CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
2.3 PRECISE DEFINITION OF A LIMIT
56. (a) 1œ lim œ œ Ê lim f(x)œ4.
xÄ c# xÄ c#
f(x)
x lim x
lim f(x) lim f(x)
# x # x
x
Ä# Ä#
Ä# %
(b) 1œ lim œ lim lim œ lim Ê lim œ 2.
xÄ c# xÄ c# xÄ c# xÄ c# xÄ c#
f(x) f(x) f(x) f(x)
x# ’ x “ ’ x"“ ’ x “ ˆ ‰#" x
57. (a) 0œ3 0† œ’ lim “ ’ lim (x2)“œ lim ’Š ‹(x2)“œ lim [f(x)5]œ lim f(x)5
xÄ # xÄ # xÄ # xÄ # xÄ #
f(x) 5 f(x) 5
x x
# #
lim f(x) 5.
Ê œ
xÄ #
(b) 0œ4 0† œ’ lim “ ’ lim (x2) “ Ê lim f(x)œ5 as in part (a).
xÄ # xÄ # xÄ #
f(x) 5
x
#
58. (a) 0œ1 0† œ’ lim “ ’ lim x“ œ’ lim “ ’ lim x “œ lim ’ †x “œ lim f(x). That is, lim f(x
xÄ ! xÄ ! xÄ ! xÄ ! xÄ ! xÄ ! xÄ !
f(x) f(x) f(x)
x# x# x#
# # # )œ0.
(b) 0œ1 0† œ’ lim “ ’ lim x“œ lim ’ †x“œ lim . That is, lim œ0.
xÄ ! xÄ ! xÄ ! xÄ ! xÄ !
f(x) f(x) f(x) f(x)
x# x# x x
59. (a) lim x sin 0
xÄ ! "x œ
(b) Ÿ1 sin "x Ÿ1 for xÁ0:
x0 Ê Ÿ x x sin "x Ÿx Ê lim x sin "x œ0 by the sandwich theorem;
xÄ !
x0 Ê x x sin "x x Ê lim x sin "x œ0 by the sandwich theorem.
xÄ !
60. (a) lim x cos 0
xÄ ! # ˆ ‰x"$ œ
(b) Ÿ1 cos ˆ ‰x"$ Ÿ1 for xÁ0 Ê Ÿ x# x cos # ˆ ‰x"$ Ÿx # Ê lim x cos # ˆ ‰x"$ œ0 by the sandwich
xÄ !
theorem since lim x 0.
xÄ ! #œ
2.
Step 1: kx Ê Ê # 2k $ $ x 2 $ $ x $ 2 Step 2: œ$ 2 1 Ê œ$ 1, or $ œ2 7 Ê œ$ 5.
The value of which assures x$ k Ê2k $ 1 x 7 is the smaller value, $œ1.
3.
Step 1: xk ( 3)k Ê $ Ê $ $ x $ $ 3 x $ 3 Step 2: œ Ê œ$ 3 7# $ #", or $ $ œ Ê œ#" $ 5#.
The value of which assures x$ k ( 3)k Ê $ 7# x #" is the smaller value, $œ#". 4.
Step 1: x¸ ˆ 3#‰¸ Ê Ê $ $ x 3# $ $ 3# x $ 3# Step 2: œ Ê œ #$ 3# 7# $ , or $ œ Ê œ3# #" $ 1.
The value of which assures x$ ¸ ˆ 3#‰¸ Ê $ 7# x #" is the smaller value, $œ ". 5.
Step 1: x¸ "#¸ Ê Ê $ $ x "# $ $ "# x $ "# Step 2: œ$ "# 49 Ê œ$ 18", or $ œ"# 47 Ê œ$ 14".
The value of which assures x$ ¸ "#¸ Ê$ 49 x 47 is the smaller value, $œ18". 6.
Step 1: kx Ê Ê 3k $ $ x 3 $ $ 3 x $ 3
Step 2: $ œ$ 2.7591 Ê œ$ 0.2409, or $ $ œ3.2391 Ê œ$ 0.2391.
