chapter 5: introduction to limits - achsprecalc · 2019-08-31 · cpm educational program © 2012...

CPM Educational Program © 2012 Chapter 5: Pre-Calculus with Trigonometry

Chapter 5: Introduction to Limits Lesson 5.1.1 5-1. 3. Decreases 4. Decreases 5. y = 1

x 5-3. a. y = kx2 b. 3 = k !22

34 = k

y = 34 x

2

c. y = 34 ! ("3)

2 # y = 34 !9 =

274

Review and Preview 5.1.1 5-4. f (x) = 1

x f (x) = 2x

5-5. f (x) = k(x2 ! 4)

2 = k(32 ! 4)2 = 5k25 = k

f (4) = 25 (4

2 ! 4) = 25 (16 ! 4) =

25 "12 =

245 = 4.8

5-6. y = k

x3

0.35 = k!0.753

0.35 " !0.9086 = !0.318 = k

y = !0.318x3

y = !0.31817.113 = !0.318

2.5768 = !0.123


5-7. a. sin ! "

3( ) = ! sin "3 = ! 3

2

cos ! "3( ) = cos "3 = 1

2

tan ! "3( ) = ! tan "

3 = ! 31 = ! 3

csc ! "3( ) = 1

sin(! " 3)= ! 1

32

= ! 23= ! 2 3

3

sec ! "3( ) = 1

cos ! "3( ) =

112= 2

cot ! "3( ) = 1

tan(! " 3)= ! 1

3= ! 3

3

b. sin !3"( ) = ! sin 3" = !0 = 0cos !3"( ) = cos 3" = !1tan !3"( ) = ! tan 3" = !0 = 0

csc !3"( ) = 1sin(!3" )

= 10# und.

sec !3"( ) = 1cos(!3" )

= 1!1

= !1

cot !3"( ) = 1tan(!3" )

= 10# und.

5-8. Length of interval = 3! (!1) = 4

h(x)!1

3

" dx = f (x)!1

3

" dx + 5 # 4 = 17 + 20 = 37

5-9. Volume of Florida grapefruit = 4

3 ! " r3

Diameter of Texas grapefruit = 2r where r = diameter of Florida grapefruit.

Volume of Texas grapefruit = 43 ! 2r( )3 = 4

3 !8r3 = 8 " 43 ! " r3( )

# Texas grapefruit is 8 times as big as Florida grapefruit, but only costs 7 times as much, thus making it the better deal.

5-10. a. 7!

6 ,11!6 b. 5!

6 ,11!6

5-11. a. (x+2)(x+2)(x+2)

(x+2) = (x + 2)(x + 2) = (x + 2)2 x ! "2

b. (x!2)(x!3)(x!2)(x!2) =

x!3x!2 x " 2

c. 2(x+y)(x!y)(x+y) = 2 x ! y( ) x " !y

5-12. y = k

x + a!!!!!x = length of tube!!!!!a = cross-section area of tube


Lesson 5.1.2 5-13. Numerator and denominate must be polynomials. x and x are not polynomials. 5-14. a. 20

17 = 1317 b. 19

7 = 14+57 = 14

7 + 57 = 2

57

5-15. a. f (x) = 1

x b. f (x) = 1x+2 + 3

This is the graph of f (x) shifted 2 units to the left and 3 units up. 5-16. g(x) = 1+ 3

x!1 a. b. Graph b. h(x) = 2x!5

x!3 = 2(x!3)+1x!3 = 2(x!3)

x!3 + 1x!3 = 2 +

1x!3

5-17. It is the graph of y = 1

x shifted up 2 units and right 3 units. y = 1x!3 + 2

5-18. Multiply the numerator and denominator by xy2 . 5-19.

a. x2

x2! x"1y2 +yx"2y"1+xy2

= yy !

xy2 +x2yy"1+x3y2

= xy3+x2y2

1+x3y3 b. y3

y3! x2y3+y"3

y"3+x= x2y6 +1

1+xy3


5-20.

a. u = 1x +

xy2

= y2

xy2+ x2

xy2= x2 +y2

xy2 b. xy2

x2 +y2

c. x2 +y2

xy2!"#

$%&'2

= xy2

x2 +y2!"#

$%&2= (xy2 )2

(x2 +y2 )2= x2y4

(x2 +y2 )2

5-21.

