2.8 The Derivative as a Function

Transcription

1 SECTION 2.8 THE DERIVATIVEASA FUNCTION D The Derivative as a Functin 1. It appears that I is an dd functin, s l' will be an even functin- that is, t ' (-a) = l'(a). (a) l'(- 3) ~ 1.5 (b ) 1' ( - 2 ) ~1 (c) 1'(- 1) ~ 0 (d) 1'(0) ~ - 4 (e) 1' (1) ~ 0 (f) 1' (2) ~ 1 (g) l'(3) ~ Nte: Yur answers ma var depending n ur estimates. B estimating the slpes ftangent lines n the graph f I, it appears that (a) l'(0) ~ -3 (b) 1'(1) ~ 0 (c) 1'(2) ~ 1.5 ( d) 1'(3 ) ~2 (e) 1' (4) ~ 0 (t) t ' (5) ~ (a)' = II, since frm left t right, the slpes f the tangent s t graph (a) start ut negative, becme 0, then psitive, then 0, then negative again. The actual functin values in graph II fllw the same pattern. (b)' = IV, since frm left t right, the slpes f the tangents t graph (b) start ut at a fied psitive quantit, then suddenl becme negative, then psitive again. The discntinuities in graph IV indicate sudden changes in the slpes f the tangents. (c)' = l, since the slpes fthe tangents t graph (c) are negative fr < 0 and psitive fr > 0, as are the functin values f graph 1. (d)' = III, since frm left t right, the slpes f the tangents t graph (d) are psitive, then 0, then negative, then 0, then psitive, then 0, then negative again, and the functin values in graph III fllw the same pattern. Hints fr Eercises 4-11: First plt -intercepts nthe graph f l' fran hrizntaltangents n thegraph f I. Lkfr an crnersnthe graph f I - therewill be a discntinuit n the graph f1'. On an intervalwherei has a tangentwith psitive (rnegative) slpe, the graph f l' will be psitive(r negative). If the graph fthe functin is linear, the graph fl' will be a hrizntal line.

2 130 D CHAPTER 2 LIMITS AND DERIVATIVES i i 9. i i' 0- i' i i' ~L 12. The slpes f the tangent lines n the graph f == P (t) are alwas psitive, s the -values f == P/(t) are alwas psitive. These values start ut relativel small and keep increasing, reaching a maimum at abut t == 6. Then the -values f == P/(t) decrease and get clse t zer. The =P'(t) graph f p/ tells us that the east culture grws mst rapidl after 6 hurs t and then the grwth rate declines. 13. It appears that there are hrizntal tangents n the graph f M fr t == 1963 and t == Thus, there are zers fr thse values f t n the graph f M'. The derivative is negative fr the ears 1963 t =M'(t) See Figure 1 in Sectin 3.3.

3 SECTION 2.8 THE DERIVATIVE ASA FUNCTION D = f() = ln = f'() The slpe at 0 appears t be 1 and the slpe at 1 appears t be 2.7. As decreases, the slpe gets clser t O. Since the graphs are s similar, we might guess that f' () = e", As increases tward 1, f' () decreases frm ver large numbers t 1. As becmes large, f' () gets clser t O. As a guess, f'(x) = 1/ 2 r f'(x) = l/ makes sense. 17. (a) B zming in, we estimate that f' (0) = 0, f' (~) = 1, f' (1) = 2, 2.5 and f' (2) = 4. (b) B smmetr, f'(-x) = _f'(x). S f'(-~) = -1, f'(-l) = -2, andf'(-2) = -4. (c) It appears that f'(x) is twice the value f, s we guess that f'(x) = 2. (d) j'() = lim f( + h) - f() = lim ( + h? - 2 h--+o h h--+o h = lim ( + 2h + h ) - = lim 2h + h 2 = lim h(2 + h) = lim (2:; + h) = 2 h--+o h n-:«h h--+o h h--+o 18. (a) B zming in, we estimate that f'(o) = 0, f' (~) ~ 0.75, (b) B smmetr, f' (-) = f' (). S f' (-~) ~ 0.75, f' (1) ~ 3, f' (2) ~ 12, and f' (3) ~ 27. f'( -1) ~ 3, I'!-2) ~ 12, and f'( -3) ~ 27. (c) (d) Since f' (0) = 0, it appears that f' ma have the frm f' () = a 2 Using f' (1) = 3, we have a = 3, s f' () = h (e) j'() = lim f( + h) - f() = lim ( + h? - 3 = lim ( h + h ) - h--+o h u:«h h--+o h 2h _ r 3 + 3h + h _ li h(3 + 3h + h ) _ r ( h + h 2 ) h~ h - h~ h - h~ X X - X