The value of which assures x$ k Ê3k $ 2.7591 x 3.2391 is the smaller value, $œ0.2391.
7. Step 1: kx Ê Ê 5k $ $ x 5 $ $ 5 x $ 5
Step 2: From the graph, œ$ 5 4.9 Ê œ$ 0.1, or $ œ5 5.1 Ê œ$ 0.1; thus $œ0.1 in either case.
8. Step 1: kx ( 3)k Ê Ê $ $ x 3 $ $ 3 x $ 3
Step 2: From the graph, œ $ 3 3.1 Ê œ$ 0.1, or $ œ 3 2.9 Ê œ$ 0.1; thus $œ0.1.
9. Step 1: kx Ê Ê 1k $ $ x 1 $ $ 1 x $ 1
Step 2: From the graph, œ$ 1 169 Ê œ$ 167, or $ œ1 2516 Ê œ$ 169; thus $œ167. 10. Step 1: xk Ê Ê 3k $ $ x 3 $ $ 3 x $ 3
Step 2: From the graph, œ$ 3 2.61 Ê œ$ 0.39, or $ œ3 3.41 Ê œ$ 0.41; thus $œ0.39.
11. Step 1: xk Ê Ê 2k $ $ x 2 $ $ 2 x $ 2
Step 2: From the graph, œ$ 2 È3 Ê œ $ 2 È3¸0.2679, or $ œ2 È5 Ê œ$ È5 ¸2 0.2361;
thus $œÈ52.
12. Step 1: xk ( 1)k Ê Ê $ $ x 1 $ $ 1 x $ 1
Step 2: From the graph, œ $ 1 È#5 Ê œ$ È5 2# ¸0.1180, or $ œ 1 È#3 Ê œ$ 2#È3 ¸0.1340;
thus $œ È5 2# .
13. Step 1: xk ( 1)k Ê Ê $ $ x 1 $ $ 1 x $ 1
Step 2: From the graph, œ $ 1 169 Ê œ ¸$ 79 0.77, or $ œ 1 1625 Ê 259 œ0.36; thus $œ259 œ0.36.
14. Step 1: x¸ "#¸ Ê Ê $ $ x "# $ $ "# x $ "#
Step 2: From the graph, œ$ "# 2.011 Ê œ $ 12 #.01" ¸0.00248, or $ œ"# 1.991 Ê œ$ 1.991 ¸"# 0.00251;
thus $œ0.00248.
15. Step 1: (xk 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01 Step 2: xk Ê Ê Ê œ4k $ $ x 4 $ $ 4 x $ 4 0.01.$
16. Step 1: (2xk 2) ( 6)k0.02 Ê k2x 4k 0.02 Ê 0.022x 4 0.02 Ê 4.022x 3.98
Ê 2.01 x 1.99
Step 2: xk ( 2)k Ê Ê Ê œ$ $ x 2 $ $ 2 x $ 2 0.01.$
17. Step 1: ¹Èx " 1 ¹ 0.1 Ê 0.1Èx " 1 0.1 Ê 0.9Èx 1 1.1 Ê 0.81 x 1 1.21
Ê 0.19 x 0.21
Step 2: kx Ê 0k $ $ x $. Then, œ !Þ"* Ê œ !Þ"*$ $ or $œ !Þ#"; thus, $œ0.19.
18. Step 1: ¸Èx"#¸0.1 Ê 0.1Èx "# 0.1 Ê 0.4Èx0.6 Ê 0.16 x 0.36 Step 2: x¸ "4¸ Ê Ê B $ $ x "4 $ $ "4 $ "4.
Then, œ$ "4 0.16 0.09 Ê œ$ or $ œ"4 0.36 0.11; Ê œ$ thus 0.09.$œ
19. Step 1: ¹È19 $ " Ê " x ¹ È19 $ x 1 Ê 2È19 % Êx 419 x 16
x 19 16 15 x 3 or 3 x 15
Ê % Ê
Step 2: xk 10k Ê $ $ x 10 Ê $ $ 10 x $ 10.
