Differential and integral calculus by feliciano and uy PDF

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ALARCON CHAPTER 1: LIMITS EXERCISE 1.1

1. I f f x = x 2 – 4 x, f ind f

a

f –1 =

–5

2

y2+1

a f –5 b

c f x+ x

d f x + 1 –f x – 1

–4 –5

= 25 + 20 f –1 = 45 b f y2 +1 =

y2 +1

2

y2+1

–

4

= y 4+ 2 y 2 + 1 – 4 y 2 – 4 = y 4– 2 y 2 – 3 f y2+1 =

y2–

y2+1

3 2

c f x+ x = x+ x = x+ x

–4 x + x

x+ x –4

f x+ x = x+ x

x+ x–4

d f x + 1 –f x – 1 =

x +1

2

–4 x+1

x–1

2

–4 x–1

–

=

x 2 + 2 x+ 1 – 4 x – 4 –

= x2 –2x–3 – x 2

2

x2 –2x+1 –4x+4

– 6x + 5

2

= x –2 x–3 – x +6x –5 f x + 1 – f x – 1 = 4 x – 8 or 2 x – 4 x2 +3 , f ind x as a fu nc tio n of y 2. I f y = y

=

x

x2+3 x

x 2 + 3 = xy x 2 – xy + 3 = 0 Find the value of x by using the quadratic equation.

x = –b 

x= x

b 2 – 4 ac

2a

1,

, a=b

= – y, c = 3

– – y – y 2 – 4 1 3 2 1

=

y

y 2 – 12

2

3 If y= tan x + π , find x as a function of y. y = tan x + π x+ π = arctan y x = arctan y – π

4. Express the distance D travaled in t hr by a car whose speed is 60 km/hr. Distance = Rate  Time D = 60 t 5. Express the area A of an equilateral triangle as a function of its side x. 1

A=

bh

2

a +b2=c2 2

1

x

2

+h2= x2

3

Let b = x, h =

A= 1

1

x

3

x

2

x

x

4

h 2 =x2 –

1

2

x

2 2

–x

3 x

x

3

2

4

4 4x

A=

1 2

2

2

2

h = 4 2

h = h=

3x 4 3

2

2

x

1 2

x

1 2

x

6. The stiffness of a beam of rectangular cross section is proportional to the breadth and the cube of the depth. If the breadth is 20 cm, express the stiffness as a function of the depth.

Let S = stiffness, b = breadth, d = depth

S=bd3 S = 20 d 3 D

d

b 7. A right circular cylinder, radius of base x, height y, is incribed in a right circular cone, radius of base r and height h. Express y as function of x (r and h are constant)

Using ratio and proportion of two similar triangle  BCD and  ACE

E

y h r– x = r

D

ry=h r–x y=

h

h r –x r

y B A x

= x 2 1, find 8. If f x

+

f x + h –f x h

= = =

=

r

f x +h – f x , hs0 h

x+h 2+1 – x 2+1 h

x 2 + 2 xh + h 2 + 1 – x 2 – 1 h 2 xh + h

2

h h 2 x+ h h

f x +h – f x =2 x+h h 9. If f 3 x 2 – 4 x + 1 , find

f h+ 3 –f 3

, h s0

h f h+3 –f 3

=

= =

=

3 h +3

2

–4 h+3 + 1– 3 3

h h 2 3 h + 6 h + 9 – 4 h – 12 + – 27 – 12 + 1 1 h 2 3 h + 18 h + 27 – 4 h – 12 + 1 – 27 + 12 – 1

