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

CHAPTER 7 LIMITS AND CONTINUITY

Focus on Exam 7

-x - 3, x < -3, 1 (a) |x + 3| = { x + 3, x ≥ -3.

For x < -3, f (x) = (x + 1)(-x - 3)

x + 3 = -x - 1

For x ≥ -3, f (x) = (x + 1)(x + 3)

x + 3 = x + 1 Hence, in the non-modulus form,

-x - 1, x < -3, f (x) = { x + 1, x ≥ -3.

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

−1

−1

−2

−2−3

1

2

yy = −x − 3

y = x + 1

Ox

(c) limx → -3-

f (x) = 2

limx → -3+

f (x) = -2

(d) limx → -3

f (x) does not exist because limx → -3

f (x) ≠ limx → -3+

f (x).

2 (a) limx → -1

h(x) = 2

-1 + p = 2

p = 3

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

Chap-07-FWS.indd 1 10/18/2012 9:48:51 AM

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

ACE AHEAD Mathematics (T) Second Term2

(b) Since limx → -3

h(x) exists,

limx → -3-

h(x) = limx → -3+

h(x)

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

x→0 h(x) exists,

limx→0-

h(x) = limx→0+

h(x)

0 + 3 = e0 - q

ln 3 = -q q = -ln 3 = ln 3-1

= ln 13

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

x2 - 9, x < -3,

h(x) = {x + 3, -3 ≤ x < 0,

3ex, x ≥ 0.

ex-q = ex

eq

= ex

eln 1

3

= ex

13

= 3ex

3 (a) f o g = f [g(x)]

= f 1 1x - 3 2

= 213 + 1

1 1x - 3 22

= 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.

x2 - k

x + 3

ex - q

1

1−1−2−3−4 2

2

3

x

y

y = 3e x

y = x

+ 3

x 2 − 9y =

(1, 8.2)

O

Chap-07-FWS.indd 2 10/18/2012 9:48:52 AM

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

Fully Worked Solution 3

O 3

6

x

y = 2x

y

(c) limx→3-

fg(x) = 2(3)

= 6 and

limx→3+

fg(x) = 2(3)

= 6

Since limx→3+

f g(x) = limx→3+

f g(x)

= 6

then limx→3

fg(x) = 6

4 In the non-modulus form,

x2 - 1, x < -1,

f (x) = {-x2 + 1, -1 ≤ x < 1,

(x - 2)(x - 3), x ≥ 1.

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

1−1 2

2

1

3 4x

y

y = x 2 − 1

O

y = −x 2 + 1

y = (x − 2)(x − 3)

(b) (i) limx→-1-

f (x) = 12 - 1

= 0

limx→-1+

f (x) = -12 + 1

= 0

f(-1) = -12 + 1

= 0

Chap-07-FWS.indd 3 10/18/2012 9:48:52 AM

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

ACE AHEAD Mathematics (T) Second Term4

Since limx→-1-

f (x) = limx→-1+

f (x)

= f (-1)

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

(ii) limx→1-

f (x) = -12 + 1

= 0

limx→1+

f (x) = (1 - 2)(1 - 3)

= 2

Since limx→1-

f (x) ≠ limx→1+

f (x), then limx→1

f (x) does not exist.

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

5 (a) In the non-modulus form,

x2

-x, x < 0,

f (x) = { x2

x, x ≥ 0.

-x, x < 0,

f (x) = { x, x ≥ 0.

In the non-modulus form,

-x + 3x2

, x < 0,

g(x) = { x + 3x2

, x ≥ 0.

x, x < 0,

g(x) = { 2x, x ≥ 0.

(b) g(-x), x < 0,

g f (x) = { g(x), x ≥ 0.

-x, x < 0,

g f (x) = { 2x, x ≥ 0.

Chap-07-FWS.indd 4 10/18/2012 9:48:52 AM

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

Fully Worked Solution 5

(c) limx→0-

f (x) = -0

= 0 lim

x→0+ f (x) = 2(0)

= 0

f (0) = 2(0) = 0

Since limx→0-

f (x) = limx→0+

f (x)

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

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

y = 2x

y = −x

xO 2−2

2

4

y

6 f (x) = 3x + 1x + 2

As f (x) → ±∞, the denominator of f (x) → 0 x + 2 → 0 x → -2

Hence, x = -2 is the vertical asymptote.

limx → ±∞

f (x) = limx → ±∞ 13x + 1

x + 2 2

= limx → ±∞1

3xx

+ 1x

xx + 2

x2

= limx → ±∞13 + 1

x

1 + 2x2

= 3 + 01 + 0

= 3

Hence, y = 3 is the horizontal asymptote.

Chap-07-FWS.indd 5 10/18/2012 9:48:52 AM

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

ACE AHEAD Mathematics (T) Second Term6

The graph of y = f (x) = 3x + 1x + 2

is as shown below.

Ox

y

3

12−

13

−2 −

x 0 - 13

f(x)12

0

Let y = f -1(x)

f (y) = x

3y + 1y + 2

= x

3y + 1 = xy + 2x 3y - xy = 2x - 1 y(3 - x) = 2x - 1

y = 2x - 13 - x

∴ f -1(x) = 2x - 13 - 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

O

1

−1

−1 1x

y

y = −x − 1y = x − 1

y = x + 1

Chap-07-FWS.indd 6 10/18/2012 9:48:53 AM

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

Fully Worked Solution 7

(a) limx → -1-

f (x) = -(-1) - 1 Substituting x = -1 into -x - 1.

= 0

limx → -1+

f (x) = -1 + 1 Substituting x = -1 into x + 1.

= 0

limx → 1-

f (x) = 1 + 1 Substituting x = 1 into x + 1.

= 2

limx → 1+

f (x) = 1 - 1 Substituting x = 1 into x - 1.

= 0

(b) f (x) is continuous at x = -1 because limx → -1-

f (x) = limx → -1+

f (x) = f (-1)

= 0. f (x) is not continuous at x = 1 because lim

x → 1- f (x) ≠ lim

x → 1+ f (x).

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