Then 10$ œ3 7, Ê œ$ or 10$ œ15 5; Ê œ$ thus 5.$œ
20. Step 1: ¹Èx 7 4¹1 Ê " Èx % 7 1 Ê 3Èx 7 5 Ê 9 x 7 25 Ê 16 x 32 Step 2: xk 23k Ê $ $ x 23 Ê $ $ 23 x $ 23.
Then 23$ œ16 7, Ê œ$ or 23$ œ32 9; Ê œ$ thus 7.$œ
21. Step 1: ¸"x"4¸0.05 Ê 0.05 "x "4 0.05 Ê 0.2 "x 0.3 xÊ 10# 103 or x103 5.
Step 2: xk Ê Ê 4k $ $ x 4 $ $ 4 x $ 4.
Then % œ$ 103 or $œ23, or $ œ4 5 or $œ1; thus $œ23.
22. Step 1: xk # ! Ê 3k .1 0.1x# 3 0.1 Ê 2.9x# 3.1 Ê È2.9 x È3.1 Step 2: x¹ È3¹ Ê $ $ x È3 Ê $ $ È3 x $ È3.
Then $ È3œÈ2.9 Ê œ$ È3È2.9¸0.0291, or $È3œÈ3.1 Ê œ$ È3.1È3¸0.0286;
thus $œ0.0286.
23. Step 1: kx# 4k 0.5Ê 0.5x# 4 0.5Ê3.5x# 4.5ÊÈ3.5k kx È4.5 Ê È4.5 x È3.5, for x near 2.
Step 2: xk ( 2)k Ê Ê # $ $ x 2 $ $ x $ 2.
Then # œ $ È4.5 Ê œ$ È4.5 # ¸0.1213, or $ # œ È3.5 Ê œ # $ È3.5¸0.1292;
thus $œÈ4.5 ¸2 0.12.
24. Step 1: ¸" ( 1)¸ 0.1 0.1 " 1 0.1 " x or x .
x x 10 x 10 11 9 9 11
11 9 10 10 10 10
Ê Ê Ê
Step 2: xk ( 1)k Ê Ê " "$ $ x 1 $ $ x $ .
Then " œ $ 109 , Ê œ$ 9" or $ " œ 1011 Ê œ$ 11"; thus $œ11".
25. Step 1: xka #5b11k " Ê kx#16k1 Ê " x#161 Ê 15x#17 Ê È15 x È17.
Step 2: kx Ê Ê % %4k $ $ x 4 $ $ x $ .
Then 15 % œ$ È Ê œ % 15$ È ¸0.1270, or 17 $ % œÈ Ê œ$ È17 % ¸0.1231;
thus $œÈ17 ¸4 0.12.
26. Step 1: ¸120 5¸ 120 1 4 120 6 x 30 x 20 or 20 x 30.
x " Ê " x & Ê x Ê 4" 120 6" Ê Step 2: xk 24k Ê $ $ x 24 Ê $ $ 24 x $ 24.
Then $ 24œ20 4, Ê œ$ or 24$ œ30 6; Ê œ$ thus 4.Ê œ$
27. Step 1: mxk 2mk0.03 Ê 0.03mx2m0.03 Ê 0.032mmx0.032m Ê 20.03m x 2 0.03m .
Step 2: kx Ê Ê # #2k $ $ x 2 $ $ x $ .
Then # œ # $ 0.03m Ê œ$ 0.03m , or $ # œ # 0.03m Ê œ$ 0.03m . In either case, $œ0.03m . 28. Step 1: mxk 3mk Ê c c mx3m Ê c c 3mmx c 3m Ê 3 mc x 3 mc
Step 2: kx Ê Ê $ B $3k $ $ x 3 $ $ $ .
Then $ œ $ $ mc Ê œ$ mc, or $ $ œ $ mc Ê œ$ mc. In either case, $œ mc.