a. u = x + y!1 = x + 1y =

xy+1y

u!1 = 1u =

1xy+1 y =

yxy+1

b. u = ab!1 + a!2b = ab +

ba2

= a3+b2

a2b

u!1 = 1u =

1a3+b2 a2b

= a2ba3+b2

c. u = ab!1 + a!2b = ab +

ba2

= a3+b2

a2b

u!1 = 1u2

= 1

a3+b2

a2b"#$

%&'2 = a2b

a3+b2( )2 = a4b2

(a3+b2 )2

Review and Preview 5.1.2 5-22. Example:

Vertical asymptote at x = !7 " y = 1x+7

Horizontal asymptote at x = 5" y = 1x+7 + 5

5-23.

2x(3x!1)(3x!1) ! 1(3x!1)(3x!1) +

3(3x!1) =

6x2 !2x!3x+1+3(3x!1) = 6x2 !5x+4

(3x!1)

5-24. a. Any angle in the 4th quadrant will satisfy this. Examples: 11!

6 , " !3

b. sin x = ! 12 " x = 7#

6 ,11#6

c. Any angle in the 3rd quadrant will satisfy this. Examples: 7!6 ,5!4

d. Approximately 6!7 e. Not possible, it does not satisfy the Fundamental Pythagorean Identity.

cos2 ! + sin2 ! = 1

(0.9)2 + (0.8)2 = 0.81+ 0.64 = 1.45 " 1


5-25. a. y = k

x2 !3

1 = k!2

!2 = k

y = !2x2 !3

b. f (x) = !2x2 !3

f (!3) = !2(!3)2 !3

= !26 = ! 1

3

f (2) = !222 !3

= !21 = !2

f 12( ) = !2

(1 2)2 !3= !21 4!12 4 =

!2!11 4 = !2 " ! 4

11( ) = 811

f ! 1a( ) = !2

(!1 a)2 !3= !21 a2 !3a2 a2

= !2 " ! a2

1!3a2( ) = !2a2

1!3a2

5-26. a. A(d) = k

d2 b. A(d) = k

(3d )2= k9d2

= 19 !

kd2

It decreases 19 as much.

5-27. a. sec! = "2

1cos! = "2

1 = "2 cos!

" 12 = cos!

! = 2#3 ,

4#3

b. cot! = 31tan! = 3

1 = 3 tan!13= tan!

! = "6 ,

7"6

5-28.

a. 3 x3 + 4x2x !1 x2

= 3+ 4x4 x3

2x3 !1 x2= 3+ 4x

4

x3" x2

2x3 !1= 3+ 4x

4

2x4 ! x

b. 1 x2 + y2

x2 !1 y2= 1+ x

2y2 x2

x2y2 !1 y2= 1+ x

2y2

x2" y2

x2y2 !1= y2 + x2y4

x4y2 ! x2

c. 1 x4y2 +1 x3y1 x !1 x3

= 1+ xy x4y2

x2 !1 x3= 1+ xyx4y2

" x3

x2 !1= 1+ xyx3y2 ! xy2


Lesson 5.1.3 5-29. See graph at right. The x-intercept of the first function is the x-value of the

vertical asymptote of the second (x = 3) . 5-30. See graph at right. a. x = 2, x = !2 b. Yes, y = 0 c. x < !2 or x > 2 d. !2 < x < 2 5-31. See graph at right. f (x) = sec(x)! a. (2!n,1) and (! + 2!n, "1), n an integer .

b. x ! "2 + "n, n an integer

c. Range: cos x! ["1,1]; sec x! y # 1 or y $ "1 5-32. See graph at right. 1

f (x)

5-33. See graph below. Cosecant: Domain: x ! "n , n an integer Range: y ! "1 or y # 1


Review and Preview 5.1.3 5-34. a. T = 0.0725P b. k = 0.0725 5-35. a. f (100) = 2(100)+3

100+1 = 203101 = 2.0099

f (1000) = 2(1000)+31000+1 = 2003

1001 = 2.0010

f (10000) = 2(10000)+310000+1 = 20003

10001 = 2.0001

b. y = 2

c. Yes, since x ! "1, x = "1 is the vertical asymptote. d. f (x) = 2x+3

x+1 = 2 + 1x+1 This is the graph of 1

x , shifted one unit left and two units up . 5-36. Possible answer: g(x) = (x +1)(x ! 2) 5-37. 3

