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1 A Topological Definition of Limits for Use in Elementary Calculus Charles L. Cooper, Phd Michael S. McClendon, PhD Department of Mathematics & Statistics University of Central Oklahoma 00 N. University Dr. Edmond, OK 700 Introduction. When Sir Isaac Newton invented the calculus as a vehicle for his development of the physics of motion, he more or less informally utilized what we now realize is the fundamental concept in the calculus: limits. In the mid- 9 th century the idea was formalized by Cauchy and Weierstrass. The definition they developed is, of course, what we now commonly call the δ ε definition in the case of limits as the variable approaches a fixed number and the limit itself is a finite number. This definition, as given in most all calculus texts, is: Definition: lim f (x) = L iff for all ε > 0, there exists a number δ > 0 such that if 0 < x a < δ, then f (x) l < ε. Geometrically what this definition requires is that if the values of f (x) are to be in an interval centered about L in the range of f (x) then one can always find an interval centered about x = a within the domain that guarantees this condition. More specifically given any ε > 0 and the open interval L ε, L + ε ( ) about L in the ( ) so range, lim f (x) = L if and only if one can always find an open interval a δ, a + δ that any value of the variable inside this interval not equal to a will give a value for the function inside the interval L ε, L + ε A Topologically Based Definition for Limits. Topologists have recognized that the demand for intervals centered about the limit and about the number approached within domain need not be this rigid. The more general topological viewpoint is that we only need to require that the relevant quantities be inside open intervals, regardless of whether those intervals are centered or not about those quantities. Thus we can reformulate this definition topologically as:

2 Definition (Topological): lim f (x) = L iff for all open intervals, p, q L within the range, there is an open interval, c, d such that for all x c, d ( ), containing ( ), about x = a within the domain ( ) such that x a, f (x) ( p, q). Using this definition, the algebra involved in verifying limits, in most cases, becomes much simpler. It is in the articulation of the algebra involved in utilizing the δ ε definition to verify limit statements that students often get lost in the forest of the algebraic manipulations required to meet the δ ε criteria of intervals centered about the relevant quantities. Example. Verify that x + 5 =. lim x Let ( p, q) be any interval containing on the y- axis. Let c, d be an ( ) = p 5, q 5 interval on the x- axis. p Since p < < q, < < q. Subtracting 5 from this latter inequality yields: Thus belongs to the interval c, d p 5 < < q 5. ( ) = p 5, q 5 Now suppose that x is any number different from in c, d. ( ) = p 5, q 5 Then p 5 < x < q 5. Multiplying by yields p 5 < x < q 5. Adding 5 then gives p < x 5 < q. Therefore, f (x) = x + 5 will belong to the interval p, q ( ), f (x) will necessarily be a member of the ( p, q). Hence by the topological definition, we conclude that lim Thus if x is any number in c, d interval. x x + 5 =. Notice that for linear functions the articulation of this definition will be about as simple as it would by using the δ ε definition. This is understandable since for a linear function if you start with a interval I centered about the limit on the y- axis, this inversely works back to an interval centered about the number approached

3 on the x- axis as its pre- image. Where the topological definition shows its true colors is with limits of non- linear functions. This is shown in the next example. Example. Verify that lim x x = 8. Start with an arbitrary interval ( p, q) about 8 on the y- axis. Since p < 8 < q, taking cube roots implies x- axis to be: p < < q. This suggests we take the interval about on the ( c, d) = p, q If x p, ( q ) and x, then p < x < q. Thus by cubing this latter inequality we see that p < x < q and so f (x) p, q definition. Thus lim x = 8. (See the figure shown below.) x ( ), satisfying the condition in the topological Most calculus instructors would never present the verification of the limit in the last example by use of the δ ε definition. The analysis of finding the δ given the arbitrary ε alone is an algebraic challenge, much less the verification that the choice of δ works. All that is required to use this topologically based definition is to determine the inverse image of the interval ( p, q). Furthermore, the verification that the choice of interval c, d ( ) works is then almost automatic from this. The limit theorems given in elementary calculus can be proven with the same amount of work, if not less, as they are using the δ ε definition. This is shown for the Sum Rule and the Constant Multiple Rule. Theorem. The Sum Rule for Limits: If f (x) and g(x) are functions and lim f (x) = L and lim g(x) = M then:

