Practice Problem : Evaluating Limits - Notes for Exam | MATH 234, Study notes of Calculus

Material Type: Notes; Professor: Cho; Class: Calculus for Business I; Subject: Mathematics; University: University of Illinois - Urbana-Champaign; Term: Unknown 1989;

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Math 234
Practice Problems: Evaluating Limits
1. Evaluate the given limits or note that they do not exist
a.
lim
xโ†’1
x2โˆ’1
xโˆ’1
lim
xโ†’1
x2โˆ’1
xโˆ’1= lim
xโ†’1
(x+ 1)(xโˆ’1)
xโˆ’1= lim
xโ†’1(x+ 1) = 1 + 1 = 2
b.
lim
xโ†’0
x2+x
x
lim
xโ†’0
x2+x
x= lim
xโ†’0
x(x+ 1)
x= lim
xโ†’0(x+ 1) = 0 + 1 = 1
c.
lim
xโ†’2
x4โˆ’16
xโˆ’2
lim
xโ†’2
x4โˆ’16
xโˆ’2= lim
xโ†’2
(x2+ 4)(x2โˆ’4)
xโˆ’2
= lim
xโ†’2
(x2+ 4)(x+ 2)(xโˆ’2)
xโˆ’2
= lim
xโ†’2(x2+ 4)(x+ 2)
= (22+ 4)(2 + 2)
= (4 + 4) โˆ—4 = 8 โˆ—4 = 32
d.
lim
xโ†’1
x2+ 5xโˆ’6
2xโˆ’2
lim
xโ†’1
x2+ 5xโˆ’6
2xโˆ’2= lim
xโ†’1
(x+ 6)(xโˆ’1)
2(xโˆ’1)
= lim
xโ†’1
x+ 6
2
=1+6
2=7
2
1
pf3
pf4
pf5

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Math 234

Practice Problems: Evaluating Limits

  1. Evaluate the given limits or note that they do not exist

a.

lim xโ†’ 1

x

2 โˆ’ 1

x โˆ’ 1

lim xโ†’ 1

x^2 โˆ’ 1

x โˆ’ 1

= lim xโ†’ 1

(x + 1)(x โˆ’ 1)

x โˆ’ 1

= lim xโ†’ 1

(x + 1) = 1 + 1 = 2

b.

lim xโ†’ 0

x^2 + x

x

lim xโ†’ 0

x^2 + x

x

= lim xโ†’ 0

x(x + 1)

x

= lim xโ†’ 0

(x + 1) = 0 + 1 = 1

c.

lim xโ†’ 2

x^4 โˆ’ 16

x โˆ’ 2

lim xโ†’ 2

x

4 โˆ’ 16

x โˆ’ 2

= lim xโ†’ 2

(x

2

  • 4)(x

2 โˆ’ 4)

x โˆ’ 2

= lim xโ†’ 2

(x^2 + 4)(x + 2)(x โˆ’ 2)

x โˆ’ 2

= lim xโ†’ 2

(x

2

  • 4)(x + 2)

2

  • 4)(2 + 2)

= (4 + 4) โˆ— 4 = 8 โˆ— 4 = 32

d.

lim xโ†’ 1

x

2

  • 5x โˆ’ 6

2 x โˆ’ 2

lim xโ†’ 1

x^2 + 5x โˆ’ 6

2 x โˆ’ 2

= lim xโ†’ 1

(x + 6)(x โˆ’ 1)

2(x โˆ’ 1)

= lim xโ†’ 1

x + 6

2

1

e.

lim xโ†’ 3

2 x + 4

x โˆ’ 3

Note that

lim xโ†’ 3 +

2 x + 4

x โˆ’ 3

= lim xโ†’ 3 +

x โˆ’ 3

= lim xโ†’ 3 +

x โˆ’ 3

However, also note that similarly

lim xโ†’ 3 โˆ’

2 x + 4

x โˆ’ 3

= lim xโ†’ 3 โˆ’

x โˆ’ 3

= lim xโ†’ 3 โˆ’

x โˆ’ 3

Therefore,

lim xโ†’ 3

2 x + 4

x โˆ’ 3 does not exist

f.

lim xโ†’ 3 โˆ’

3 x + 2

x โˆ’ 3

lim xโ†’ 3 โˆ’