h 2 3 h + 14 h

h

2

–4 3 +1

C

h 3 h + 14 h

=

f h+3 – f 3 = 3 h + 14 h

4

10 If f x =

x 3 and g x = x 2 – 3, find f g x and g f x 4 +

=

f g x

x2 –3+3

=

f g x

4

x2

g f x

=

2

4

x+3

=

g f x

–3

16

3 x2 +6x+9

g f x

=

g f x g f x

2 + 18 x + 27 x2 +6x+9

16 – 3 x

=

16 – 3 x 2 – 18 x – 27 x2 +6x+9

=

– 3 x 2 – 18 x – 11 x2 +6x+9

=

– 3 x 2 – 18 x – 11 x+3 2

g f x

EXECRCISE 1.2 Evaluate each of the following. 1. lim

x2 –4x +3

x→2 2

= 2 –4 2 +3 =4 – 8 + 3 lim

x2 – 4 x + = – 1

x→2

3 2. lim 3 x + 2 x→ 3 x + 4

=

3 3 +2 3+4

=

9+2 7 3 x+

2 lim

x→3

=

11

x+4

7

3. lim tan x + sin x π x→

4

= tan

+ sin

π

π 4

4

=1+

2

2

lim tan x + sin x π

x→

4

sin 2 x sin x

4. lim π x→

3

sin 2 π

=

3

π

sin sin

=

3 2π 3

π

sin 3 3

=

2

=

2

2+ 2

3

2

lim

x→ π

3

sin 2 x sin x

=1

3

5. lim 2 x + 3

=2 8 +

4

x

x→8

8 –4

= 16 + 2 – 4 3

lim 2 x +

x

x→8

– 4 = 14

6. lim 4 x – 3 x→2

x2+5

= 4 2 –3

2

2

= 8–3 = 5

+5

4+5

9

lim 4 x – = 45

3 x→2

3x +

7. lim

= =

3 3 3+1

3

9

3

4

lim

3x x+1

x

x→3

2

x –2x+4

x→0

=

1 2

=

3x+2

8. lim

=

x 1

x

x→3

3 0 +2 2 –2 0 +4

0 2 4

3x + 2 1 =2 x 2 – 2 x +4

lim0 x→

EXERCISE 1.3 Evaluate each of the following. x 3 – 64 1.

lim x→ 4

– 4x

= =

x 2 – 16 x 2 + 4 x + 16

x – 4 x+ 4 x 2 + 4 x + 16

\

x+4 4

=

2

+ 4 4 + 16 4+4

=

16 + 16 + 16 8

=

48 8 x

3

– 64

lim x 2 – 16

=6

x→4

2.

lim x→2

x2 +2x– 8 3 x–6

= =

x +4 x – 2 3 x– 2 x +4 3

=

2+4 3

=

6 3 x2 +2x–8 3x – 6

lim

=2

x→2

3.

lim x→ 3

=

x– 3

x– 3

x

x 3 – 14 x +

3

x2 +3x–4

15

x2 +3x–5 – 1 3

x + 1 2

x 2 + 3 x –4

= x2 +3x–5

=

2

3

=

+3 3 – 4 +3 3 – 5

2

3

9+ 9 – 4 9 +9 – 5 x 3 – 13 x + 12

lim

14

=

x 3 14 x + 15

x→3

13

– 4.