4 x2 5-38. c +1 = 25 !10c + c2

0 = c2 !11c + 240 = (c ! 8)(c ! 3)c = 3, 8

Check 3+1 = 5 ! 3 = 2

8 +1 = 3 " 5 ! 8 = !3Solution : c = 3

5-39. a(x) = 1

h(x) , b(x) =1f (x)

5-40. a. d = (5 ! (!2))2 + (!2 ! 5)2

d = 72 + 72 = 98 = 7 2

b. slope = m = !2!55!(!2) =

!77 = !1

Point-Slope Form: y ! 5 = !(x ! (!2))y ! 5 = !(x + 2)

OR y ! (!2) = !(x ! 5)y + 2 = !(x ! 5)

Slope-Intercept Form: y + 2 = !(x ! 5)

y + 2 = !x + 5y = !x + 3


Lesson 5.2.1 5-41. See graph at right. f (x) = 1

x!2 + 3 lim

x!"f (x) = 3

5-42. a. He thinks that he will get to 3. b. She thinks that she will get to 1. 5-43. a. She thinks she is getting infinitely far up. b. + means approaching from the right. – means approaching from the left. c. lim

x!3"f (x) = 1 , lim

x!3+f (x) = 3

5-44. See graph at right. f (x) = 2

x!4 !1 a. lim

x!"f (x) = #1

b. limx!4"

f (x) = "#

c. limx!4+

f (x) = +"

5-45.

a. limx!"

3x3( ) = 0 b. lim

x!" 5x4( ) = " c. lim

x!" # 9

x( ) = 0

d. limx!"

sin x No limit.

sin x oscillates between –1 and 1, but does not approach any specific number. e. An example is f (x) = cos x . 5-46. a. g(x) = 2x+4

x+3 = 2(x+3)!2(x+3) = !2

x+3 + 2 b. lim g(x)

x!"= 2,!lim g(x)

x!#3#= ",!lim g(x)

x!#3+= #"

5-47. x must approach ! from the left and must approach !" from the right, therefore we do

not need extra notation. x cannot approach ! from the right.


Review and Preview 5.2.1 5-48. See graph at right. lim

x!2f (x) = 22 + 3 = 7

5-49. In the first, the limit is approaching –2, and in the second, the limit is approaching 2 from

the left side. 5-50. y = k

x2

limx!"

1x2

= 0

As x!" , 1x2

gets smaller and smaller.

5-51. a. If u = sin! " u3#4u

u2= 3 b. u3 ! 4u = 3u2

u3 ! 3u2 ! 4u = 0u(u2 ! 3u ! 4) = 0u(u ! 4)(u +1) = 0

u = !1, 0, 4But u = sin" # 0So u = !1, 4

c. sin! = "1 or sin! = 4

! = 3#2 , !!!!!!!! sin! $ 4

5-52. a. 16

3 b. 3n2

10n2 !5

5-53. a. S(r) = kr2 b. 16! = k " (2)2

16! = 4kk = 4!

S(r) = 4! " r2

SA = S(3) = 4! " 33 = 36! 5-54. a. 2 b. 5.152


Lesson 5.2.1 5-55. f. If the procedures are followed accurately, all the last acute angles should be very close to 60º. g. They should say 60º. Once this angle is reached, the obtuse angle will be 120º. This is, of

course, interesting since theoretically this angle is never reached. When it is bisected the resulting angle will equal 60º. Do not simply accept as justification that the pattern continues.

5-56. a. They are alternate interior angles. b. 180! ! 20! = 160!; a2 = 160!

2 = 80! c. d. 60.000 is obtained by n = 18.

Students can repeat this process by using the ANS key on their calculator. Enter 20 and push Í. At the next line use (180 – Z)/2. Repeatedly hitting Í will generate the sequence.

e. lim

x!"an = 60 f. The measure approaches 60º.

5-57.

n Angle an Measure of angle a1 Error: 60 – angle measure

1 a1 20 40

2 a2 80 –20

3 a3 50 10

4 a4 65 –5

5 a5 57.5 2.5

6 a6 61.25 –1.25

7 a7 59.375 0.625

8 a8 60.3125 –0.3125

n Measure of an

1 20 2 80 3 50 4 65 5 57.5 6 61.25 7 59.375 8 60.3125


Review and Preview 5.2.2 5-58.

g(x) = ax 10 << a 1>a

Example Sketch

limx! "

g(x) 0 ! limx! "#

g(x) ! 0 5-59. See graph at right. f (x) = 2x ! 3

limx!"

