Section 1.5 Formal definitions of limits

Transcription

1 Section.5 Formal definitions of limits (3/908) Overview: The definitions of the various tpes of limits in previous sections involve phrases such as arbitraril close, sufficientl close, arbitraril large, and sufficientl large. For instance, according to Definition of Section., lim f() = L with a number L if f() is arbitraril close to L for all a a sufficientl close to a. These qualitative formulations of the definitions are all that are needed in most of calculus. More quantitative formulations of the definitions are required, however, in dealing with difficult eamples and in proofs. In these formal definitions, phrases such as arbitraril close and sufficientl close are made precise b using inequalities. In this section we discuss the formal definition of two-sided finite limits lim f() = L. Formal definitions of lim f() = L and lim f() = ± are presented in a the contet of infinite sequences in Section.. Other definitions are similar and are studied in advanced courses. Topics: The ɛ-definition of a finite two-sided limit Using approimate inequalities Two short proofs Using graphs with finite limits The ɛ-definition of a finite two-sided limit To convert Definition in Section. of lim a f() = L into a formal definition, we replace the phrase f() is arbitraril close to L with f() L < ɛ for an arbitraril small positive number ɛ, and we replace the phrase for all a sufficientl close to a with for all with 0 < a < for a sufficientl small positive number. We obtain the following: Definition (Finite two-sided limits) Suppose that = f() is defined on open intervals (b, a) and (a,c) on both sides of a. Then lim f() = L with a number L if and onl if for ever positive ɛ, there is a a positive such that f() L < ɛ for all with 0 < a <. () ɛ and are the Greek letters epsilon and delta. 5

2 p. 5 (3/908) Section.5, Formal definitions of limits Statement () means that the portions of the graph = f() for a < < a and for a < < a+ are between the lines = L + ɛ and = L ɛ as is illustrated in Figure. = f() f() L < ɛ = L + ɛ L = L ɛ ɛ ɛ for 0 < a < FIGURE a a a + To see the geometric meaning of this Definition, imagine that the positive parameter ɛ in Figure decreases toward zero, so that the lines = L ɛ and = L+ɛ approach the line = L. If the definition is satisfied, then for each ɛ, there is a such that the portion of the graph for a < < 0 on the left of a and the portion for a < < a + on the right of a lie between those lines. Since ɛ can be arbitraril small, this causes the value f() of the function to approach L as approaches a from both sides. Notice that if a has been found to satisf Definition for a particular ɛ, then an smaller positive would also work for that ɛ and the same would serve for an larger ɛ. We can often show that Definition is satisfied b giving a rule that defines a suitable for ever positive ɛ, as in the following eample. Eample Figure shows the graph of the function = F() defined b F() = 4 for. Use Definition to prove that lim F() = 8. = F() = F() ɛ 8 8 ɛ ɛ 3 lim F() = 8 F() 8 < ɛ for 0 < < FIGURE FIGURE 3

3 Section.5, Formal definitions of limits p. 53 (3/908) We need to show that for each positive number ɛ there is a positive number such that F() 8 < ɛ for all with 0 < < () (Figure 3). We do this in three steps. We begin with preliminar calculations. Net, we use these calculations to define a rule that provides a suitable for each ɛ. Then we verif that the rule works. (Preliminar calculations) For, F() equals 4 and F() 8 < ɛ 4 8 < ɛ 4 < ɛ < 4 ɛ. (3) Eample (The rule) Given ɛ > 0, set = 4ɛ so that the last epression in (3) is the same as the last epression in (). (Verification) With this definition of, statement () is valid because if satisfies 0 < <, then it satisfies the last inequalit in (3) and therefore F() 8 < ɛ. Show, using Definition, that lim 0 3 = 0. Since 3 0 = 3 and 0 =, we need a rule defining a positive for each positive ɛ (Figure 4) such that 3 < ɛ for all with 0 < <. (4) (Preliminar calculations) We solve the first inequalit in (4) for b taking cube roots of both sides: 3 < ɛ < ɛ /3 (5) (The rule) Given ɛ > 0, define = ɛ /3 so that the last epressions in (4) and (5) are the same. (Verification) Statement (4) holds with this definition of because if 0 < <, then < ɛ /3 and 3 < ɛ b (5). = 3 = ɛ 3 < ɛ for 0 < < = ɛ FIGURE 4