x

lim

=

–x –2

2

x– 2 x 2 + x + 1 x –2 2 x 2 –x +3

=

=

x

3

2

2 x – 5 x +5 x – 6

x→2

=

3–

x 2 +x + 1 2 2 x –x + 3 2

2

+2+1 –2+3

2

2 2

4+2+ 1 8 –2 + 3

x3–x 2–x –2

lim

=

2 3x–52x+5 x 6 – x +3 2– 9 x→2

5. lim

=

x→02 x x +3 2– 3

2

2 x

x+ 3 – 3

= = = =

x +3 + 3

2x

x x+ 6 2x x +6 2 6 2

x +3

2

–9

2x

lim

=3

x→0

6. lim

x + 16 – 4

x→0

= =

x

x + 16 – 4

=

x

*

x + 16 + 4 x + 16 + 4

x + 16 – 16 x + 16

x

+4

x x + 16

x

+4

1

= x + 16 + 4 1

= =

16 + 4

1 4+4

lim x→0

7 li

x + 16 – 4

x

=

1 8

1

7 9

x 3

x→1

x– 1

=

x+3 – 2

x–1

=

2

x +3 + 2 x + 3 +2

*

x +3 + 2

x + 3–4 x +3

x–1

=

+2

x –1

= x+3 +2 =

1+3

=

4

+2

=2 +2 lim

+2

x–1

x +3

x→1

8. li m x →8

3 x –2

=

x –8

=4 2

3 x –2 x –8 3 x2 +23 x +4

* 3

x2 +23 x +4

x–8

=

x–8

x2

3

3

+2

x +4

1

= 3 x2 +23 x +4 1

=

3

=

3

=

82 + 2 3 8 + 4

1 64 + 2 2 + 4

1 4+4+4

1 12

=

3 x –2

lim x – 8 x→8

1 – 1 x4

9. lim

x –4

x→4 4 –x 4x

= x –4 =

4– x 4 x x–4

x –4 4 x x– 4 1

=– =–

4x 1

=–

4 4 1 – 1 x4

lim

16

x –4

x→4

10. lim x

3

–8

x

2

–4

x→2

x–

2 =

=

1

=–

x2 +2x+4 x– 2 x+ 2

x2 +2x+4 x+2 2

2

=

+2 2 +4 2+2

=

4+4+ 4 4

=

12 4

x3–8

lim

2

=3

x –4

x→2

11. lim

x→3

x –3 x– 24––x = x –3 * x –2 + x – 2 – 4 – xx – 2 +

= x – 3x – 2 + x –2 – 4–x = x – 3x – 2 +

4–x 4–x

2x –6

= x – 3x – 2 +

4–x

2 x –3

=

x –2 +

4–x

2

3–2 +

= = =

1+1 22

4–3

2

2

x– 3

lim

x –2 –

x→3

12. lim x→0

1

x

3

=

1 3

x

–

x+9 – 3

1

= x

=1 4 –x

1

3

x+9 x +9 –

3

x+9

1

x+9

4–x 4–x

x+9 +3 x+9 +3

=

1 3

x

x +9 –9 x+9

x+9 +3

x

1

=

3 xx + 9 + 3

=

x +9

x x+ 9 + 3 x + 9

3 x

=

1 3 x + 9 +9

=

1 3 0 + 27 + 9

=

1 0 + 27 + 27 1

lim

1

x→0

3

x+9 0+9

1

–

x+9

=

1 54

x –9 2

x 13. lim

x– 3

x→ 3

x 2– 9 x– 3

=

x 2 –9

*

x 2 –9

x2– 9

=

x – 3x 2 – 9 x –3 x +3 = x – 3x 2 – 9 x+ 3

=

x 2–9 3+3

=

32–9

6 0

=

=  The Limit does not exist x 2–9 x –3

lim x→3

tan 2 x sec 2 x

14. limπ x→

4

sin 2 x cos 2 x

=

1

cos 2 x

=

sin 2 x cos 2 x cos 2 x

= sin 2 x = sin 2 π

= sin

π 4

2

tan 2 x sec 2 x

lim

x→ π

=1

4

15. lim

sin 3 x sin x – tan x sin x

x→0

3

= sin

=

sin x – cos x

sin 3 x sin x cos x – sin x cos x

sin 3 x cos x

= sin x cos x – sin x sin 3 x cos x

= sin x cos x – 1 sin 2 x cos x

= cos x – 1 = =

1 – cos 2 x cos x cos x – 1 1 – cos x

1 + cos x cos x cos x – 1

= – 1 + cos x cos x

= – 1 + cos 0 cos 0 =– 1+1 1 sin 3 x

lim sin x – tan x x→0

= –2 1 – cos

16. lim

2

x

1 + cos x

x→0

1 – cos x 1 + cos x 1 + cos x

=

= 1 – cos x = 1 – cos 0 =1 – 1 lim x→0

1 – cos

2

x

1 + cos x

17. lim

=0

sin x sin 2 x 1 – cos x

x→0

=

sin x 2 sin x cos x 1 – cos x

=

2 sin x cos x 1 – cos x

=

2 cos x 1 – cos 1 – cos x

2

=

2

x

1 – cos x 1 +cos x 1 – cos x

2 cos x

= 2 cos x

1 + cos x

= 2 cos 0 1 + cos 0 =2 1

1+1

=2 2 lim x→0

sin x sin 2 x 1 – cos x

=4

18.