(2x # 3) = " # 3 = "

limx! #"

(2x # 3) = 0 # 3 = #3

5-60. The limit is 60; it will never actually equal 60 unless the original acute angle was 60º. 5-61. a. S = kp

m b. 60 c. The score would be infinite.

5-62.

sin 60! = 0.866

sin !3 = 0.866

They are the same because 60! = !

3 .

5-63. a. 4 and 5 should be multiplied, not added. log 4 + log 5 = log 20 b. Base should not change. log3 7 + log3 7 = log3 49 c. Log of a product is the SUM of the logs. log(4 !5) = log 4 + log 5 d. No rule for log(4 + 5) . 5-64. f (x) = 1

x!2 ! 3 , asymptotes at x = 2 and y = !3


5-65. a. Closest that Jade can be is when sin! = "1 . ! d = 25("1) + 35 = 10 feet. Farthest away

that Jade can be is when sin! = 1 . ! d = 25(1) + 35 = 60 feet b. 35 = 25 sin ! "t

15( ) + 350 = 25 sin ! "t

15( )0 = sin ! "t

15( )sin#1 0 = sin#1 sin ! "t

15( )( )2! = ! "t

15

30! = ! " tt = 30 seconds

c. 60 = 25 sin ! "t15( ) + 35

25 = 25 sin ! "t15( )

sin#1 1 = sin#1 sin ! "t15( )( )

15! " !2 = ! "t

15 " 15!

t = 152 = 7.5 seconds

Second time = 30 sec + 7.5 sec= 37.5 sec

d. 5-66.

a. x2y3 +1 x2

1 y3 + x= x4y3 +1 x2

1+ xy3 y3= x4y3 +1

x2! y3

1+ xy3= x4y6 + y3

x2 + x3y3

b. 1 x + x2

1 x2 + x= 1+ x3 x1+ x3 x2

= 1+ x3

x! x2

1+ x3= x


Lesson 5.2.3 5-67. a. He thinks the height will be y = 2 . b. She thinks she will approach y = 2 . Benny: lim

x!3"f (x) = 2 ; Bertha: lim

x!3+f (x) = 2

c. f (x) =!x2 + 4 x " 32x ! 3 x > 3

#$%&

limx!1+

f (x) = 3

5-68. Right- and left-hand limits must be equal for a limit to exist. 5-69.

a. f (x) = x2 ! 2 for x " 2

! 12 (x ! 2) + 4 for x > 2

#$%

&% b. i. !" , ii. ! , iii. 4, iv. 2

c. It does not exist since the left hand limit and right hand limit are not the same. 5-70.

Example: f (x) =!x2 + 4 x " 32x ! 3 x > 3

#$%&

5-71. a. lim

x! 3"f (x) = 6.2, lim

x! 3+f (x) = 6.2

b. The limit exists since the limits as x approaches 3 from both the left and right sides are equal.

Review and Preview 5.2.3 5-72. lim

x!3"f (x) = "2, lim

x!3+f (x) = "1

limx!3

f (x) does not exist as the limits from the left and right are not equal.

5-73. See graph at right. f (x) = ! 1

x!3 + 2

limx!"#

f (x) = limx!"#

" 1x"3 + 2 = "0 + 2 = 2

x

y

f(x)


5-74. See graph at right. 5-75.

limx!"

1x2 # 4

= 0

5-76.

x + x!3

x!2 + x!1"#$

%&'

1

= x +1 x3

1 x2 +1 x= x4 +1 x3

1+ x x2= x4 +1

x3( x2

1+ x= x4 +1x + x2

5-77.

42b6 = a, 15 = 42

b6 !b9

15 = 42b3, 514 = b3

b = 514

3 " 0.709, a = 425

143( )6

" 329.28

5-78. a. Possible estimate: lim

x!1g(x) " 0.42062

b. See graph at right. limit ≈ 0.42061984 5-79. See graph at right. a. lim

x!3x2 " 9 = 0

b. limx!3+

1x2 "9

= # c. limx!3"

1x2 "9

= "# d. limx!3

1x2 "9

Limit does not exist as left and right limits are not equal.


