MTH 1125 (11am) - Test #1 - Solutions
Spring, 2005
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. Compute: limx→1
x2 −x−6
3x2 +1
=
1. Try plugging in:
limx→1
x2 −x−6
3x2 +1
i.e., limx→1
2. Compute: limx→5
(1)2 −(1)−6
3(1)2 +1
=
x2 −x−6
3x2 +1
x2 −2x−15
x2 −8x+15
= − 32
= − 32
=
1. Try plugging in:
limx→5
x2 −2x−15
x2 −8x+15
=
(5)2 −2(5)−15
(5)2 −8(5)+15
=
0
0
No Good! Division by zero!
2. Try factoring and plugging in:
limx→5
x2 −2x−15
x2 −8x+15
i.e., limx→5
3. Compute: limx→2
= limx→5
x2 −2x−15
x2 −8x+15
√
7+x−3
x−2
(x−5)(x+3)
(x−5)(x−3)
(x+3)
(x−3)
= limx→5
=
((5)+3)
((5)−3)
=4
=4
=
1. Try plugging in:
√
limx→2 7+x−3
x−2
=
√
7+(2)−3
(2)−2
=
0
0
No Good! Division by zero!
2. Try factoring and plugging in:
√
limx→2 7+x−3
x−2
limx→2
=
√
√
√7+x+3
limx→2 7+x−3
·
x−2
7+x+3
(x−2)
√
(x−2)[ 7+x+3]
i.e., limx→2
√
7+x−3
x−2
= limx→2
=
1
6
1
√
7+x+3]
[
=
√
2
7+x) −(3)2
limx→2 (x−2) √7+x+3
[
]
(
= h√
1
i
7+(2)+3
=
√1
9+3
=
= limx→2
1
6
(7+x)−9
√
(x−2)[ 7+x+3]
=
4. Compute: limx→−3
x2 −3x−9
x2 −2x−15
=
1. Try plugging in:
limx→−3
x2 −3x−9
x2 −2x−15
=
(−3)2 −3(−3)−9
(−3)2 −2(−3)−15
=
9
0
No Good! Division by zero!
2. Since the numerator is not zero, Step 2 will not work. Go to Step 3.
3. Compute the one-sided limits as x → −3.
limx→−3−
x2 −3x−9
x2 −2x−15
= limx→−3−
x2 −3x−9
(x+3)(x−5)
=
9
(−ε)(−8)
=
(− 98 )
x2 −3x−9
(x+3)(x−5)
=
9
(+ε)(−8)
=
(− 98 )
(−ε)
= +∞
x → −3−
⇒ x < −3
⇒ x+3<0
limx→−3+
x2 −3x−9
x2 −2x−15
= limx→−3+
(+ε)
= −∞
x → −3+
⇒ x > −3
⇒ x+3>0
Since the one-sided limits are not equal, “THE LIMIT” does not exist.
i.e., limx→−3
5. f (x) =
4x2 −3x+2
x2 +4x+3
x2 −3x−9
x2 −2x−15
Does Not Exist
Find the asymptotes and graph.
Verticals Find x-values that cause division by zero.
⇒ x2 + 4x + 3 = 0
⇒ (x + 3) (x + 1) = 0
⇒ x = −3; x = −1 possible vertical assymptotes.
Compute one-sided limits as x → −3; and x → −1
limx→−3−
4x2 −3x+2
x2 +4x+3
=
limx→−3−
4x2 −3x+2
(x+3)(x+1)
=
47
(−ε)(−2)
=
(− 472 )
(−ε)
= +∞
x → −3−
⇒ x < −3
⇒ x+3<0
limx→−3+
4x2 −3x+2
x2 +4x+3
=
limx→−3+
- Infinite limits indicate
that x = −3 IS a
vertical asymptote
4x2 −3x+2
(x+3)(x+1)
x → −3+
⇒ x > −3
⇒ x+3>0
2
=
47
(ε)(−2)
=
(− 472 )
(ε)
= −∞
.
limx→−1−
4x2 −3x+2
x2 +4x+3
=
limx→−1−
4x2 −3x+2
(x+3)(x+1)
=
9
(2)(−ε)
=
( 92 )
(−ε)
= −∞
x → −1−
⇒ x < −1
⇒ x+1<0
limx→−1+
4x2 −3x+2
x2 +4x+3
limx→−1+
- Infinite limits indicate
that x = −1 IS a
. vertical asymptote
4x2 −3x+2
(x+3)(x+1)
=
9
(2)(ε)
=
( 92 )
(ε)
= +∞
x → −1+
⇒ x > −1
⇒ x+1>0
Horizontals Compute the limits as x → −∞ and as x → ∞
2
−3x+2
= limx→−∞
limx→−∞ 4x
x2 +4x+3
2
−3x+2
= limx→+∞
limx→+∞ 4x
x2 +4x+3
Graph f (x) =
4x2
x2
4x2
x2
= limx→−∞ 4 = 4
= limx→+∞ 4 = 4
4x2 −3x+2
x2 +4x+3
y=4
(!2, !24)
x = !3
6. ~
x
1.0
1.5
1.9
1.99
1.999
f (x)
6.25
96.789
788.78
12768.12
45877.79
x
3.0
2.5
2.1
2.01
2.001
f (x)
6.25
96.789
788.78
12768.12
45877.79
3
x = !1
- Finite, constant limits indicate
that y = 4 IS a
. horizontal asymptote
(a) limx→2− f (x) = +∞
(b) limx→2+ f (x) = +∞
(c) Sketch a graph of f (x)
x=2
7. Compute, using the properties of limits. Document each step.
limx→1 [(3x2 − 2x) (x2 − 5x + 3)] =
=
∙
|
³
´¸ ∙
2
lim 3x − 2x
x→1
³
´¸
2
lim x − 5x + 3
x→1
{z
}
Limit of a product equals the product of the limits
µ
=
|
2
¶µ
lim 3x − lim 2x
x→1
x→1
2
¶
lim x − lim 5x + lim 3
x→1
{z
x→1
x→1
}
Limit of a sum or difference equals the sum or difference of the limits
=
µ
|
2
¶µ
3 lim x − 2 lim x
x→1
x→1
2
¶
lim x − 5 lim x + lim 3
x→1
x→1
{z
x→1
}
Limit of a constant times a function equals the constant times the limit of the function
=
µ
|
¶µ
3 lim (1)2 − 2 lim (1)
x→1
x→1
x→1
{z
limx→c x=c
limx→c xn =cn
= −1
4
¶
lim (1)2 − 5 lim (1) + 3
x→1
}