= =

1 – cos 2 x 1 + cos x 1 – cos x 1 + cos x 1 + cos x

= 1 – cos x = 1 – cos π =1– –1 =2 sin 2 x lim + x→π 1 cos x If f x = x , find; f x –f 4

19. lim

x–4

x→4

= x –x 4– = = = = = =

4

x –2 x– 4 x –2

x– 4



x +2 x +2

x –4 x–4

+2

x

1

x +2 1 4 +2

1 2+2

f x –f 4

limx→4x – 4 20. lim

x→0

=

=

1 4

f 9+x –f 9 x

9+x –

9

x

=

x +9 –3

x

=

x +9 – 3

x



x+ 9+3 x+9 +3

= =

x+ 9 – 9 +3

x +9

x

x +3

x +9

x

1

= x +9 +3 1

=

0+9 +3

=

9 +3

=

1

1 3+3

f 9 + x –f 9 x

lim x→0

f x –f 2 x –2

21. lim x→2

x 2 –2 x +3 – 2 x–2

=

1 6

= x 2 – 2 x + 3, find;

If f x

=

=

2

–2 2 + 3

x 2 – 2x + 3 – 4+4 – 3 x–2 x2 – 2x

= x–2 =

x x –2 x– 2

=x f x –f 2 x –2

lim x→2

=2

f x +2 –f 2 x 2 x +2 – 2 x +2 +3 – 2 2 – 2 2 + 3 = x x 2+4x +4–2x –4+3 –4+4–3

22. lim x→0

= = =

x x 2 +2 x x x x+ 2 x

=x +2 =0 +2 f x+ 2 –f 2 x

lim x→0

=2

EXERCISE 1.4 Evaluate each of the following. 6x 3+4x 2+5 1. lim 8x 3+7x –3 x→ 1

=

3

6x3 +4x2 +5

x→

x

 1

3 8 x + 7 x– 3

x3 3

6x

3

8x

x

4x

+

+

3

6+

4

x

7

= 8+

–

x2

3

5

x3

–

5

3

x3

3

x3

=

6+0+ 0 8 +0 – 0

=

6 8

lim

+

7x

x +

2

6x 3+4x 2+5

x→

2. lim x→

8x 3+7x –3 3x 2+x +2

x 3+8x +1

REMINDER: 1 =0 lim

=

3 4

x

1 2

3x +x+2 x = x3 +8x+1 1 2

3x

x3

x

+

3

2

+

x3 8x 1 + + x3 x3 x 3 3

=

=

1

+

x 1+

2

+

x8 2

x13

+

x2 x3 0+0+ 0 1+0+0

lim

3x2 +x+ 2

x→

=0

x3 +8x+1 4x+5 3. lim

x2 +1

x→

1

=

4 x +5 x2 +1

x2

1

x2

4x

5

+

x2 1 + 2 x x2 4

=

x

5

+

x1 2

1+

x2

=

0+0 1+0 4 x+ 5

lim

=0 x2 + 1 x3+x+2 4. lim x→ x 2 – 1 x→

1

=

x3 +x+2 x3 1 x2 –1 x3

= 3

2

+

3

x

x

x3

1 x2 – x 13 x23 + 1+

x2

=

1

x

=

x3

x

+

x3 1

–

3

x

1+0+ 0 0 –0

x3 +x+2

=

lim x→

x 2 –1 8 x–5 5. lim x→

4x

8x –5

=

2

+3 1

1

4x 2+3



x

x 8x

x

=

–

5

x

4x2 + 3

x 2x 2 5

8–

x

=

4+ 3 x2

8 –0

= =

4+0

8 2

lim x→ 

8 x –5 4x 2+3

=4

x3

6. lim

2x –1

x→

x

=

2

3

1

4 x2 – 4 x + 1

x3

· =

1

x3

x

2 4 x –4 x+ 1

3

x3 x3

= 4x2

=

1

+

3

x

x3

1 4

–

x

=

4x

–

x3

4

x2

1

+

x

3

2x –1

2

1 0+0+0

x3