Chemistry and Biology

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.

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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

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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