4 Proof. To verify this limit, let p, q lim f (x) + g(x) [ ] = lim f (x) + lim g(x) = L + M. ( ) be any interval on the y- axis that contains ( ) containing a such if x ( c, d), x a, First, let e = ( L + M ) p and let e = q ( L + M ). Note that ( ) L + M. We must find an interval c, d then f (x) + g(x) p, q e and e are both positive and they measure the distance the interval p, q extends either side of the number L + M. Now consider the interval p, q ( ) = L e, L + e on the y- axis. Since lim f (x) = L, the topological definition guarantees the existence of an interval c, d for all x ( c, d ), x a, f (x) ( p, q ). Similarly, consider the interval p, q ( ) = M e ( ) such that, M + e. Since lim g(x) = M, the definition gives an interval ( c, d ) such that if x ( c, d ), x a, then g(x) ( p, q ). Let ( c, d) = ( c, d ) ( c, d ). If x (c, d), x a, then x (c, d ) and x (c, d ). Since x (c, d ), f (x) belongs to p, q ( ) = L e, L + e, that is, L e < f (x) < L + e. Since x (c, d ), then g(x) p, q is, M e < g(x) < M + e. Adding these inequalities: L e Simplifying this we get: ( ) = M e + M e < f (x) + g(x) < L + e + M + e ( L + M ) e < f (x) + g(x) < ( L + M ) + e ( L + M ) ((L + M ) p) < f (x) + g(x) < ( L + M ) + q ( L + M ) p < f (x) + g(x) < q, M + e, that ( ) Hence if x (c, d), x a, f (x) + g(x) (p, q). Thus we have satisfied the criteria of the topological definition of limits and therefore lim f (x) + g(x) [ ] = L + M. 4

5 Theorem. The Constant Multiple Rule. Suppose lim f (x) = L. Then lim k f (x) = k lim f (x) = k L. Proof. Suppose ( p, q) is an interval containing k L. Then p < k L < q. Case : k > 0. Dividing both sides of the above inequality by k, p k < L < q k. Thus L is in the open interval p k, q k. Since lim f (x) = L the topological definition of limits guarantees the existence of an interval c, d ( ) about x = a so that for all values of x ( c, d ), x a, f (x) p k, q k Multiplying by k, p < k L < q and thus k f (x) ( p, q) as required by the definition., hence p k < f (x) < q k. Case : k < 0. Dividing both sides of the inequality by k, p k > L > q k, or q k < L < p k. Thus L is in the open interval q k, p k. Since lim f (x) = L the ( ) topological definition of limits guarantees the existence of an interval c, d about x = a so that for all values of x c, d ( ), x a, f (x) q k, p k, hence q k < f (x) < p. Multiplying by k < 0, q > k L > p, or, p < k L < q and thus k k f (x) p, q ( ) as required by the definition. Case : k = 0. Observe that k f (x) = 0 f (x) = 0 for all values of x in the ( ) about 0 L = 0, let ( c, d) be any ( ), k f (x) = 0 f (x) = 0, which is clearly ( ) as required by the definition. domain of f. Given any interval p, q interval about x = a. Then if x c, d in the interval p, q Hence, in all cases the topological definition implies that lim k f (x) = k L. Observe that the argument in Case can be modified to produce an argument of the limit property that the limit of a constant is that constant, lim k = k. It is left to the reader to formulate the precise argument. This is just a sampling of the limit theorems discussed in elementary calculus as viewed from the topological standpoint but the others are proved similarly. 5

6 The Equivalence of the Definitions. One can easily show that both definitions are equivalent. To do so, suppose lim f (x) = L and that the topological criterion is satisfied. Then given ε > 0 the intervals p, q ( ) = ( L ε, L + ε ) and the topological condition for existence of the interval ( c, d) about x = a implies we can let δ = min{ a c, d a}. Then ( a δ, a + δ ) ( c, d) and thus for any value of x ( a δ, a + δ ), x ( c, d) and thus the topological definition then implies f (x) ( p, q) = ( L ε, L + ε ). Hence the δ ε definition is satisfied. Conversely, Suppose lim f (x) = L and that the δ ε criterion is satisfied. Then given p, q ( ) an interval in the range containing L let ε = min{ L p, q L} and ( ) ( p, q). Thus given this ε - interval about L, the δ ε ( ) then ( ) ( p, q). Thus if we let the interval ( c, d) be ( a δ, a + δ ), the observe that L ε, L + ε criterion implies there is a number δ > 0 such that if x a δ, a + δ f (x) L ε, L + ε topological criterion is satisfied. Hence the two definitions are in fact equivalent. Using Topological Criteria to Define Other Types of Limits. A topological definition can be utilized with respect to other types of limits as well. Several examples of these are given below. Definitions: a. lim f (x) = L iff for all open intervals, p,q + an open interval, a,d ( ), containing L within the range, there is ( ), within the domain such that for all x a,d b. lim f (x) = iff for all open intervals, p, about x = a within the domain, ( c,d), such that for all x c,d c. lim f (x) = L iff for all open intervals, p,q x open interval, ( c, ), in the domain such that if x c, d. lim f (x) = iff for all open intervals,,q x open interval, ( c, ), within the domain such that if x c, ( ), f (x) p,q ( ), within the range, there is an interval ( ), x a, f (x) ( p, ). ( ), containing L in the range, there is an ( ), then f (x) ( p,q). ( ), within the range, there is an interval ( ) then f (x) (,q). The reader can easily formulate the topological definitions of the other types of limits not represented here. Below we give a few examples of the uses of these definitions. 6