Lesson 5.2.4 5-80. a. Yes, the limit exists, since the limit from the left and right are equal. lim

x!3 g(x) = 2

b. limx!3

h(x) = 2

c. h(3) = 3 d. No, the limit is the number f(x) approaches as x gets closer and closer to a. 5-81. a. No, f(–3) and the limit do not exist. b. Continuous c. No, limit ≠ f(2). d. No, f(4) does not exist. 5-82. a. b. c. 5-83. a. x ! 0 b. See graph at right. c. lim

x!0sin x = 0, lim

x!0x = 0

The numerator and denominator of f(x) are both approaching 0. d. lim

x!0sin xx = 1

5-84. a. n! , n " 0 b. f (x) = sin x

x oscillates as x!" but gets closer to y = 0 . c. The line y = 0 IS a horizontal asymptote. d. lim

x!0sin xx = 1

x

y

a x

y

a x

y

a


Review and Preview 5.2.4 5-85. See graph at right. 1

f (x) =1

x3!3x!2 x " !1, 2

Vertical Asymptotes of f (x) # x = !1, x = 2

5-86. lim

x!3x

x2 +1= 332 +1

= 310

5-87. lim

x!" 2sin xx = sin" 2

" 2 = 1" 2 =

2"

5-88.

k10 = x

10!h

k(10 ! h) = 10xk(10!h)

10 = x (radius of cross-section)

Area of cross-section = ! k(10"h)10( )2

= !100 k

2(10 " h)2

5-89. a. cos 60

! = 12 b. sin 3!4 = 2

2

c. sin 3!2 = "1 d. cos 90! = 0


5-90. a. csc2 x + sec2 x ! tan2 x ! cot2 x = 2

(csc2 x ! cot2 x) + (sec2 x ! tan2 x) = 21+1 = 22 = 2

b. sec x!11!cos x = sec x

sec x!11!cos x

1+cos x1+cos x( ) = sec x

sec x+1!1!cos x1!cos2 x

= sec x

sec x!cos xsin2 x

= sec x

1!cos2 xcos xsin2 x

= sec x

sin2 xcos xsin2 x

= sec x

1cos x = sec x

5-91. log2 A + log2 B = 4 ! log2(AB) = 4

log3 A " log3 B = 2 ! log3AB( ) = 2

24 = AB 32 = AB( ) = 16

B2( )AB = 16 9 = 16

B2

A = 16B B = 4

3 A = 164/3 = 12

5-92. a. A = kw

5!GPA

14.40 = 10k5!3.5 =

10k1.5 = 20

3 k" k = 2.16

A = 2.16w5!GPA

b. 20 = 2.16(15)5!GPA

5 !GPA = 2.16(15)20 = 1.62

GPA = 5 !1.62 ! 3.38

5-93. For x = –2: 2(!2) + a = (!2)2 ! 2

!4 + a = 2a = 6

For x = 3: 32 ! 2 = !3+ b7 = !3+ b10 = b


Lesson 5.2.5 5-94. i. lim

x!"#f (x) = 0 ii. lim

x!"2"f (x) = 2 iii. lim

x!"2+f (x) = 3

iv. limx!"2

f (x) = no limit v. limx!2

f (x) = 3 vi. limx!"

f (x) = #"

5-95. lim

x!"#f (x) = 0 , lim

x!"4"f (x) = # , lim

x!"4+f (x) = # , lim

x!"4f (x) does not exist

limx!"1

f (x) = 1 , limx!1"

f (x) = "1 , limx!1+

f (x) = 2 , limx!1

f (x) does not exist

5-96. When x is very small, AC and AB! are almost the same length and parallel. Therefore as

x changes, the lengths of AC and AB! change at similar rates. 5-97. a. cos(x) approaches 1 while x approaches 0, so this is 1

little = big . b. x approaches 0 while 2x approaches 1, so this is

1big =little .

5-98. a. x ≠ 0 b. Both approach 0. c. See graph at right. lim

x!0f (x) = 0

d. Yes, y = 0 . Review and Preview 5.2.5 5-99. See graph at right.

limx!0

1" cos xx2

= 0.5

5-100. See graph at right. a. lim

x!2+f (x) = 3

b. limx!2"

f (x) = 1

c. limx!2

f (x)

The limit does not exist because the limits from the left and the right sides are different.


