MTH 1125 - Test #1 - Solutions
Summer 2007
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. Compute: limx→2
x2 −x−2
x−2
=
1. Try Plugging in:
limx→2
x2 −x−2
x−2
=
(2)2 −(2)−2
(2)−2
=
0
0
Division By Zero - No Good!
2. Try factoring and cancelling:
limx→2
x2 −x−2
x−2
i.e., limx→2
2. Compute: limx→1
x2 +5
x2 −3
= limx→2
x2 −x−2
x−2
(x−2)(x+1)
x−2
= limx→2 (x + 1) = (2 + 1) = 3
=3
=
1. Try Plugging in:
limx→1
x2 +5
x2 −3
=
i.e., limx→1
3. Compute: limx→−2
(1)2 +5
(1)2 −3
x2 +5
x2 −3
= −3
= −3
x2 +5
x2 −3x−10
=
1. Try Plugging in:
limx→−2
Good!
x2 +5
x2 −3x−10
=
(−2)2 +5
(−2)2 −3(−2)−10
=
9
0
Division By Zero - No
2. Try factoring and cancelling:
No Good! Factoring and cancelling only works when Step 1 yields 00 .
3. Compute the one-sided limits
limx→−2−
x2 +5
x2 −3x−10
x → −2−
⇒ x < −2
⇒ x+2<0
= limx→−2−
x2 +5
(x+2)(x−5)
=
9
(−ε)(−7)
=
9
(ε)(7)
=
( 97 )
ε
= +∞
limx→−2+
x2 +5
x2 −3x−10
= limx→−2+
x2 +5
(x+2)(x−5)
=
9
(ε)(−7)
=
(− 97 )
ε
= −∞
x → −2+
⇒ x > −2
⇒ x+2>0
Since the one-sided limits are not equal, limx→−2
x2 +5
x2 −3x−10
Does not exist
4. f (x) = 4x2 + 3x − 6. Compute f 0 (x) using the definition of derivative (i.e., using the
“limiting process”).
f 0 (x) = lim∆x→0
= lim∆x→0
= lim∆x→0
f (x+∆x)−f (x)
∆x
= lim∆x→0
[4(x+∆x)2 +3(x+∆x)−6]−[4x2 +3x−6]
∆x
[4(x2 +2x∆x+∆x2 )+3(x+∆x)−6]−[4x2 +3x−6]
∆x
[8x∆x+4∆x2 +3∆x]
∆x
= lim∆x→0
= lim∆x→0
∆x[8x+4∆x+3]
∆x
[4x2 +8x∆x+4∆x2 +3x+3∆x−6]−[4x2 +3x−6]
= lim∆x→0 (8x + 4∆x + 3) = 8x + 3
i.e., f 0 (x) = 8x + 3
5. f (x) =
6x+4
.
3x−2
Find all vertical and horizontal asymptotes. Graph f (x) .
Verticals Find the x-values that cause division by zero.
3x − 2 = 0
⇒ 3x = 2
⇒ x = 23 possible vertical asymptote
limx→ 2 −
3
6x+4
3x−2
= limx→ 2 −
3
6x+4
3(x− 23 )
=
8
(3)(−ε)
6x+4
3(x− 23 )
=
8
(3)(ε)
=
( 83 )
(−ε)
= −∞
−
x → 23
⇒ x < 23
⇒ x − 23 < 0
limx→ 2 +
3
6x+4
3x−2
= limx→ 2 +
3
=
( 83 )
(ε)
= +∞
+
x → 23
⇒ x > 23
⇒ x − 23 > 0
Infinite limits tell us that x = 23 IS a vertical asymptote.
2
∆x
Horizontals Let x → −∞ and let x → +∞
limx→−∞
6x+4
3x−2
= limx→−∞ 6x
= limx→−∞ 2 = 2
3x
limx→+∞
6x+4
3x−2
= limx→+∞ 6x
= limx→+∞ 2 = 2
3x
Finite, constant limits tell us that y = 2 IS a horizontal asymptote
Graph: f (x) =
6x+4
3x−2
y=2
x=b
6. Compute: limx→5
√
x+11−4
x−5
=
1. Try Plugging in: √
√
(5)+11−4
limx→5 x+11−4
= (5)−5 =
x−5
0
0
Division By Zero - No Good!
2. Try factoring and cancelling:
√
limx→5 x+11−4
x−5
= limx→5
=
√
limx→5 x+11−4
x−5
x+11−16
√
(x−5)[ x+11+4]
= limx→5
1
8
i.e., limx→5
√
x+11−4
x−5
=
1
8
3
·
√
√x+11+4
x+11+4
=
√
2
x+11) −(4)2
√
limx→5 (x−5) x+11+4
[
]
(x−5)
√
(x−5)[ x+11+4]
(
= limx→5
1
√
x+11+4]
[
= h√
1
i
(5)+11+4
=
7. ~
x
1.0
1.4
1.49
1.499
1.4999
f (x)
6.25
96.789
788.78
12768.12
45877.79
x
2.0
1.6
1.51
1.501
1.5001
f (x)
6.25
96.789
788.78
12768.12
45877.79
(a) limx→1.5− f (x) = ∞
(b) limx→1.5+ f (x) = ∞
(c) Sketch a graph of f (x)
x = 1.5
4