7 Example. Show that lim x 0 + x = 0. Let ( p,q) be an open interval about 0 on the y- axis and consider the interval ( 0, q ) on the x- axis. If x ( 0, q ), then 0 < x < q. The square root function is an increasing function and this interval lies in the domain of f (x) = x, therefore 0 < x < q and thus 0 < x < q = q, since q > 0. Since p < 0, p < f (x) < q and so f (x) p,q Therefore, lim x 0 + x = 0 by part (a) of the above definition. Example 4. Show that lim x 0 x =. Let ( p, ) be an interval on the y- axis. Observe that since the function f (x) = has positive x values we can assume that p > 0. Consider the interval p, p on the x- axis and note that 0 p, p. Thus if x p, p, x 0, then p < x < p. In the case that 0 < x < p, squaring we get, 0 < x < p and taking the reciprocal of the right side of this inequality we get x > p. Similarly, in the case p < x < 0, squaring we also get, x >. Once again taking the reciprocal of this last inequality we now see that p x > p. In either case if x p, p the definition given above, lim x 0 x =., then f (x) ( p, ). Hence by part (b) of 7

8 Example 5. Show lim( ) x =. x Let (, q) be an interval on the y- axis. Since for all positive values of x, f (x) = x will be negative, we can assume that q < 0. Consider the interval q, ( ) on the x- axis and observe that since q < 0, q > 0. Then if x q, x < q ( ), then x > q ( ) = q. Thus, whenever x q (d) of the definition, lim x x ( ) =. ( ) = q and hence (, ), f (x) = x,q. Therefore x > q ( ) and so by part The one point that is clear about the use of the topologically based definition of limits is that the algebraic manipulations required to utilize it are much simpler in most cases. Thus if used in an elementary calculus class, the students can concentrate more on what the definition means in terms of formalizing the notions of nearness and not get lost in the algebraic manipulations that are often required to utilize the δ ε formulation. To reiterate, forcing the intervals to be centered about the limit and about the number approached in the domain as required by the δ ε definition needlessly complicates both the analysis of the particular situation being addressed as well as the algebra to accomplish the successful application of the definition. Having said that, there are a couple of drawbacks to the topological definition and we would be remiss to not point them out. In some cases, as is usually true when using more general conceptualizations (as topology is in relation to analysis), some arguments may devolve into a bunch of case- by- case arguments. For example, the validation of the multiplication rule for limits by use of the topological definition, while not complicated, involves five or six cases that need to be looked at individually. The proof of Theorem is also an example of this, although the δ ε derivation of the result of the theorem also requires the same cases. Additionally, it is clearly necessary that students understand some of the more elementary set theoretic ideas such as set intersection. Given the state of high school algebra, college algebra and pre- calculus courses today, many students have not been introduced to these ideas. Thus a very short introduction to set theory may need to be given to your students prior to engaging in the utilization of a topological definition for the limit concept. 8

9 Nonetheless we feel that the advantages of the use of a topologically based definition of limits in the exposition of elementary calculus far outweigh the disadvantages. The authors have been utilizing the topological definition in their elementary calculus courses for several years now and the results of that use are better student comprehension of the formal limit idea as well as an enhanced ability of students to formulate and successfully perform a valid limit argument. The authors are currently writing a calculus text that utilizes the topological definition in its chapter on limits. REFERENCES. () Howard Anton, Irl Bivens, and Stephen Davis, Calculus, 9 th edition, 009, John Wiley and Sons, Inc. () Robert Bartle, Donald Sherbert, Introduction to Real Analysis, 4 th edition, 0, John Wiley and Sons, Inc. () James Dugundji, Toplogy, 6 th edition, 966, William C. Brown Publishing. (4) James Munkres, Topology, nd edition, 000, Pearson Publishing. (5) S. Salas, G. Etgen, E. Hille, Calculus: One and Several Variables, 0 th edition, 006, John Wiley and Sons, Inc. 9