4 p. 54 (3/908) Section.5, Formal definitions of limits Eample 3 Use Definition to prove that the statement lim 3 = is false. 0 We need to show that there is a positive ɛ such that there is no positive with the propert that 3 < ɛ for all with 0 < <. Since the limit is actuall 0, we can use an ɛ <. We will use ɛ =. There is no positive such that 3 < for all with 0 < < since 0 < 3 < and hence 3 > for 0 < <. Using approimate inequalities In Eamples and we found formulas for to satisf the definition of lim a f() = L b solving the inequalit f() L < ɛ eactl. In man cases, this inequalit is relativel difficult to solve and we use, instead, the following procedure in which this inequalit is replaced b one that is easier to deal with. (Preliminar calculations) Use factoring or other techniques to write f() L in the form f() L = g()( a) with another function = g(). Then pick a specific positive number ( = will often work) such that g() is bounded for 0 < a < and pick a positive constant M such that g() M for 0 < a <. Then f() L M a for 0 < a <. (6) (The rule) The positive number must be defined so that f() L < ɛ for all with 0 < a <. (7) Accordingl given ɛ > 0, let be the smaller of and ɛ/m. (Verification) Statement (7) is valid with this definition of because if 0 < a <, then f() L M a < since, and then f() L < ɛ since ɛ/m. Eample 4 Show that b Definition, lim = 9. 3 We need a rule that gives a positive for ever positive ɛ such that 9 < ɛ for 0 < 3 <. (8) (Preliminar calculations) We start b factoring 9 in the form Since + 3 is bounded in an finite interval, we set =. Then < < 4 for 3 < and, consequentl, + 3 < 7 for 3 <. Therefore, 9 = for 3 <. (9) (The rule) Inequalit (9) shows that we can have 9 < ɛ, as in (8), b requiring 3 < 7 ɛ so that 7 3 < ɛ and b requiring 3 <, so that (8) holds. Given ɛ > 0, we define to be the smaller of and 7 ɛ. (Verification) Statement (8) holds with this definition of because if 0 < 3 <, then 3 < so that b (9) and then < 7 7 ( 7 ɛ ) = ɛ. The procedure in Eample 4 gives, for each ɛ > 0, a that satisfies Definition, even though it is generall not the largest possible such for that ɛ. For eample, the inequalit 9 < holds for positive if 8 < < 9 or equivalentl if 8 < < 0 (Figure 5). Therefore, the largest possible for ɛ = in the definition of lim 3 = 9 is the smaller of 3 8. = 0.75 and 0 3. = 0.6. Since the latter number is smaller, the largest for ɛ = is 0 3. = 0.6, which is slightl larger than the = /7. = 0.48 obtained with the procedure of Eample 3. Either of these s or an smaller positive would serve in the definition.

5 Section.5, Formal definitions of limits p. 55 (3/908) ɛ = 0 9 = 8 FIGURE Eample 5 Show that b Definition, lim =. We need a rule that gives a positive for ever positive ɛ such that < ɛ for all with 0 < <. (0) (Preliminar calculations) We write = =. To have the quantit on the right be no greater than M with a constant M, we restrict to the interval [, 3 ] that etends one-half unit to the left and one-half unit to the right of =. (We need to keep awa from 0 since / as 0.) Because / for 3, we then have = for 3. () (The rule) Given ɛ > 0, let be the minimum of and ɛ. (Verification) Statement (0) is valid with this choice of because if 0 < <, then 3 since and then b () < ( ɛ) = ɛ. since ɛ.

