LIMITS AND CONTINUITY

Transcription

1 LIMITS AND CONTINUITY Joe McBride/Stone/Gett Imges Air resistnce prevents the velocit of skdiver from incresing indefinitel. The velocit pproches it, clled the terminl velocit. The development of clculus in the seventeenth centur b Newton nd Leibniz provided scientists with their first rel understnding of wht is ment b n instntneous rte of chnge such s velocit nd ccelertion. Once the ide ws understood conceptull, efficient computtionl methods followed, nd science took quntum lep forwrd. The fundmentl building block on which rtes of chnge rest is the concept of it, n ide tht is so importnt tht ll other clculus concepts re now bsed on it. In this chpter we will develop the concept of it in stges, proceeding from n informl, intuitive notion to precise mthemticl definition. We will lso develop theorems nd procedures for clculting its, nd we will conclude the chpter b using the its to stud continuous curves.. LIMITS (AN INTUITIVE APPROACH) The concept of it is the fundmentl building block on which ll clculus concepts re bsed. In this section we will stud its informll, with the gol of developing n intuitive feel for the bsic ides. In the net three sections we will focus on computtionl methods nd precise definitions. Tngent t P = f() P( 0, 0 ) Mn of the ides of clculus originted with the following two geometric problems: the tngent line problem Given function f nd point P( 0, 0 ) on its grph, find n eqution of the line tht is tngent to the grph t P (Figure..). Figure.. the re problem Given function f, find the re between the grph of f nd n intervl [,b] on the -is (Figure..). Trditionll, tht portion of clculus rising from the tngent line problem is clled differentil clculus nd tht rising from the re problem is clled integrl clculus. However, we will see lter tht the tngent line nd re problems re so closel relted tht the distinction between differentil nd integrl clculus is somewht rtificil. 67

2 68 Chpter / Limits nd Continuit Figure.. b = f() TANGENT LINES AND LIMITS In plne geometr, line is clled tngent to circle if it meets the circle t precisel one point (Figure..). Although this definition is dequte for circles, it is not pproprite for more generl curves. For emple, in Figure..b, the line meets the curve ectl once but is obviousl not wht we would regrd to be tngent line; nd in Figure..c, the line ppers to be tngent to the curve, et it intersects the curve more thn once. To obtin definition of tngent line tht pplies to curves other thn circles, we must view tngent lines nother w. For this purpose, suppose tht we re interested in the tngent line t point P on curve in the -plne nd tht Q is n point tht lies on the curve nd is different from P. The line through P nd Q is clled secnt line for the curve t P. Intuition suggests tht if we move the point Q long the curve towrd P, then the secnt line will rotte towrd iting position. The line in this iting position is wht we will consider to be the tngent line t P (Figure..). As suggested b Figure..b, this new concept of tngent line coincides with the trditionl concept when pplied to circles. () (b) Tngent line Tngent line Figure.. (c) P Q Secnt line P Q Secnt line Figure.. () (b) Emple P(, ). Find n eqution for the tngent line to the prbol = t the point Solution. If we cn find the slope m tn of the tngent line t P, then we cn use the point P nd the point-slope formul for line (Web Appendi G) to write the eqution of the tngent line s = m tn ( ) () To find the slope m tn, consider the secnt line through P nd point Q(, ) on the prbol tht is distinct from P. The slope m sec of this secnt line is Wh re we requiring tht P nd Q be distinct? m sec = () Figure.. suggests tht if we now let Q move long the prbol, getting closer nd closer to P, then the iting position of the secnt line through P nd Q will coincide with tht of the tngent line t P. This in turn suggests tht the vlue of m sec will get closer nd closer to the vlue of m tn s P moves towrd Q long the curve. However, to s tht Q(, ) gets closer nd closer to P(, ) is lgebricll equivlent to sing tht gets closer nd closer to. Thus, the problem of finding m tn reduces to finding the iting vlue of m sec in Formul () s gets closer nd closer to (but with = to ensure tht P nd Q remin distinct).