3 x + 2

x โˆ’ 3

= lim xโ†’ 3 โˆ’

x โˆ’ 3

= lim xโ†’ 3 โˆ’

x โˆ’ 3 = โˆ’โˆž

g.

lim xโ†’ 2 +

40 x

2

  • 1

x โˆ’ 2

lim xโ†’ 2 +

40 x^2 + 1

x โˆ’ 2

= lim xโ†’ 2 +

40(2)^2 + 1

x โˆ’ 2

= lim xโ†’ 2 +

x โˆ’ 2 = โˆž

h.

lim xโ†’ 0 โˆ’

x^2 โˆ’ x

x

lim xโ†’ 0 โˆ’

x

2 โˆ’ x

x

= lim xโ†’ 0 โˆ’

x(x โˆ’ 1)

x

= lim xโ†’ 0 โˆ’

(x โˆ’ 1) = 0 โˆ’ 1 = โˆ’ 1

m.

lim xโ†’ 2

x โˆ’ 2 โˆš x + 2 โˆ’ x

lim xโ†’ 2

x โˆ’ 2 โˆš x + 2 โˆ’ x

= lim xโ†’ 2

x โˆ’ 2 โˆš x + 2 โˆ’ x

x + 2 + x โˆš x + 2 + x

= lim xโ†’ 2

(x โˆ’ 2)(

x + 2 + x)

(

x + 2)^2 โˆ’ x^2

= lim xโ†’ 2

(x โˆ’ 2)(

x + 2 + x)

โˆ’x^2 + x + 2

= lim xโ†’ 2

(x โˆ’ 2)(

x + 2 + x)

โˆ’(x โˆ’ 2)(x + 1)

= lim xโ†’ 2

x + 2 + x)

โˆ’(x + 1)

n.

lim xโ†’ 3

3 x โˆ’

8 x^2 + 9

x โˆ’ 3

lim xโ†’ 3

3 x โˆ’

8 x^2 + 9

x โˆ’ 3

= lim xโ†’ 3

3 x โˆ’

8 x^2 + 9

x โˆ’ 3

3 x +

8 x^2 + 9

3 x +

8 x^2 + 9

= lim xโ†’ 3

(3x)^2 โˆ’ (

8 x^2 + 9)^2

(x โˆ’ 3)(3x +

8 x^2 + 9)

= lim xโ†’ 3

9 x

2 โˆ’ (8x

2

(x โˆ’ 3)(3x +

8 x^2 + 9)

= lim xโ†’ 3

x^2 โˆ’ 9

(x โˆ’ 3)(3x +

8 x^2 + 9)

= lim xโ†’ 3

(x โˆ’ 3)(x + 3)

(x โˆ’ 3)(3x +

8 x^2 + 9)

= lim xโ†’ 3

x + 3

3 x +

8 x^2 + 9)

8(3)^2 + 9

o.

lim xโ†’ 1

x โˆ’ 1

5

x โˆ’ 5

lim xโ†’ 1

x โˆ’ 1

5

x โˆ’ 5

= lim xโ†’ 1

x โˆ’ 1

5

x โˆ’ 5

x + 5

5

x + 5

= lim xโ†’ 1

(x โˆ’ 1)(

x + 5)

(

x)^2 โˆ’ 52

= lim xโ†’ 1

(x โˆ’ 1)(5)(

x + 1)

25 x โˆ’ 25

= lim xโ†’ 1

5(x โˆ’ 1)(

x + 1)

25(x โˆ’ 1)

= lim xโ†’ 1

x + 1

5

p.

lim xโ†’ 0

x + 25 โˆ’ 5

x

lim xโ†’ 0

x + 25 โˆ’ 5

x

= lim xโ†’ 0

x + 25 โˆ’ 5

x

x + 25 + 5 โˆš x + 25 + 5

= lim xโ†’ 0

x + 25)^2 โˆ’ 52

x(

x + 25 + 5)

= lim xโ†’ 0

(x + 25) โˆ’ 25

x(

x + 25 + 5)

= lim xโ†’ 0

x

x(

x + 25 + 5)

= lim xโ†’ 0

x + 25 + 5

q.

lim xโ†’โˆ’ 1 +

x + 1 โˆš x + 1

lim xโ†’โˆ’ 1 +

x + 1 โˆš x + 1

= lim xโ†’โˆ’ 1 +

x + 1)

2 โˆš x + 1

= lim xโ†’โˆ’ 1 +

x + 1 =