Chapter 7 Limits and Continuity

CHAPTER 7 LIMITS AND CONTINUITY

Focus on Exam 7 1 (a) |x + 3| =

{

-x - 3, x < -3, x + 3,

x ≥ -3.

(x + 1)(-x - 3) x+3 = -x - 1 (x + 1)(x + 3) For x ≥ -3, f (x) = x+3 =x+1 Hence, in the non-modulus form, -x - 1, x < -3, f (x) = x + 1, x ≥ -3. For x < -3, f (x) =

{

(b) The graph of f(x) is as shown below. y = −x − 3

y

2

y=x+1

1 −3

−2

−1

x

O −1 −2

(c) lim  f (x) = 2 x → -3-



lim  f (x) = -2

x → -3+

(d) lim  f (x) does not exist because lim f (x) ≠ lim  f (x). x → -3

2 (a) lim  h(x) = 2 x → -1



-1 + p = 2 p = 3 

x → -3

x → -3+

x = -1 is in the range -3 ≤ x < 0, so the part of the function x + p is used.

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 1

10/18/2012 9:48:51 AM

2

ACE AHEAD Mathematics (T) Second Term

(b) Since lim  h(x) exists, x → -3



lim  h(x) = lim  h(x)

x → -3-

x → -3+

x2 - k



(-3)2 - k = -3 + 3 k = 9  Since lim h(x) exists,



lim  h(x) = lim  h(x)



x+3

x→0

x→0-

x→0+

x+3

0+3=e

0 - q

e x - q

ln 3 = -q q = -ln 3 = ln 3-1 = ln 1 3 (c) The graph of y = h(x) is as shown below.

{

x < -3, -3 ≤ x < 0, x ≥ 0.

x 2 - 9, x + 3, 3e x,

=

ln

y = 3e x 1

e 3 ex = 1 3 = 3e x



3 (a) f o g = f [g(x)] 1 = f  x-3

1

3

3

+



1

ex eq ex

y = x2 − 9

x



(1, 8.2)

2

=

e x-q =

y

y

h(x) =

1

−4 −3 −2 −1 O

x 1

2

2

2

1 1 x-3 = 2(3 + x - 3) = 2x The domain of f o g is the same as the domain of g, i.e. {x : x ∈ R, x ≠ 3}. Because the domain cannot take the value 3, the range of f o g cannot take the value 2x = 2(3) = 6. Hence, the range of f o g is {y : y ∈ R, y ≠ 6}. (b) The graph of y = f g(x) = 2x, x ≠ 3 is as shown below.

=2 3+

1

2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 2

10/18/2012 9:48:52 AM

Fully Worked Solution

3

y

6

x

O

3

y = 2x

(c) lim fg(x) = 2(3) x→3 =6 and lim fg(x) = 2(3) -

x→3+

=6



Since lim f g(x) = lim f g(x) x→3+

x→3+



=6



then lim fg(x) = 6 x→3

4 In the non-modulus form, f (x) =

{

x 2 - 1, -x2 + 1, (x - 2)(x - 3),

x < -1, -1 ≤ x < 1, x ≥ 1.

The graph of y = f(x) is as shown below. y

2 2

y=x −1

y = (x − 2)(x − 3)

1 −1 O

1

2

3

4

x

y = −x 2 + 1

(b) (i) lim  f (x) = 12 - 1 x→-1 =0 lim  f (x) = -12 + 1 x→-1 =0 f(-1) = -12 + 1 =0 -

+

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 3

10/18/2012 9:48:52 AM

4

ACE AHEAD Mathematics (T) Second Term

Since lim  f (x) = lim  f (x) x→-1-



x→-1+

= f (-1) = 0, then f (x) is continuous at x = -1.



(ii) lim  f (x) = -12 + 1 x→1-

=0



lim  f (x) = (1 - 2)(1 - 3) x→1+

=2



Since lim f (x) ≠ lim f (x), then lim f (x) does not exist. x→1-



x→1+

x→1

Hence, f (x) is not continuous at x = 1.

5 (a) In the non-modulus form, f (x) =

{

f (x) =

{



x2 , x < 0, -x x2 , x ≥ 0. x

-x, x < 0, x,

x ≥ 0.

In the non-modulus form,

g(x) =

{



{

-x + 3x , x < 0, 2 x + 3x , x ≥ 0. 2 x,

g(x) =

(b) g f (x) = g f (x) =

x < 0,

2x, x ≥ 0.

{ {

g(-x), x < 0,

g(x),

x ≥ 0.

-x, x < 0, 2x, x ≥ 0.

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 4

10/18/2012 9:48:52 AM

Fully Worked Solution

5

(c) lim f (x) = -0 x→0 =0 lim f (x) = 2(0) x→0 =0 f (0) = 2(0) =0 -

+

Since lim f (x) = lim f (x) x→0-

x→0+

= f (0) = 0, then f (x) is continuous at x = 0.



(d) The graph of y = gf (x) is as shown below. y 4

2 y = 2x y = −x

−2

O

2

x

3x + 1 x+2 As f (x) → ±∞, the denominator of f (x) → 0 x+2→0 x → -2 Hence, x = -2 is the vertical asymptote. 3x + 1 lim f (x) = lim x → ±∞ x → ±∞ x + 2 6 f (x) =

1



2

1 2 1 2

3x 1 + x x = lim x → ±∞ x 2 + x x

1 x = lim x → ±∞ 2 1+ x 3+0 = 1+0 =3 Hence, y = 3 is the horizontal asymptote. 3+

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 5

10/18/2012 9:48:52 AM

6

ACE AHEAD Mathematics (T) Second Term

The graph of y = f (x) =

3x + 1 is as shown below. x+2 y

3

−2

x

0

f(x)

1 2

1 − 2 −1 O 3

x

1 -  3

0 Let   y = f -1(x) f (y) = x 3y + 1 =x y+2 3y + 1 = xy + 2x 3y - xy = 2x - 1 y(3 - x) = 2x - 1 2x - 1 y= 3-x 2x - 1 ∴ f -1(x) = 3-x The domain of f -1 is the same as the range of f, i.e. {x : x ∈R, x ≠ 3}. The range of f -1 is the same as the domain of f, i.e. {f -1(x) : f -1(x) ∈R, f -1(x) ≠ -2}. 7

y y=

y = −x − 1

x+1

y=x−1 1

−1

O

1

x

−1

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 6

10/18/2012 9:48:53 AM

Fully Worked Solution

(a) lim  f (x) = -(-1) - 1 x → -1 = 0  -



lim  f (x) = -1 + 1 

x → -1+

lim  f (x) = 1 + 1 

x → 1-

Substituting x = -1 into x + 1. Substituting x = 1 into x + 1.

= 2



Substituting x = -1 into -x - 1.

=0



7

lim  f (x) = 1 - 1  =0

x → 1+

Substituting x = 1 into x - 1.

(b) f (x) is continuous at x = -1 because lim  f (x) = lim  f (x) = f (-1) x → -1 x → -1 = 0. f (x) is not continuous at x = 1 because lim  f (x) ≠ lim  f (x). -

x → 1-

+

x → 1+

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Chap-07-FWS.indd 7

10/18/2012 9:48:53 AM