6 p. 56 (3/908) Section.5, Formal definitions of limits Two short proofs The net two eamples illustrate how Definition is used in proofs. Eample 6 Use Definition to prove that if lim g() = 0, then lim g() = We need a rule that gives a positive for ever positive ɛ such that g() 0 < ɛ for all with 0 < 8 <. () Eample 7 Since the inequalit on the left of () can be written g() 0 < ɛ and this is equivalent to g() 0 < ɛ, we appl Definition to lim g() = 0 with 8 ɛ in place of ɛ. According to that definition, there is a > 0 such that g() 0 < ɛ for 0 < 8 <, and this shows that () holds. Suppose that lim f() = L and lim g() = M with numbers L and M. Prove, using 0 0 Definition, that lim [f()+g()] = L+M. (This is the first statement in Theorem 0 of Section..) We need to show that for ever ɛ > 0 there is a > 0 such that [f() + g()] (L + M) < ɛ for all 0 < 0 <. (3) (Preliminar calculations) Because f() tends to L and g() tends to M, f() L and g() M are arbitraril small for all 0 sufficientl close to 0. We use the triangle inequalit A + B A + B (4) which is valid for an numbers A and B, to write [f() + g()] (L + M) = [f() L] + [g() M] f() L + g() M. (5) (The rule) Let ɛ be an arbitrar positive number. The number on the right of (5) is less than ɛ if f() L < ɛ and g() M < ɛ. From Definition with ɛ replaced b ɛ and applied to f and g, there are positive numbers and such that f() L < ɛ for 0 < 0 < g() M < ɛ for 0 < 0 <. (6) We defiine to be the smaller of and. (Verification) With this definition of, statement (3) holds because if 0 < 0 <, then inequalities (6) hold and b (5), [f() + g()] (L + M) f() L + g() M < ɛ + ɛ = ɛ. Inequalit (4) is valid because A + B = A + B if A and B are both 0 or are both 0, and A + B < A + B if one of A and B is positive and the other is negative. This result is called the triangle inequalit because it is the statement PR PQ + QR that the length of one side of a triangle PQR is no greater than the sum of the lengths of the other two sides in the special case where P = ( A, 0), Q = (0, 0), and R = (B, 0) and PQR is a line segment.

7 Section.5, Formal definitions of limits p. 57 (3/908) Using graphs with finite limits Suitable values of in Definition for specific ɛ s can sometimes be found from graphs, as in the net eample. Eample 8 According to Definition applied to the limit lim ( + 3 ) = with ɛ = 0.5, there is a positive such that ( + 3 ) < 0.5 for all with 0 < <. (7) Use the graph of = + 3 in Figure 6 to find the approimate value of the largest that could be used in (7). = + 3 FIGURE 6 Since 0.5 =.75 and =.5, (7) is the same as the statement.75 < + 3 <.5 for all with 0 < <. (8) We draw the horizontal lines =.75 and =.5 and then vertical lines from their intersections with the graph to the -ais, as in Figure 7. We see that the graph lies between these lines for a < < b with a 0.73 and b.33. The largest possible that could be used in (8) is the smaller of the numbers a 0.73 = 0.7 and b.33 = The largest possible is 0.7. = FIGURE

8 p. 58 (3/908) Section.5, Formal definitions of limits Interactive Eamples.5 Interactive solutions are on the web page http// ashenk/.. Use Definition to verif the limit lim 0 3 = Verif Definition for lim 3 = (a) Use the graph in Figure 8 to find the approimate value of the largest that can be used in Definition applied to lim graphing calculator or computer = with ɛ = 0.4. (b) Check our answer with a = FIGURE 8 Eercises.5 A Answer provided. CONCEPTS: O Outline of solution provided. C Graphing calculator or computer required.. Figure 9 shows the graph of a function = g() such that lim 3 g() = 0 (a) Which of the values = 0., = 0., = 0.3, and = would satisf Definition for this limit with ɛ = 5? (b) For which the values ɛ =, ɛ =,ɛ = 5, and ɛ = 0 would = 0.5 work? = g() FIGURE In the published tet the interactive solutions of these eamples will be on an accompaning CD disk which can be run b an computer browser without using an internet connection.