29. Step 1: (mx¸ b) ˆm# b‰¸ - Ê c mx Ê m# c c m# mx c m# Ê #" mc x #" mc. Step 2: ¸x"#¸ Ê Ê $ $ x "# $ $ "# x $ "#.
Then œ $ "# "# mc Ê œ$ mc, or $ œ "# "# mc Ê œ$ mc. In either case, $œmc. 30. Step 1: (mxk b) (mb)k0.05 Ê 0.05mx m 0.05 Ê 0.05 m mx0.05m
1 x .
Ê 0.05m " 0.05m
Step 2: kx Ê Ê " "1k $ $ x 1 $ $ x $ .
Then " œ " $ 0.05m Ê œ$ 0.05m , or $ " œ " 0.05m Ê œ$ 0.05m . In either case, $œ0.05m . 31. lim (3 2x) 3 2(3) 3
xÄ3 œ œ
Step 1: 3ka 2xb ( 3)k0.02 Ê 0.02 6 2x0.02 Ê 6.02 2x 5.98 Ê 3.01 x 2.99 or
2.99 x 3.01.
Step 2: 0 Ê Ê $ $kx 3k $ $ x 3 $ $ x $ .
Then $ œ$ 2.99 0.01, Ê œ$ or $ $ œ3.01 0.01; Ê œ$ thus 0.01.$œ 32. lim ( 3x ) ( 3)( 1) 2 1
xÄ c1 # œ œ
Step 1: ( 3xk 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99.
Step 2: xk ( 1)k Ê Ê " $ $ x 1 $ $ x $ 1.
Then " œ $ 1.01 0.01, Ê œ$ or $ " œ 0.99 0.01; Ê œ$ thus 0.01.$œ
33. lim lim lim (x 2) 4, x 2
xÄ # xÄ # xÄ #
x 4
x (x 2)
(x 2)(x 2)
#
# œ œ œ # # œ Á
Step 1: ¹Šxx#24‹4¹0.05 Ê 0.05(x(x2)(x2)2) % 0.05 Ê 3.95 x 2 4.05, xÁ2 1.95 x 2.05, x 2.
Ê Á
Step 2: kx Ê Ê # 2k $ $ x 2 $ $ x $ 2.
Then # œ$ 1.95 0.05, Ê œ$ or $ # œ2.05 0.05; Ê œ$ thus 0.05.$œ
34. lim lim lim (x 1) 4, x 5.
xÄ c& xÄ c& xÄ c&
x 6x 5
x 5 (x 5)
(x 5)(x 1)
#
œ œ œ Á
Step 1: ¹Šx#x6x55‹ ( 4)¹0.05 Ê 0.05(x(x5)(x5) ") 4 0.05 Ê 4.05 x 1 3.95, xÁ 5
Ê 5.05 x 4.95, xÁ 5.
Step 2: xk ( 5)k Ê Ê & &$ $ x 5 $ $ x $ .
Then & œ $ 5.05 0.05, Ê œ$ or $ & œ 4.95 0.05; Ê œ$ thus 0.05.$œ
35. lim 1 5x 1 5( 3) 16 4
xÄ c$È œÈ œÈ œ
Step 1: ¹È15x4¹0.5 Ê 0.5È15x 4 0.5 Ê 3.5È15x4.5 Ê 12.25 1 5x20.25 Ê 11.25 5x 19.25 Ê 3.85 x 2.25.
Step 2: xk ( 3)k Ê Ê $ $$ $ x 3 $ $ x $ .
Then $ œ $ 3.85 Ê œ$ 0.85, or $ $ œ 2.25 Ê 0.75; thus $œ0.75.
36. lim 2
xÄ #
4 4
x œ œ#
Step 1: ¸4x2¸0.4 Ê 0.4 4x 2 0.4 Ê 1.6 4x 2.4 Ê 1016 x4 1024 xÊ 104 106 or x53 52. Step 2: kx Ê Ê # #2k $ $ x 2 $ $ x $ .