3 . Limits (An Intuitive Approch) 69 = P(, ) = Figure..5 A A A A A Figure..6 We cn rewrite () s m sec = = ( )( + ) ( ) = + where the cncelltion of the fctor ( ) is llowed becuse =. It is now evident tht m sec gets closer nd closer to s gets closer nd closer to. Thus, m tn = nd () implies tht the eqution of the tngent line is = ( ) or equivlentl = Figure..5 shows the grph of = nd this tngent line. AREAS AND LIMITS Just s the generl notion of tngent line leds to the concept of it, so does the generl notion of re. For plne regions with stright-line boundries, res cn often be clculted b subdividing the region into rectngles or tringles nd dding the res of the constituent prts (Figure..6). However, for regions with curved boundries, such s tht in Figure..7, more generl pproch is needed. One such pproch is to begin b pproimting the re of the region b inscribing number of rectngles of equl width under the curve nd dding the res of these rectngles (Figure..7b). Intuition suggests tht if we repet tht pproimtion process using more nd more rectngles, then the rectngles will tend to fill in the gps under the curve, nd the pproimtions will get closer nd closer to the ect re under the curve (Figure..7c). This suggests tht we cn define the re under the curve to be the iting vlue of these pproimtions. This ide will be considered in detil lter, but the point to note here is tht once gin the concept of it comes into pl. Figure..7 () b (b) b (c) b Jmes Okle/Alm This figure shows region clled the Mndelbrot Set. It illustrtes how complicted region in the plne cn be nd wh the notion of re requires creful definition. DECIMALS AND LIMITS Limits lso rise in the fmilir contet of decimls. For emple, the deciml epnsion of the frction is = () in which the dots indicte tht the digit repets indefinitel. Although ou m not hve thought bout decimls in this w, we cn write () s = 0....= () which is sum with infinitel mn terms. As we will discuss in more detil lter, we interpret () to men tht the succession of finite sums 0., , , ,... gets closer nd closer to iting vlue of s more nd more terms re included. Thus, its even occur in the fmilir contet of deciml representtions of rel numbers.

4 70 Chpter / Limits nd Continuit LIMITS Now tht we hve seen how its rise in vrious ws, let us focus on the it concept itself. The most bsic use of its is to describe how function behves s the independent vrible pproches given vlue. For emple, let us emine the behvior of the function f() = + for -vlues closer nd closer to. It is evident from the grph nd tble in Figure..8 tht the vlues of f()get closer nd closer to s vlues of re selected closer nd closer to on either the left or the right side of. We describe this b sing tht the it of + iss pproches from either side, nd we write ( + ) = (5) f() = f() = + f() f() Figure..8 Left side Right side This leds us to the following generl ide... its (n informl view) If the vlues of f() cn be mde s close s we like to L b tking vlues of sufficientl close to (but not equl to ), then we write f() = L (6) Since is required to be different from in (6), the vlue of f t, or even whether f is defined t, hs no bering on the it L. The it describes the behvior of f close to but not t. which is red the it of f() s pproches is L or f() pproches L s pproches. The epression in (6) cn lso be written s f() L s (7)

5 . Limits (An Intuitive Approch) 7 Emple Use numericl evidence to mke conjecture bout the vlue of (8) Solution. Although the function f() = (9) TECHNOLOGY MASTERY Use grphing utilit to generte the grph of the eqution = f() for the function in (9). Find window contining = in which ll vlues of f() re within 0.5 of = nd one in which ll vlues of f() re within 0. of =. is undefined t =, this hs no bering on the it. Tble.. shows smple -vlues pproching from the left side nd from the right side. In both cses the corresponding vlues of f(), clculted to si deciml plces, pper to get closer nd closer to, nd hence we conjecture tht = This is consistent with the grph of f shown in Figure..9. In the net section we will show how to obtin this result lgebricll. Tble = f() = f() Left side Right side Emple Use numericl evidence to mke conjecture bout the vlue of sin 0 (0) Figure..9 Use numericl evidence to determine whether the it in () chnges if is mesured in degrees. Solution. With the help of clculting utilit set in rdin mode, we obtin Tble... The dt in the tble suggest tht sin = () 0 The result is consistent with the grph of f() = (sin )/ shown in Figure..0. Lter in this chpter we will give geometric rgument to prove tht our conjecture is correct. (rdins) ±.0 ±0.9 ±0.8 ±0.7 ±0.6 ±0.5 ±0. ±0. ±0. ±0. ±0.0 Tble.. = sin Figure..0 f() 0 As pproches 0 from the left or right, f() pproches. = f () = sin SAMPLING PITFALLS Numericl evidence cn sometimes led to incorrect conclusions bout its becuse of roundoff error or becuse the smple vlues chosen do not revel the true iting behvior. For emple, one might incorrectl conclude from Tble.. tht 0 sin ( π ) = 0