4 132 D CHAPTER 2 LIMITS AND DERIVATIVES 19. j'() = lim f( + h) - f() = lim [~( + h) - ~] - G - ~) = lim ~ + ~h - ~ - ~ + ~ h h h lh = lim L = lim 1 = 1 h 2 2 Dmain f f = dmain f f' = JR. 20. f'(x) = lim f(+h)-f() = lim [m(+h)+bj-(m+b) = lim m+mh+b-m-b h h h mh li = 1Im-= Imm=m h Dmain f f = dmain f f' = JR. 2) '( ). f(t + h) - f(t). [5(t + h) - 9(t + h)2j - (5t - 9t 21. f t = lim h = Iim h 2) = lim 5t + 5h - 9(t 2 + 2th + h - 5t + 9t = lim 5t + 5h - 9t - 18th - 9h - 5t + 9t h h 2 = lim 5h - 18th - 9h = lim h(5-18t - 9h) = lim (5 _ 18t - 9h) = 5-18t h h Dmain f f = dmain f f' = JR. '(). f( + h) - f(). [1.5( + h)2 - ( + h) + 3.7J - ( X + 3.7) 22. f = lim h = lim h = lim h + 1.5h 2 - X - h X = lim 3h + 1.5h 2 - h h h = lim ( h - 1) = 3-1 Dmain f f = dmain f f' = JR. '(). f( + h) - f(). [( + h)3-3( + h) + 5J - ( ) 23. f = lim h = lim h ( h + 3h 2 + h - 3-3h + 5) - ( ). 3 2 h + 3h 2 + h 3-3h = lim = lim h h h( h + h 2-3) 2 2 = lim h = lim ( h + h - 3) = 3-3 Dmain f f = dmain f f' = JR. 24. j'() = lim f( + h) - f() = lim ( + h + VX+h) - ( + v) h h = lim (!!. + VX+h - v.vx+h + v) = lim [1 + ( + h) - ] h h vi + h + VX h (vi + h + VX) = lim (1 + 1 ) = = 1 + _1_ vi + h + VX VX + VX 2 VX Dmain f f = [0,00), dmain f f' = (0,00).

5 SECTION 2.8 THE DERIVATIVE AS A FUNCTION D 133 '() -1' g(+h)-g() -1' V1+2(+h)-V1+ 2 [V1+2(X+h)+V1+2X] m - lifl ~ h h h h J1+2(+h)+V1+ 2 == lim ( h) - (1 + 2) == lim 2 == 2 == 1 h--->o h [VI+ 2( + h) + VI + 2X] h--->o VI h + VI VI + 2 VI + 2 Dmain f 9 == [- ~,(0),dmain f g' == (- ~,(0), 3+(+h) j'() = lim f( + h) - f() = lim 1-3( + h) -~ = lim (3 + + h)(1-3) - (3 + )( h) h h h h h h(l - 3-3h)(1-3), ( h - 3h) - ( h h) == lim --: ~-' ~ h h(l - 3-3h)(1-3) == lim 10h == lim h h(l - 3-3h)(1-3) h ( h)(1-3) (1-3X)2 Dmain f f == dmain f f' == (-00, ~) U (~, (0), 4(t + h) 4t 4(t + h) (t + 1) - 4t (t + h + 1) 27. G'(t) = lim G(t + h) - G(t) = lim (t + h) t+1 = lim (t + h + 1)(t + 1) h h h h h h, (4t 2 + 4ht + 4t + 4h) - (4t 2 + 4ht + 4t), 4h == lim == lim h h(t + h + l)(t + 1) h h(t + h + l)(t + 1), 4 4 == 11m----- h (t + h + l)(t + 1) (t + 1)2 Dmain fg == dmain fg' == (-00, -1) U (-1, (0), 1 1 Vi-vt+h 28 '(t) == r g(t+ h) - g(t) == li ~ - 7t == lim vt+h Vi == lim (Vi - vt+h, Vi + vt+h). 9 h~ h h~ h h h h h vt + h Vi Vi + vt + h == lim t - (t + h) == lim - h == lim -1 h h vt + h Vi (Vi + vt + h) h h vt + h Vi (Vi + vt + h) h vt + h Vi (Vi + vt + h) Vi Vi (Vi + Vi) t (2 Vi) - 2t 3 2 / Dmain f 9 == dmain f g' == (0, 00). Dmain f f == dmain f f' == JR,

