MTH 1125 (1pm) - Test #1
Fall, 2004
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. Compute: limx→3
x2 −5x+6
x2 −x−6
=
1. Try Plugging in:
limx→3
x2 −5x+6
x2 −x−6
(3)2 −5(3)+6
(3)2 −(3)−6
=
==
0
0
No Good - Zero Divide!
2. Try Factoring and Cancelling:
limx→3
x2 −5x+6
x2 −x−6
i.e., limx→3
2. Compute: limx→2
= limx→3
x2 −5x+6
x2 −x−6
x2 −x−6
3x2 +1
=
(x−2)(x−3)
(x+2)(x−3)
= limx→3
(x−2)
(x+2)
=
3−2
3+2
=
1
5
1
5
=
1. Try Plugging in:
limx→2
x2 −x−6
3x2 +1
i.e., limx→2
3. Compute: limx→7
(2)2 −(2)−6
3(2)2 +1
=
x2 −x−6
3x2 +1
√
x+9−4
x−7
4
= − 13
4
= − 13
=
1. Try Plugging in:
√
limx→7 x+9−4
x−7
=
√
(7)+9−4
(7)−7
=
0
0
No Good - Zero Divide!
2. Try Factoring and Cancelling:
√
limx→7 x+9−4
x−7
limx→7
h√ 1
=
√
limx→7 x+9−4
x−7
(x+9)−16
√
(x−7)[ x+9+4]
i
=
(7)+9+4
i.e., limx→7
= limx→7
1
8
√
x+9−4
x−7
=
1
8
·
√
√x+9+4
x+9+4
=
x−7
√
(x−7)[ x+9+4]
√
2
x+9) −(4)2
limx→7 (x−7) √x+9+4
[
]
(
= limx→7
1
√
x+9+4]
[
=
=
4. Compute: limx→2
x2 +x−2
x2 −5x+6
=
1. Try Plugging in:
limx→2
x2 +x−2
x2 −5x+6
=
(2)2 +(2)−2
(2)2 −5(2)+6
=
4
0
No Good - Zero Divide!
2. Try Factoring and Cancelling:
No Good - Cancelling will only work when Step #1 yields 00 .
3. Evaluate the one-sided limits:
limx→2−
x2 +x−2
x2 −5x+6
= limx→2−
x2 +x−2
(x−2)(x−3)
=
4
(−ε)(−1)
limx→2+
x2 +x−2
x2 −5x+6
= limx→2+
x2 +x−2
(x−2)(x−3)
=
4
(ε)(−1)
=
4
−ε
Since the one-sided limits are not equal, limx→2
5. f (x) =
2x2
x2 −9
−4
−ε
=
= +∞
= −∞
x2 +x−2
x2 −5x+6
Does Not Exist.
Find the asymptotes and graph.
Verticals Look for those x-values that cause division by zero.
⇒ x2 − 9 = 0
⇒ (x + 3) (x − 3) = 0
⇒ x = −3 and x = 3 are possible vertical asymptotes.
Compute the one-sided limits of f (x) , as x approaches these values.
limx→−3−
limx→−3+
limx→3−
limx→3+
2x2
x2 −9
= limx→−2−
2x2
x2 −9
= limx→−2+
2x2
x2 −9
= limx→1−
2x2
x2 −9
= limx→1+
2x2
(x+3)(x−3)
2x2
(x+3)(x−3)
2x2
(x+3)(x−3)
2x2
(x+3)(x−3)
=
=
=
=
18
(−ε)(−6)
18
(ε)(−6)
18
(6)(−ε)
18
(6)(ε)
2
=
=
3
ε
=
3
ε
= +∞
=
−3
ε
3
−ε
= −∞
= −∞
= +∞
- Infinite limits indicate
that x = −3 IS a
. vertical asymptote
- Infinite limits indicate
that x = 3 IS a
. vertical asymptote
Horizontals Compute the limits as x → ±∞
2
limx→−∞ x2x
2 −9 = limx→−∞
2
limx→+∞ x2x
2 −9 = limx→+∞
Graph f (x) =
2x2
x2
2x2
x2
= limx→−∞ 2 = 2
= limx→+∞ 2 = 2
- Finite, constant limits indicate
that y = 2 IS a
. horizontal asymptote
2x2
x2 −9
y=2
x = !3
x=3
6. ~
x
1.0
1.5
1.9
1.99
1.999
f (x)
3.25
56.789
488.78
7768.12
15877.79
x
3.0
2.5
2.1
2.01
2.001
f (x)
−6.25
−96.789
−788.78
−12768.12
−45877.79
(a) limx→2− f (x) = ∞
(b) limx→2+ f (x) = −∞
3
(c) Sketch a graph of f (x)
x=2
7. Compute, using the properties of limits. Document each step.
limx→1
h
3x2 −2x
x2 −5x+3
i
limx→1 (3x2 − 2x)
limx→1 (x2 − 5x + 3)
=
|
{z
}
The limit of a quotient equals the quotient of the limits
limx→1 3x2 − limx→1 2x
lim
x2 − limx→1
5x + limx→1 3}
| x→1
{z
=
The limit of a sum or difference equals the sum or difference of the limits
3 limx→1 x2 − 2 limx→1 x
lim
x2 − 5 lim
x + limx→1 3}
| x→1
{zx→1
=
The limit of a constant times a function equals the constant times the limit of the function
3 limx→1 x2 − 2 limx→1 x
lim
x2 − {z
5 limx→1 x + 3}
| x→1
=
The limit of a constant is the constant itself
=
3 (1)2 − 2 (1)
(1)2 − 5 (1) + 3
|
{z
limx→c x=c
limx→c xn =cn
}
= −1
4