5-101. See graph at right. Example (one of many): 5-102. a. 22 = 4

limx!2

f (x) = 4Therefore the graph is continuous at x = 2.

b. limx!5

f (x) = 4

5-103. See graph at right. Find the area under the curve. Distance traveled in the first five seconds: feet50510 =! Distance traveled in the t – 5 seconds: 1

2 (10 + 2t)(t ! 5) = 200 feet(10 + 2t)(t ! 5) = 40010t ! 50 + 2t2 !10t = 4002t2 = 450t2 = 225t = 15

5-104. a. y = 2 + 1

2x!3 b. y = 2, x = 32 c. lim

x!"4x /x#5/x2x /x#3/x = limx!"

42 = 2

d. Vertical asymptote because x ! 1.5 . We know the function will be approaching positive or negative infinity. f (1.51) = 4(1.51)!5

2(1.51)!3 = 52!!"!! limx#1.5+

= +$

5-105. a. See graph at right. b. x ! 1 c. x ! "1, 1, 3 d. lim

x!"#h(x) = 0

5-106. g(x) is the graph of f (x) with a vertical shift of 3 units. Therefore Bertha thinks that she

is getting closer to Benny’s height + 3, or 2 + 3 = 5 units. 5-107.


a. h = kdg b. 2.4 = k(70!50)

0.30.72 = 20k0.036 = k

c. h = 0.036(70!30)1.5

= 0.036(40)1.5

= 1.441.5 = 0.96 BTU


Closure Problems CL 5-108. a. lim

x!1"f (x) # 0.999+2

0.999"1 = "$!!!!! limx!1+

f (x) # 61.00012 "1

= $ Therefore limx!1

f (x) does not exist.

b. limx!2

f (x) = 622 "1

= 63 = 2

c. limx!"#

f (x) = x /x+2/xx /x"1/x = 1

1 = 1

CL 5-109. Yes, for all. CL 5-110. y = x2 ! 2x ! 8 = (x ! 4)(x + 2) This is a parabola opening up with x-intercepts at 4 and –2. Therefore f (x) will have vertical asymptotes

at x = 4 and x = !2 CL 5-111.

limx!0

f (x)x =

11"sin x "

11+sin xx =

(1+sin x)"(1"sin x)(1"sin x)(1+sin x)

x =2 sin x1"sin2 x

x = 2 sin xx cos2 x

= sin xx # 2

cos2 x= 1 # 2

cos2 (0)= 2

CL 5-112.

a. 2 b. 2 c. –2 d. 3 e. ∞ f. –1

CL 5-113.

m = kpqt!!!!!7 = k(5)(10)

4!!!!!k = 14

50 =725 21 =

725 p(2 p)

9

63 = 1425 p

2

63!2514 = p2

9!252 = p

3!52= 15 2

2 = p

CL 5-114. a. f (0.0001) = f (!0.0001) = !0.1111

b. f (x) =13+x !

13

x =3!(3+x)3(3+x)x =

!x3(3+x)x = ! 1

9+3x

c. limx!0

" 19+3(0) = " 1

9


CL 5-115. They will have to agree that they think they are headed to the same place. CL 5-116. a. 3(!1)y!1 ! 2(!1) + 2y!1 = 1

!3y!1 + 2 + 2y!1 = 1!y!1 = !1

1y = 1

y = 1

b. 3xy !

2x +

2y = 1

3! 2y + 2x = xy3+ 2x = xy + 2y3+ 2x = y(x + 2)2x + 3x + 2

= y

CL 5-117. See graph at right. a. x ≠ 3 b. 3x =

c. limx!"

#5x + 2x # 3

= #5

y = #5

CL 5-118. a. See graph at right below. b. Find the area under the curve from 0 to 2t t= = . c. He is approximating the area under the curve by using right-endpoint rectangles with

width = 0.1.

d. 6(0.1k)k=1

20

!

e. 0.5[6(0)2 + 12(0.1)2 + 12(0.2)2 + .... + 12(1.9)2 + 6(2)2] f. Right-endpoint rectangles give 17.22 mi, left=14.82, trap=16.02

x

y

ft/sec

seconds 2 1

25

chapter 5: introduction to limits - achsprecalc · 2019-08-31 · cpm educational program © 2012...

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