6 134 0 CHAPTER 2 LIMITSAND DERIVATIVES 30. (a) Y=h =.)-( - 6) =.) ) - 1 (b) Nte that the third graph in part (a) has small negative values fr its slpe, 1' ; but as -+ 6-, l' See the graph in part (d). ( c ) f ' ( ) = li f ( + h) h~ h = lim )6 u-:» f ( ) ( + h) h!i3=x [)6- ( + h) +!i3=x ] )6 ( + h) +!i3=x - 1 (d) = lim [6 - ( + h)] (6 - ) h ---> O h [)6 ( -i- h) +!i3=x] = lim -;-;----r;;===-=;h~~==\ t.-,» h( '6 - h +!i3=x) -1 = lim ----;,:= = ::::;=--:-;:;:;=:= IHO '6 - - h +!i3=x - 1 2!i3=X Dmain f f = (- 00,6 ], dmain f l' = (- 00,6). 31. (a) 1' ( ) = lim f ( + h) h ---> O h f ( ) = lim [( + h)4 + 2( + h)] u-:«h ( 4 + 2). 4 3 h + 6 2h2 + 4h 3 + h 4 + 2h. h( h + 4h 2 + h 3 + 2) = lim = lim --' , ' h --->O h h ---> O h = lim ( h + 4h 2 + h 3 + 2) = h--+ O (b) Ntice that l'() = 0 when f has a hrizntal tangent, l'() is psitive when the tangents have psitive slpe, and l'() is negative when the tangents have negat ive slpe. - 2 f----\ i"' : (a) 1' (t) = lim f(t + h) - f (t) = lim [(t + h)2 - vt+h] - (e - 0) h ---> O h h--->o h 2 e + 2ht+ h2 - vt+h - e +0 I' = (2ht +h 0 -vt+h) I Hfl. = lhl + -'------,----'- h ---> O h h --->O h h = lim (h(2t + h) vt+h. 0 + vt+h) h ---> O h h 0 + 't + h = lim (2t + h + t - (t + h) ) = lim (2t + h + - h ) h--->o h(0 + 't + h ) h ---> O h(0 + 't + h ) = lim (2t + h + r: - ~) = 2t _ 1;; h ---> O V t + V t + h 2 V t

7 SECTION 2.8 THE DERIVATIVE ASA FUNCTION (b) Ntice that 1"' (t) = 0 when f has a hrizntal tangent, 1"' (t ) is psitive when the tangents have psitive slpe, and J'(t) is negative when the tangents have negative slpe f~ (a) V ' (t) is the rate at which the unemplment rate is changing with respect t time. Its units are percent per ear.. V(t + h ) - V( t) Vet + h) - V(t) (b) T find Vi (t), we use lim h ~ I fr small values f h. h_ O t F,'1 993' Vi (1993) ~ V (1994) - V( 1993) = = Fr 1994: We estimate V i(1994) b using h = - 1 and h = 1, and then average the tw results t btain a final estimate. h = - 1 => V '( 1994) ~ V( 1993) - V( 1994) = = ' ' h = 1 V '( 1994) ~ V( 1995) - V( 1994) = = => S we estimate that VI (1994) ~ ~ [(- 0.80) + (- 0.50)] = t V'( t) (a) p i(t ) is the rate at which the percentage f Americans under the age f 18 is changing with respect t time. Its units are percent per ear (%jr). I. P (t + h) - P (t ) P (t + h) - P( t) (b) T find P (t), we use Inn h ~ h fr small values f h. h _ O F ' 1950 P I (1950) ~ P (1960) - P (1950) = = _ Fr 1960 : We estimate p i(1960) b using h = - 10 and h = 10, and then average the tw results t btain a final estimate. h = -10 P I (1960) ~ P (1950) - P (1960) = = 0.46 => h = 10 => P I (1960) ~ P (1~;~6 =~9~~60 ) = 34.0 ~ 35.7 = S we estimate that pi (1960 ) ~ ~ [ (- 0.17)] = t P I(t )

