claim that
xlim→101/[[x]] = 1/10.
(a) Find the largest δ that represents a proper response to the challenge of ǫ= 1/2.
(b) Find the largestδthat represents a proper response toǫ= 1/50.
(c) Find the largest ǫ challenge for which there is no suitable δ response possible.
4.2. Functional Limits 121 Exercise 4.2.5. Use Definition4.2.1to supply a proper proof for the following limit statements.
(a) limx→2(3x+ 4) = 10.
(b) limx→0x3= 0.
(c) limx→2(x2+x−1) = 5.
(d) limx→31/x= 1/3.
Exercise 4.2.6. Decide if the following claims are true or false, and give short justifications for each conclusion.
(a) If a particularδhas been constructed as a suitable response to a particular ǫ challenge, then any smaller positiveδwill also suffice.
(b) If limx→af(x) =Landahappens to be in the domain off, thenL=f(a).
(c) If limx→af(x) =L, then limx→a3[f(x)−2]2= 3(L−2)2.
(d) If limx→af(x) = 0, then limx→af(x)g(x) = 0 for any function g (with domain equal to the domain off.)
Exercise 4.2.7. Letg:A→Rand assume thatf is a bounded function onA in the sense that there exists M >0 satisfying|f(x)| ≤M for allx∈A.
Show that if limx→cg(x) = 0, then limx→cg(x)f(x) = 0 as well.
Exercise 4.2.8. Compute each limit or state that it does not exist. Use the tools developed in this section to justify each conclusion.
(a) limx→2|x−2| x−2
(b) limx→7/4|x−2| x−2
(c) limx→0(−1)[[1/x]]
(d) limx→0√3x(−1)[[1/x]]
Exercise 4.2.9 (Infinite Limits). The statement limx→01/x2=∞certainly makes intuitive sense. To construct a rigorous definition in the challenge–
response style of Definition 4.2.1 for an infinite limit statement of this form, we replace the (arbitrarily small) ǫ > 0 challenge with an (arbitrarily large) M >0 challenge:
Definition: limx→cf(x) =∞means that for all M >0 we can find aδ >0 such that whenever 0<|x−c|< δ, it follows thatf(x)> M.
(a) Show limx→01/x2=∞in the sense described in the previous definition.
(b) Now, construct a definition for the statement limx→∞f(x) = L. Show limx→∞1/x= 0.
122 Chapter 4. Functional Limits and Continuity (c) What would a rigorous definition for limx→∞f(x) =∞ look like? Give
an example of such a limit.
Exercise 4.2.10 (Right and Left Limits). Introductory calculus courses typically refer to the right-hand limit of a function as the limit obtained by
“lettingxapproachafrom the right-hand side.”
(a) Give a proper definition in the style of Definition4.2.1for the right-hand and left-hand limit statements:
xlim→a+f(x) =L and lim
x→a−f(x) =M.
(b) Prove that limx→af(x) = L if and only if both the right and left-hand limits equalL.
Exercise 4.2.11 (Squeeze Theorem). Letf, g,andhsatisfyf(x)≤g(x)≤ h(x) for allxin some common domainA. If limx→cf(x) =Land limx→ch(x) = Lat some limit pointcofA, show limx→cg(x) =L as well.