9 Section.5, Formal definitions of limits p. 59 (3/908) BASICS:. O According to Definition applied to lim 4 (5 6) = 4, there is, for each ɛ > 0, a > 0 such that (5 6) 4 < ɛ for 0 < 4 <. Find the largest possible such for ɛ = 0.0. Verif Definition or for the limits in Eercises 3 through O lim = 0 4. A lim 3 (6 + ) = 0 5. lim ( + ) = 3 6. O According to Definition applied to the limit lim positive such that ( + ) ( + ) < 0.5 for all with 0 < <. ( ) = with ɛ = 0.5, there is a (a) Use the graph of = + in Figure 0 to find the approimate value of the largest that could be used in ( ). C (b) Check the result of part (a) with a graphing calculator or computer. = + 3 FIGURE 0 Verif Definition for the limits in Eercises 7 through O lim 5 = 5 8. A 6 lim 0 + = 3 9. lim (3 + 6) = lim ( + 3) = 7. A 5 lim 5 = 3. O 8 lim = 3. A lim = lim + =

10 p. 60 (3/908) Section.5, Formal definitions of limits In Eercises 5 through 8, (a) use the graph to find the approimate value of the largest that can be used in Definition with the given value of ɛ. C (b) Check our result with a graphing calculator or computer. 5. O lim ( ) = with ɛ = 0.4; Figure 5 = FIGURE 6. A lim ( + ) = with ɛ = 0.5; Figure = + 3 FIGURE 7. A lim 4 ( + ) = 8 with ɛ = 4; Figure 3 30 = FIGURE 3 3 4

11 Section.5, Formal definitions of limits p. 6 (3/908) lim = 0 with ɛ = ; Figure 4 + = FIGURE 4 3 EXPLORATION: 9. O (a) The width w of a cube is measured to be centimeters and this value is used to calculate the volume V = w 3 = 8 (cubic centimeters) of the cube. How accuratel must the width be measured in order that the resulting error in the calculated volume be 0. cubic centimeters? (b) What does this calculation have to do with the definition of a limit? 0. A (a) A hunk of cla weighs eactl 00 grams, and its volume is measured to be 50 cubic centimeters. This value is used to calculate the densit ρ = 00/V = (grams per cubic centimeter) of the cla. How accuratel must the volume be measured in order that the resulting error in the calculated densit be 0.05 grams per cubic centimeter? (b) What does this calculation have to do with the definition of a limit?. (a) The area A of a square is measured to be 9 square meters and this value is used to calculate its width w = A = 9 = 3. How accuratel must the area be measured in order that the resulting error in the calculated width be 0. meters? (b) What does this calculation have to do with the definition of a limit? C In Eercises and 3, use a graphing calculator or computer to find the approimate value of the the largest that could be used in Definition with the given value of ɛ in the definition of the limit C. A C 3. lim (4 + 4 ln ) = 4 with ɛ = 0.5 lim [ + sin( π)] = 3 with ɛ = Does = 0. work with ɛ = 0. in Definition applied to lim =? Use Definition to prove that the statements in Eercises 5 through 8 are false. 5. O lim 3 = A lim = 0 6. O lim = 8. lim ( + 5) = Define G() = for rational numbers and G() = 0 for irrational. Use Definition to show that (a) lim G() = 0 and (b) the two-sided finite limit lim G() does not eist for Suppose that lim 0 f() = L and lim 0 g() = M with numbers L and M. Use Definition to show that (a O f() ) lim [f()g()] = LM and (b) lim 0 0 g() = L M (End of Section.5) if M 0.