6 7 Chpter / Limits nd Continuit The fct tht this is not correct is evidenced b the grph of f in Figure... The grph revels tht the vlues of f oscillte between nd with incresing rpidit s 0 nd hence do not pproch it. The dt in the tble deceived us becuse the -vlues selected ll hppened to be -intercepts for f(). This points out the need for hving lterntive methods for corroborting its conjectured from numericl evidence. Tble.. c f () = sin c = sin c _ + = ± = ±0. = ±0.0 = ±0.00 = ± ± c ±0c ±00c ±000c ±0,000c. sin(±c) = 0 sin(±0c) = 0 sin(±00c) = 0 sin(±000c) = 0 sin(±0,000c) = 0. Figure.. Figure.. = ONE-SIDED LIMITS The it in (6) is clled two-sided it becuse it requires the vlues of f() to get closer nd closer to L s vlues of re tken from either side of =. However, some functions ehibit different behviors on the two sides of n -vlue, in which cse it is necessr to distinguish whether vlues of ner re on the left side or on the right side of for purposes of investigting iting behvior. For emple, consider the function f() = = {, > 0, < 0 which is grphed in Figure... As pproches 0 from the right, the vlues of f() pproch it of [in fct, the vlues of f() re ectl for ll such ], nd similrl, s pproches 0 from the left, the vlues of f()pproch it of. We denote these its b writing = nd = () With this nottion, the superscript + indictes it from the right nd the superscript indictes it from the left. This leds to the generl ide of one-sided it. () As with two-sided its, the one-sided its in () nd (5) cn lso be written s f() L s + nd respectivel. f() L s.. one-sided its (n informl view) If the vlues of f() cn be mde s close s we like to L b tking vlues of sufficientl close to (but greter thn ), then we write f() = L () + nd if the vlues of f() cn be mde s close s we like to L b tking vlues of sufficientl close to (but less thn ), then we write f() = L (5) Epression () is red the it of f() s pproches from the right is L or f() pproches L s pproches from the right. Similrl, epression (5) is red the it of f() s pproches from the left is L or f() pproches L s pproches from the left.

7 . Limits (An Intuitive Approch) 7 THE RELATIONSHIP BETWEEN ONE-SIDED LIMITS AND TWO-SIDED LIMITS In generl, there is no gurntee tht function f will hve two-sided it t given point ; tht is, the vlues of f() m not get closer nd closer to n single rel number L s. In this cse we s tht f() does not eist Similrl, the vlues of f() m not get closer nd closer to single rel number L s + or s. In these cses we s tht + f() does not eist or tht f() does not eist In order for the two-sided it of function f() to eist t point, the vlues of f() must pproch some rel number L s pproches, nd this number must be the sme regrdless of whether pproches from the left or the right. This suggests the following result, which we stte without forml proof... the reltionship between one-sided nd two-sided its The twosided it of function f() eists t if nd onl if both of the one-sided its eist t nd hve the sme vlue; tht is, f() = L if nd onl if f() = L = f() + Emple does not eist. Eplin wh 0 Solution. As pproches 0, the vlues of f() = / pproch from the left nd pproch from the right [see ()]. Thus, the one-sided its t 0 re not the sme. Emple 5 For the functions in Figure.., find the one-sided nd two-sided its t = if the eist. Solution. The functions in ll three figures hve the sme one-sided its s, since the functions re identicl, ecept t =. These its re f() = nd f() = + In ll three cses the two-sided it does not eist s becuse the one-sided its re not equl. = f () = f () = f() Figure..

8 7 Chpter / Limits nd Continuit Emple 6 For the functions in Figure.., find the one-sided nd two-sided its t = if the eist. Solution. As in the preceding emple, the vlue of f t = hs no bering on the its s, so in ll three cses we hve f() = nd f() = + Since the one-sided its re equl, the two-sided it eists nd f() = = f() = f() = f() Figure.. INFINITE LIMITS Sometimes one-sided or two-sided its fil to eist becuse the vlues of the function increse or decrese without bound. For emple, consider the behvior of f() = / for vlues of ner 0. It is evident from the tble nd grph in Figure..5 tht s -vlues re tken closer nd closer to 0 from the right, the vlues of f() = / re positive nd increse without bound; nd s -vlues re tken closer nd closer to 0 from the left, the vlues of f() = / re negtive nd decrese without bound. We describe these iting behviors b writing =+ nd = The smbols + nd here re not rel numbers; the simpl describe prticulr ws in which the its fil to eist. Do not mke the mistke of mnipulting these smbols using rules of lgebr. For emple, it is incorrect to write (+ ) (+ ) = 0. = Increses without bound = Decreses without bound ,000 0, Figure..5 Left side Right side