8 136 D CHAPTER 2 LIMITS AND DERIVATIVES (c) P(t) P'(t) (d) We culd get mre accurate values fr p' (t) b btaining data fr the mid-decade ears 1955, 1965, 1975, 1985, and f is nt differentiable at == -4, because the graph has a cmer there, and at == 0, because there is a discntinuit there. 36. f is nt differentiable at == 0, because there is a discntinuit there, and at == 3, because the graph has a vertical tangent there. 37. f is nt differentiable at == -1, because the graph has a vertical tangent there, and at == 4, because the graph has a cmer there. 38. f is nt differentiable at == -1, because there is a discntinuit there, and at == 2, because the graph has a cmer there. 39. As we zm in tward (-1, 0), the curve appears mre and mre like a 2 straight line, s f () == + M is differentiable at == -1. But n matter hw much we zm in tward the rigin, the curve desn't straighten ut-we can't eliminate the sharp pint (a cusp). S f is nt differentiable at == :::lII~--+---~ As we zm in tward (0, 1), the curve appears mre and mre like a straight line, s f is differentiable at == O. But n matter hw much we zm in tward (1,0) r (-1,0), the curve desn't straighten ut-we can't eliminate the sharp pint (a cusp). S f is nt differentiable at == ±1. -2 t t------f f a == f, b == i', c == [", We can see this because where a has a hrizntal tangent, b == 0, and where b has a hrizntal tangent, c == 0. We can immediatel see that c can be neither f nr f', since at the pints where c has a hrizntal tangent, neither a nr b is equal t 0.

9 SECTION 2.8 THE DE RIVATIVEASA FUN CTION D Where d has hrizntal tangents, nl c is 0, s d' = c. c has negative tangents fr < 0 and b is the nl graph that is negative fr < 0, s c' = b. b has psitive tangents n IR (ecept at = 0), and the nl graph that is psit ive n the same dmain is a, s b' = a. We cnclude that d = f, c = 1', b = f",and a = j'", 43. We can immediate l see that a is the graph fthe acceleratin functin, since at the pints where a has a hrizntal tangent, neither c nr b is equal t O. Net, we nte that a = 0 at the pint where b has a hriznta l tangent, s b must be the graph f the velcit functin, and hence, b' = a. We cnclude that c is the graph fthe psitin functin. 44. a must be the jerk since nne fthe graphs are 0 at its high and lw pints. a is 0 where b has a maimum, s b' = a. b is 0 where c has a maimum, s c' = b. We cnclude that d is the psitin functin, c is the velcit, b is the acceleratin, and a is the jerk. 2) ' ( ). f ( + h ) - f ( ). [1 + 4( + h ) - ( + h )2] - ( f = Iim h = lim I h ~O h ~O L = lim ( h h - h ) - ( ) = lim 4h - 2h - h = lim (4 _ 2 _ h ) = 4 _ 2 h~ O h h ~ O h h~ O 1"( ) = lim 1' ( + h ) - 1'( ) = lim [4-2( + h )] - (4-2 ) = lim - 2h = lim (- 2) = - 2 h ~ O h h~ O h h ~ O h h ~O We see frm the graph that ur answers are reasnab le because the graph f - 6 t /t------'l:--\-->--+--i 10 l' is that f a linear functin and the graph f I" is that f a cnstant fn functin '( ) = lim f ( + hl - f ( ) = lim + h = lim - ( + h ) = lim - h = lim - 1 h~ O L h~ O h h ~O h( + h ) h ~ O h ( + h ) u-:» ( + h ) 1"( ) = lim 1' ( + h ) - 1' ( ) = lim - (: h? - ( - ~ ) h~ O h h~ O h = lim 2 + h h ~O 2 ( + h)2 2 2 f =-- We see frm the graph that ur answers are reasnable because the graph f l' is that f an even functin and is negative fr all =I- 0, and the graph f 1" is that f an dd functin (negative fr < 0 and psitive fr > 0). - 5