Then # œ$ 53 Ê œ , $ 3" or $ # œ 5# Ê œ ; $ "# thus .$œ3"
37. Step 1: k(9 Ê Ê Ê % Ê % x) 5k % % 4 x % % 4 x % 4 % x 4 % % x 4 %. Step 2: kx Ê Ê % %4k $ $ x 4 $ $ x $ .
Then œ Ê œ$ 4 % 4 $ %, or $ % œ % Ê œ% $ %. Thus choose $œ%. 38. Step 1: (3xk Ê 7) 2k % % 3x Ê 9 % 9 % 3x * Ê % 3 3% x 3 %3. Step 2: xk Ê Ê 3k $ $ x 3 $ $ 3 x $ 3.
Then œ $ $ 3 %3 Ê œ$ %3, or $ œ 3 3 3% Ê œ$ %3. Thus choose $œ3%.
39. Step 1: ¹Èx 5 2¹ Ê % % Èx # Ê # 5 % % Èx # Ê # 5 % ( %)# # x 5 ( %)#
( ) x ( ) 5.
Ê # %# & # % #
Step 2: xk Ê Ê 9k $ $ x 9 $ $ 9 x $ 9.
Then * œ$ %# % * Ê œ % % $ % %#, or $ * œ%# % * Ê œ % % $ % %#. Thus choose the smaller distance, $œ % % %#.
40. Step 1: ¹È4 x 2¹ Ê % % È4 # Ê # x % % È4 # Ê # x % ( %)# % # x ( %)#
( ) x 4 ( ) ( ) x ( ) .
Ê # % # # %# Ê # % # % # % # % Step 2: xk Ê 0k $ $ x $.
Then œ # $ ( %)# œ % Ê œ % 4 %# % $ % %#, or $œ # ( %)# œ4 4% % #. Thus choose the smaller distance, $œ4% % #.
41. Step 1: For xÁ1, kx# Ê 1k % % x# " Ê " % % x# " Ê% È1 % k kx È1% Ê È" % x È1%near B œ ".
Step 2: kx Ê Ê " "1k $ $ x 1 $ $ x $ .
Then " œ$ È1 Ê œ " % $ È1%, or $ œ " Ê œ " 1 È % $ È % 1. Choose min 1 1 , that is, the smaller of the two distances.
$œ š" È ß% È "% ›
42. Step 1: For xÁ 2, xk # Ê 4k % % x# Ê 4 % 4 % x# Ê4 % È4 % k kx È4%
Ê È4 % x È4%near B œ 2.
Step 2: xk ( 2)k Ê Ê $ $ x 2 $ $ 2 x $ 2.
Then œ % Ê œ % #$ 2 È % $ È % , or $ # œ % Ê œ # % È % $ È %. Choose min$œ šÈ% #ß # % % È %›.
43. Step 1: ¸"x1¸ Ê " Ê " " Ê% % "x % % "x % 1"% x 1"%. Step 2: kx Ê Ê " " 1k $ $ x 1 $ $ x $.
Then " œ$ " "% Ê œ " $ " "% œ" %%, or " œ$ " "% Ê œ$ " "% " œ " %%. Choose $œ " %%, the smaller of the two distances.
44. Step 1: ¸x"#"3¸ Ê % % x"# Ê"3 % "3 % x"# Ê"3 % 133% x"# 1 $3 % Ê " $3 % x# " $3 %
x , or x for x near .
Ê É 3 k kÉ 3 É 3 É 3 È$
1 $% " $% " $% "$%
Step 2: ¹xÈ3¹ Ê $ $ x È3 Ê$ È3 $ x È3$.
Then 3È œ$ É" $3 % Ê œ$ È3É" $3 %, or 3È œ$ É" $3 % Ê œ$ É" $3 % È3.
Choose $œminšÈ3É" $3 %ßÉ" $3 %È3 .›
45. Step 1: ¹Šxx 3#*‹ ( 6)¹ Ê % % (x 3) 6 %, xÁ Ê Ê $ $3 % x 3 % % x % . Step 2: xk ( 3)k Ê Ê $ $ $ x 3 $ $ x $ 3.
Then $ œ $ Ê œ$ % $ %, or $ $ œ $ Ê œ% $ %. Choose $œ%. 46. Step 1: ¹Šxx 1#1‹2¹ Ê % % (x 1) 2 %, xÁ1 Ê " " % x %.
