# Chemistry and Biology

Topics: Asymptote, Limit of a function, Asymptotic curve Pages: 19 (1190 words) Published: March 16, 2013
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,

x+3

x→0

lim h(x) = lim h(x)
x→0-

x→0+

x+3

0+3=e

0-q

ex - q

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

{

ex
eq
ex
ln

y = 3e x
1

e3
ex
=
1
3
= 3e x

3 (a) f o g = f [g(x)]
1
=f
x-3

1

1

+

3

3
y = x2 − 9

x

=

(1, 8.2)

2

=

e x-q =

y

y

h(x) =

x < -3,
-3 ≤ x < 0,
x ≥ 0.

x 2 - 9,
x + 3,
3e 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 f g(x) = 2(3)
x→3
=6
and
lim f g(x) = 2(3)
-

x→3+

=6
Since lim f g(x) = lim f g(x)
x→3+

x→3+

=6
then lim f g(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

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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) =

{

x2 , x < 0,
-x