10 138 D CHAPTER 2 LIMITS AND DE RIVATI VES '(). f ( + h ) - f( ). [2( + h? - ( + h? ] - (2 2-3 ) 47. f = lim } = lim } h ~O L h ~ O L 2 2) = lim h (4 + 2h - 3-3h - h = lim (4 + 2h _ 3 2 _ 3h _ h2) = 4 _ 3 2 u-: «h h ~O 2 "(). 1'( +h) -1'( ). [4(+ h ) - 3( + h )2] - (4-3 ). h(4-6 -3h) f = Inn = Inn = lim ----'------:----' h ~O h h ~ O h h ~O h = lim ( h ) = 4-6 h ---+ O j"'( ) = lim j"( + h ) - j"() = lim [4-6( + h )] - (4-6) = lim - 6h = lim (- 6) = - 6 h - >O h h ~O h h~ O h h ~ O f (4)(X) = lim j"'( + h) - j"'( ) = lim (-6) = lim Q = lim (0) = 0 h ~ O h h ~O h h~ O h h~ O f\ If" - 4 ;~ 6 I f' 1-7 \\\ f ill I. The graphs are cnsis tent with the gemetric interpretatins f the derivatives because l' has zers where f has a lcal minimum and a lcal maimum, f" has a zer where l' has a lcal maimum, and f"' is a cnstant functin equal t the slpe f f". 48. (a) Since we estimate the velcit t be a maimum at t = 10, the acceleratin is 0 at t = 10. v 50 a (b) Drawing a tangent line at t = 10 nthe graph f a, a appears t decrease b 10 ft/s 2 ver a perid f 20 s..> S at t = 10 s, the jerk is apprimatel - 10/ 20 = (ft/s 2) /s r ft /s". 49. (a) Nte that we have factred - a as the difference f tw cubes in the third step '(a) = lim f ( ) - f(a) = lim / - a / = lim X / - a / ~ a X - a ~a X - a ~ a (X 1 / 3 - a 1 / 3) ( 2 / / 3 a l / 3 + a 2 / 3) = lim 1 = _ 1_ r la- 2 / 3 ~ a X2/3 + 1 / 3a l / 3 + a2/3 3a 2/3 3 (b) 1'(0 ) = lim f (O+ h~ - f(o ) = lim ijhh- 0 = lim h:/ 3 ' This functin increases withu t bund, s the limit des nt h ---+ O h ---+ O h - O eist, and theref re 1'(0) des nt eist. (c) lim 1'() I = lim ;/3= 00 and f is cntinuus at = 0 (rt functin), s f has a vertical tangent at = O. ---+O ---+ O (a) g'(o ) = lim g() - g(o ) = lim X / - 0 = lim 1/ 3 ' which des nt eist O X - 0 -to X ---+O X

11 SECTION 2.8 THE DERIVATIVE AS A FUNCTION D 139 () () 2/3 2/3. (X 1/3 _ a 1/3)(1/3 + a1/3) (b) g'(a) = lim 9-9 a = lim - a = lim ~ -.:...-: :...- X-7a X - a X-7a X - a X-7a (X 1/3 a1/3)(2/3 + 1/3a1/3 + a2/3) 1/3 + X a 1 / 3 2a 1 / 3 2 = lim - - r ~3a-l/3 X-7a X2/3 + 1/3a1/3 + a2/3-3a2/3-3a1/3 (c) g() = X 2/3 is cntinuus at = 0 and (d) 0.4 lim Ig'()I = lim -----;-/3 = 00. This shws that X-7Q X-7Q 3[l 9 has a vertical tangent line at = ' '" 0.2 X - 6 if - 6 ~ 6 { - 6 if ~ I () = I - 61 = { -( - 6) if - 6 < if < 6 S the right-hand limit is lim I () - 1(6) lim I - 61 ;; 0 = lim - 6 = lim 1 = 1, and the left-hand limit X-76+ X - 6 X X-76+ X - 6 X-76+ is lim I() - 1(6) = lim I - 6] - 0 = lim 6 - = lim (-1) = -1. Since these limits are nt equal, X-76- X - 6 X-76- X - 6 X-76- X - 6 X-76 I' (6) = lim I() - 1(6) des nt eist and I is nt differentiable at 6. X-76 X - 6 = f'() 1 if > 6 Hwever, a frmula fr I' is I' () =. 6 { -1 If < 6 Anther wa fwriting the frmula is I'() = I I () = [] is nt cntinuus at an integer n, s I is nt differentiable at n b the cntrapsitive ftherem 4. If a is nt an integer, then I is cnstant n an pen interval cntaining a, s I/(a) = O. Thus, -1 I/() = 0, nt an integer. if.> (a) I() = Il = _ 2 (b) Since I() = 2 fr ~ 0, we have I' () = 2 fr > O. { if < 0 [See Eercise 17(d).] Similarl; since I() = - 2 fr < 0, 2 we have I'() = -2 fr < O. At = 0, we have 1 /(0) = lim I() - 1(0) = lim Il = lim Il = O. X-7Q X - 0 X-7Q X X-7Q S I is differentiable at O. Thus, I is differentiable fr all. 2 if > O} (c) Frm part tbjwe have j'tz) =. - =2Il. { -2 If < 0