9 . Limits (An Intuitive Approch) 75.. infinite its (n informl view) The epressions f() =+ nd f() =+ + denote tht f() increses without bound s pproches from the left nd from the right, respectivel. If both re true, then we write Similrl, the epressions f() =+ f() = nd f() = + denote tht f() decreses without bound s pproches from the left nd from the right, respectivel. If both re true, then we write f() = Emple 7 For the functions in Figure..6, describe the its t = in pproprite it nottion. Solution (). In Figure..6, the function increses without bound s pproches from the right nd decreses without bound s pproches from the left. Thus, =+ nd + = Solution (b). In Figure..6b, the function increses without bound s pproches from both the left nd right. Thus, ( ) = + ( ) = ( ) =+ Solution (c). In Figure..6c, the function decreses without bound s pproches from the right nd increses without bound s pproches from the left. Thus, + = nd =+ Solution (d). In Figure..6d, the function decreses without bound s pproches from both the left nd right. Thus, ( ) = + ( ) = ( ) = f () = f () = ( ) f() = f () = ( ) Figure..6 () (b) (c) (d)

10 76 Chpter / Limits nd Continuit VERTICAL ASYMPTOTES Figure..7 illustrtes geometricll wht hppens when n of the following situtions occur: f() =+, f() =+, + f() =, f() = + In ech cse the grph of = f() either rises or flls without bound, squeezing closer nd closer to the verticl line = s pproches from the side indicted in the it. The line = is clled verticl smptote of the curve = f() (from the Greek word smptotos, mening nonintersecting ). Figure..8 f() = + Figure..7 = f() f() = + + f() = f() = + In generl, the grph of single function cn displ wide vriet of its. Emple 8 For the function f grphed in Figure..8, find () f() (b) (c) (d) (e) (f) f() (g) the verticl smptotes of the grph of f. + Solution () nd (b). f()= = f( ) nd + Solution (c) nd (d). f()= 0 = f(0) 0 nd 0+ Solution (e) nd ( f ). f()does not eist due to oscilltion nd + Solution (g). The -is nd the line = re verticl smptotes for the grph of f. QUICK CHECK EXERCISES. (See pge 79 for nswers.). We write f() = L provided the vlues of cn be mde s close to s desired, b tking vlues of sufficientl close to but not.. We write f() =+ provided increses without bound, s pproches from the left.. Stte wht must be true bout f() nd + in order for it to be the cse tht. Use the ccompning grph of = f()( <<) to determine the its. () f() = 0 (b) (c) (d) f() = f() = + f() = Figure E- 5. The slope of the secnt line through P(, ) nd Q(, ) on the prbol = is m sec = +. It follows tht the slope of the tngent line to this prbol t the point P is.

11 . Limits (An Intuitive Approch) 77 EXERCISE SET. Grphing Utilit C CAS 0 In these eercises, mke resonble ssumptions bout the grph of the indicted function outside of the region depicted.. For the function g grphed in the ccompning figure, find () g() (b) g() (c) g() (d) g(0). 0 = g() 9 Figure E-. For the function G grphed in the ccompning figure, find () G() (b) G() (c) G() (d) G(0). 0 5 = G() Figure E-. For the function f grphed in the ccompning figure, find () f() (b) f() + (c) f() (d) f(). = f() 0 Figure E-. For the function f grphed in the ccompning figure, find () f() (b) f() + (c) f() (d) f(). = f() Figure E- 5. For the function F grphed in the ccompning figure, find () F() (b) F() + (c) F() (d) F( ). = F() Figure E-5 6. For the function G grphed in the ccompning figure, find () G() (b) G() (c) G() (d) G(0). 0 = G() Figure E-6 7. For the function f grphed in the ccompning figure, find () f() (b) f() + (c) f() (d) f(). = f () Figure E-7 8. For the function φ grphed in the ccompning figure, find () φ() (b) φ() + (c) φ() (d) φ(). = f() Figure E-8 9. For the function f grphed in the ccompning figure on the net pge, find () f() (b) f() 0 (c) f() (d) f() 0 + (e) f() + (f ) the verticl smptotes of the grph of f.

