Exercise 4.2.3. Review the definition of Thomae's function t(x) from Section 4.1. (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 lim t(xn), lim t(yn), and lim t(zn). (c) Make an educated conjecture for lim-1 t(x), and use Definition 4.2.1B to verify the claim. (Given e > 0, consider the set of points {x = R: t(x) ≥ e}. Argue that all the points in this set are isolated.) 114 1 Chapter 4. Functional Limits and Continuity 4.2. Func converges sequence 112 112 112 1 312 22 Figure 4.3: THOMAE'S FUNCTION, t(x). 2 If c = 0, however, then these two limits are both equal to h(0) = 0. In fact, it appears as though no matter how we construct a sequence (zn) converging to zero, it will always be the case that lim h (zn) = 0. This observation goes to the heart of what we want functional limits to entail. To assert that lim h(x) = L x c But noti case, the and is The clo its den the biz discont Is t does t should imply that tob ismo no dou (Zn) → c. h(zn) L for all sequences For reasons not yet apparent, it is beneficial to fashion the definition for func- tional limits in terms of neighborhoods constructed around c and L. We will quickly see, however, that this topological formulation is equivalent to the sequential characterization we have arrived at here. To this point, we have been discussing continuity of a function at a particular point in its domain. This is a significant departure from thinking of continuous functions as curves that can be drawn without lifting the pen from the paper, and it leads to some fascinating questions. In 1875, K.J. Thomae discovered the function to be arbitr that i funct What are s funct migh on F 4.2 Con gai 391 lo sail smo the 1 if x = 0 ICT ale dans to tha t(x) = 1/n 0 if x = m/ne Q\{0} is in lowest terms with n > 0W if x Q. sec of If cЄ Q, then t(c) > 0. Because the set of irrationals is dense in R, we can find a sequence (yn) in I converging to c. The result is that lim t(yn) = 0 t(c), as 11 Justeftiboa: ow (a) bas elanoltar los+ (a) and Thomae's function (Fig. 4.3) fails to be continuous at any rational point. The twist comes when we try this argument on some irrational point in the √2. All irrational values get mapped to zero by t, so the natural thing would be to consider a sequence (xn) of rational numbers that domain such as c = is cl tl

Exercise 4.2.3. Review the definition of Thomae's function t(x) from Section 4.1. (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 lim t(xn), lim t(yn), and lim t(zn). (c) Make an educated conjecture for lim-1 t(x), and use Definition 4.2.1B to verify the claim. (Given e > 0, consider the set of points {x = R: t(x) ≥ e}. Argue that all the points in this set are isolated.) 114 1 Chapter 4. Functional Limits and Continuity 4.2. Func converges sequence 112 112 112 1 312 22 Figure 4.3: THOMAE'S FUNCTION, t(x). 2 If c = 0, however, then these two limits are both equal to h(0) = 0. In fact, it appears as though no matter how we construct a sequence (zn) converging to zero, it will always be the case that lim h (zn) = 0. This observation goes to the heart of what we want functional limits to entail. To assert that lim h(x) = L x c But noti case, the and is The clo its den the biz discont Is t does t should imply that tob ismo no dou (Zn) → c. h(zn) L for all sequences For reasons not yet apparent, it is beneficial to fashion the definition for func- tional limits in terms of neighborhoods constructed around c and L. We will quickly see, however, that this topological formulation is equivalent to the sequential characterization we have arrived at here. To this point, we have been discussing continuity of a function at a particular point in its domain. This is a significant departure from thinking of continuous functions as curves that can be drawn without lifting the pen from the paper, and it leads to some fascinating questions. In 1875, K.J. Thomae discovered the function to be arbitr that i funct What are s funct migh on F 4.2 Con gai 391 lo sail smo the 1 if x = 0 ICT ale dans to tha t(x) = 1/n 0 if x = m/ne Q\{0} is in lowest terms with n > 0W if x Q. sec of If cЄ Q, then t(c) > 0. Because the set of irrationals is dense in R, we can find a sequence (yn) in I converging to c. The result is that lim t(yn) = 0 t(c), as 11 Justeftiboa: ow (a) bas elanoltar los+ (a) and Thomae's function (Fig. 4.3) fails to be continuous at any rational point. The twist comes when we try this argument on some irrational point in the √2. All irrational values get mapped to zero by t, so the natural thing would be to consider a sequence (xn) of rational numbers that domain such as c = is cl tl

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section: Chapter Questions

Problem 35RE

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