12 140 D CHAPTER 2 LIMITS AND DERIVATIVES 54. (a) f~(4) = lim f(4 + h) - f(4) = lim 5 - (4 + h) - 1 lim -h = -1 and h----+o- h h----+o- h h----+o- h 1-1 f~(4)= lim f(4+h)-f(4) = lim 5-(4+h) = lim l-(l-h) lim _1_=1. h----+o+ h h----+o+ h h----+o+ h(l - h) h----+o+ 1 - h (b) Y At 4 we have lim f() = lim (5 - ) = 1 and = f() =5 lim f() = lim _1_ = 1, s lim f() = 1 = f(4) and f is X cntinuus at 4. Since f(5) is nt defined, f is discntinuus at 5. 0 if < 0 (c) f () = 5 - if 0 < < 4 { 1/(5 - ) if 2: 4 These epressins shw that f is cntinuus n the intervals (-00,0), (0,4), (4,5) and (5,00). Since lim f () = lim (5 - ) = 5 -I- 0 = lim f (), lim f () des nt eist, s f is discntinuus (and therefre nt ----+O O O differentiable) at. (d) Frm (a), f is nt differentiable at 4 since [': (4) -I- ['; (4), and frm (c), f is nt differentiable at 0 r (a) If f is even, then fl(-x) = lim f(-+h)-f(-) = lim f[-(-h)]-f(-) n-:«h h----+o h = lim f( - h) - f() = _ lim f( - h) - f() [let Ll = -h] h----+o h h----+-h Therefre, fl is dd. = _ lim f ( + Ll) - f () = - fl () ~----+O Ll (b) If f is dd, then fl(-x) = lim f(-+h)-f(-) = lim f[-(-h)]-f(-) h----+o h h----+o h = lim - f( - h) + f() = lim f( - h) - f() [let Ll = -h] h----+o h h----+-h Therefre, fl is even. = lim f(+ll)-f() =fl() ~----+O Ll 56. (a) T (b) The initial temperature f the water is clse t rm temperature because f the water that was in the pipes. When the water frm the ht water tank starts cming ut, dt/ dt is large and psitive as T increases t the temperature f the water

13 CHAPTER 2 REVIEW D 141 in the tank. In the net phase, dt/ dt = 0 as the water cmes ut at a cnstant, high temperature. After sme time, dt/ dt becmes small and negative as the cntents f the ht water tank are ehausted. Finall, when the ht water has run ut, dt/ dt is nce again 0 as the water maintains its (cld) temperature. (c) dt/dt In the right triangle in the diagram, let ~ be the side ppsite angle and ~ the side adjacent angle. Then the slpe f the tangent line f! is m = D../ D.. = tan. Nte that 0 < < ~. We knw (see Eercise 17) that the derivative f f () = 2 is f' () = 2. S the slpe f the tangent t the curve at the pint (1,1) is 2. Thus, is the angle between 0 and ~ whse tangent is 2; that is, = tan-1 2 ~ Review CONCEPT CHECK 1. (a) lim f() = L: See Definitin and Figures 1 and 2 in Sectin a (b) lim f () = L: See the paragraph after Definitin and Figure 9(b) in Sectin a+ (c) lim f() = L: See Definitin and Figure 9(a) in Sectin a (d) lim f () = 00: See Definitin and Figure 12 in Sectin a (e) lim f() = L: See Definitin and Figure 2 in Sectin In general, the limit f a functin fails t eist when the functin des nt apprach a fied number. Fr each f the fllwing functins, the limit fails t eist at = 2. 2 L The left- and right-hand There is an There are an infinite limits are nt equal. infinite discntinuit. number f scillatins. =2 3. (a)-(g) See the statements f Limit Laws 1-6 and 11 in Sectin 2.3.