Step 2: kx Ê Ê " " 1k $ $ x 1 $ $ x $.
Then " œ " Ê œ$ % $ %, or " œ " Ê œ$ % $ %. Choose $œ%. 47. Step 1: x1: (4l 2x) l Ê ! 2 % 2 2x% since x Þ1 Thus, 1 !%# x ;
x 1: (6xl l Ê ! Ÿ4) 2 % 6x 6 % since x 1. Thus, " Ÿ x 1 6%. Step 2: kx Ê Ê " 1k $ $ x 1 $ $ x 1 $.
Then 1 œ " $ #% Ê œ$ %#, or " œ $ 1 6% Ê œ$ 6%. Choose $œ6%. 48. Step 1: x !: 2xk Ê 0k % % 2x ! Ê %# x 0;
x 0: ¸x# ! Ê ! Ÿ #¸ % x %. Step 2: xk Ê 0k $ $ x $.
Then œ Ê œ$ #% $ #%, or $œ # Ê œ #% $ %. Choose $œ %#.
49. By the figure, x Ÿx sin "x Ÿx for all x0 and x x sin "x x for x0. Since lim ( x) œ lim xœ0,
xÄ ! xÄ !
then by the sandwich theorem, in either case, lim x sin 0.
xÄ ! "x œ
50. By the figure, x Ÿ# x sin # "x Ÿx for all x except possibly at x# œ0. Since lim x# œ lim x#œ0, then
xÄ !a b xÄ !
by the sandwich theorem, lim x sin 0.
xÄ ! # "x œ
51. As x approaches the value 0, the values of g(x) approach k. Thus for every number %0, there exists a $ ! such that x! Êk 0k $ kg(x) kk %.
52. Write xœ h c. Then ! l l Í x c $ $ x c $, xÁ Í c $ ah cb c $, h Ác c
h , h h .
Í $ $ Á ! Í ! l !l $
Thus, limf x L for any , there exists such that f x L whenever x c
xÄca bœ Í % ! $ ! la b l % ! l l $ f hÍ la l cb L %whenever ! l !l Íh $ limf ha cbœL.
hÄ!
53. Let f(x)œx . The function values do get closer to 1 as x approaches 0, but lim f(x)# œ0, not 1. The
xÄ !
function f(x)œx never gets # arbitrarily close to 1 for x near 0.
54. Let f(x)œsin x, Lœ"#, and x!œ0. There exists a value of x (namely, xœ16) for which sin x¸ "#¸% for any given %0. However, lim sin xœ0, not . The wrong statement does not require x to be arbitrarily close to
xÄ ! "#
x . As another example, let g(x)! œsin , L"x œ "#, and x!œ0. We can choose infinitely many values of x near 0 such that sin "x œ "# as you can see from the accompanying figure. However, lim sin fails to exist. The"x
xÄ !
wrong statement does not require all values of x arbitrarily close to x!œ0 to lie within %0 of Lœ"#. Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ0. If we choose %"4 we cannot satisfy the inequality sin ¸ "x "#¸ % for all values of x sufficiently near x!œ0.
55. kA * Ÿk 0.01 Ê 0.01Ÿ1ˆ ‰x# # Ÿ9 0.01 Ê 8.99Ÿ14x# Ÿ9.01 Ê 14(8.99)Ÿx# Ÿ14(9.01) 2 x 2 or 3.384 x 3.387. To be safe, the left endpoint was rounded up and the right Ê É8.991 Ÿ Ÿ É9.011 Ÿ Ÿ
endpoint was rounded down.