12 78 Chpter / Limits nd Continuit C = f() Figure E-9 0. For the function f grphed in the ccompning figure, find () f() (b) f() (c) f() + 0 (d) f() (e) f() (f ) f() (g) the verticl smptotes of the grph of f. = f() Figure E-0 (i) Complete the tble nd mke guess bout the it indicted. (ii) Confirm our conclusions bout the it b grphing function over n pproprite intervl. [Note: For the inverse trigonometric function, be sure to put our clculting nd grphing utilities in rdin mode.]. f() = e ; f() 0 f() 0.0 Tble E-. f() = sin ; f() 0 f() 0. Tble E (i) Mke guess t the it (if it eists) b evluting the function t the specified -vlues. (ii) Confirm our conclusions bout the it b grphing the function over n pproprite intervl. (iii) If ou hve CAS, then use it to find the it. [Note: For the trigonometric functions, be sure to put our clculting nd grphing utilities in rdin mode.]. () ; =,.5,.,.0,.00, 0, 0.5, 0.9, 0.99, (b) ; =,.5,.,.0,.00, (c) ; = 0, 0.5, 0.9, 0.99, 0.999, () ; =±0.5, ±0., ±0.00, 0 ± (b) ; = 0.5, 0., 0.00, (c) ; = 0.5, 0., 0.00, sin 5. () ; =±0.5, ±0., ±0.00, ± cos (b) ; = 0, 0.5, 0.9, 0.99, 0.999, +.5,.,.0,.00 tn( + ) 6. () ; = 0, 0.5, 0.9, 0.99, 0.999, +.5,.,.0,.00 sin(5) (b) ; =±0.5, ±0., ±0.00, ± sin() 7 0 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 7. If f() = L, then f() = L. 8. If f() eists, then so do f() nd + f(). 9. If f() nd + f() eist, then so does f(). 0. If + f() =+, then f() is undefined. 6 Sketch possible grph for function f with the specified properties. (Mn different solutions re possible.). (i) the domin of f is [, ] (ii) f( ) = f(0) = f() = 0 (iii) f() = f() = f() = + 0. (i) the domin of f is [, ] (ii) f( ) = f(0) = f() = 0 (iii) f() =, f() = 0, nd + 0 f() =. (i) the domin of f is (, 0] (ii) f( ) = f(0) = (iii) f() =+. (i) the domin of f is (0, + ) (ii) f() = 0 (iii) the -is is verticl smptote for the grph of f (iv) f() < 0if0<<

13 . Limits (An Intuitive Approch) (i) f( ) = f(0) = f() = 0 (ii) f() =+ nd f() = + (iii) f() =+ 6. (i) f( ) = 0,f(0) =,f() = 0 (ii) (iii) f() = 0 nd f() =+ + f() = nd f() = Modif the rgument of Emple to find the eqution of the tngent line to the specified grph t the point given. 7. the grph of = t (, ) 8. the grph of = t (0, 0) 9. the grph of = t (, ) 0. the grph of = t (, ) FOCUS ON CONCEPTS. In the specil theor of reltivit the length l of nrrow rod moving longitudinll is function l = l(v) of the rod s speed v. The ccompning figure, in which c denotes the speed of light, displs some of the qulittive fetures of this function. () Wht is the phsicl interprettion of l 0? (b) Wht is v c l(v)? Wht is the phsicl significnce of this it? Length l 0 l Figure E- l = l(v) Speed. In the specil theor of reltivit the mss m of moving object is function m = m(v) of the object s speed v. The ccompning figure, in which c denotes the speed of light, displs some of the qulittive fetures of this function. () Wht is the phsicl interprettion of m 0? c v (b) Wht is v c m(v)? Wht is the phsicl significnce of this it? Mss m 0 m Figure E- Speed m = m(v). Let f() = ( + )./ () Grph f in the window [, ] [.5,.5] nd use the clcultor s trce feture to mke conjecture bout the it of f() s 0. (b) Grph f in the window [ 0.00, 0.00] [.5,.5] nd use the clcultor s trce feture to mke conjecture bout the it of f() s 0. (c) Grph f in the window [ , ] [.5,.5] nd use the clcultor s trce feture to mke conjecture bout the it of f() s 0. (d) Lter we will be ble to show tht 0 ( + )./ Wht flw do our grphs revel bout using numericl evidence (s reveled b the grphs ou obtined) to mke conjectures bout its?. Writing Two students re discussing the it of s pproches 0. One student mintins tht the it is 0, while the other clims tht the it does not eist. Write short prgrph tht discusses the pros nd cons of ech student s position. 5. Writing Given function f nd rel number, eplin informll wh f(+ ) = f() 0 (Here equlit mens tht either both its eist nd re equl or tht both its fil to eist.) c v QUICK CHECK ANSWERS.. f(); L; ; ;. f(); ;. Both one-sided its must eist nd equl L.. () 0 (b) (c) + (d) 5.