56. VœRI I 5Ê VR œ Ê ¸VR ¸Ÿ0.1 Ê 0.1Ÿ 120R Ÿ5 0.1 Ê 4.9Ÿ120R Ÿ5.1 Ê 1049 1 0R# 1051 Ê R(120)(10)51 Ÿ Ÿ (120)(10)49 Ê 23.53Ÿ ŸR 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) $ x 1 0 Ê " $ x 1 Ê f(x)œx. Then f(x)k œ œ œ2k kx 2k 2 x 2 1 1. That is, f(x) 2 1 no matter how small is taken when x 1 lim f(x) 2.
k k "# $ " $ Ê Á
xÄ1
(b) 0 Ê " " Êx 1 $ x $ f(x)œ x 1. Then f(x)k œ1k k(x œ1) 1k k kx œ x 1. That is, f(x) 1 1 no matter how small is taken when x lim f(x) 1.
k k $ " " Ê$ Á
xÄ1
(c) ! Ê " $ x 1 $ x 1 Ê f(x)œx. Then f(x)k 1.5kœ kx 1.5kœ1.5 x 1.5 œ1 0.5.
Also, ! Êx 1 $ 1 " Êx $ f(x)œ x 1. Then f(x)k 1.5kœk(x 1) 1.5kœ kx 0.5k x 0.5 0.5 0.5. Thus, no matter how small is taken, there exists a value of x such that œ " œ $
x 1 but f(x) 1.5 lim f(x) 1.5.
$ $ k k "# Ê Á
xÄ1
58. (a) For 2 Êx 2 $ f(x)œ2 Ê f(x)k œ4k 2. Thus for %2, f(x)k 4k % whenever 2 x 2 $ no matter how small we choose $0 Ê lim f(x)Á4.
xÄ #
(b) For 2 Êx 2 $ f(x)œ2 Ê f(x)k œ3k 1. Thus for %1, f(x)k 3k % whenever 2 x 2 $ no matter how small we choose $0 Ê lim f(x)Á3.
xÄ #
(c) For 2 $ x 2 Ê f(x)œx so f(x)# k œ2k kx#2 . No matter how small k $0 is chosen, x is close to 4# when x is near 2 and to the left on the real line Ê xk #2 will be close to 2. Thus if k %1, f(x)k 2k % whenever 2 $ x 2 no mater how small we choose $0 Ê lim f(x)Á2.
xÄ #
59. (a) For 3 Ê$ x 3 f(x)4.8 Ê f(x)k 4k 0.8. Thus for %0.8, f(x)k 4k % whenever 3 $ x 3 no matter how small we choose $0 Ê lim f(x)Á4.
xÄ $
(b) For 3 Êx 3 $ f(x)3 Ê f(x)k 4.8k 1.8. Thus for %1.8, f(x)k 4.8k % whenever 3 x 3 $ no matter how small we choose $0 Ê lim f(x)Á4.8.
xÄ $
(c) For 3 $ x 3 Ê f(x)4.8 Ê f(x)k 3k 1.8. Again, for %1.8, f(x)k 3k % whenever $ $ x 3 no matter how small we choose $0 Ê lim f(x)Á3.
xÄ $
60. (a) No matter how small we choose $0, for x near 1 satisfying " " $ x $, the values of g(x) are near 1 Ê g(x)k 2 is near 1. Then, for k %œ"# we have g(x)k 2k "# for some x satisfying
x , or x 1 lim g(x) 2.
" " $ $ ! Êk k $ Á
xÄ c1
(b) Yes, lim g(x) 1 because from the graph we can find a such that g(x) 1 if x ( 1) .
xÄ c1 œ $ ! k k % ! k k$
61-66. Example CAS commands (values of del may vary for a specified eps):
: Maple
f := x -> (x^4-81)/(x-3);x0 := 3;
plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" );
L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01,
color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" );
q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q);
plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" );
s in [0.1, 0.005, 0.001 ] do # (e) for ep
q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 );
delta := abs(x0-q);
head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta );
print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head ));
end do:
(assigned function and values for x0, eps and del may vary):
Mathematica Clear[f, x]
y1:œ L eps; y2:œ L eps; x0œ1;
f[x_]:œ(3x2(7x1)Sqrt[x]5)/(x1) Plot[f[x], {x, x00.2, x00.2}]
L:œ Limit[f[x], xÄx0]
epsœ0.1; delœ0.2;
Plot[{f[x], y1, y2},{x, x0del, x0del}, PlotRangeÄ{L2eps, L2eps}]