MTH 1125 (10 am) Test #1 - Solutions
Fall 2011
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers
1. Compute: limx→3
(a)
x2 −x−6
x2 +3x−12
1. Try plugging in:
limx→3
x2 −x−6
x2 +3x−12
i.e., limx→3
2. Compute: limx→2
(a)
=
=
(3)2 −(3)−6
(3)2 +3(3)−12
x2 −x−6
x2 +3x−12
2x2 +1
x2 −3
=0
=
1. Try plugging in:
limx→2
2x2 +1
x2 −3
i.e., limx→2
=
2(2)2 +1
(2)2 −3
2x2 +1
x2 −3
=9
=9
=
0
6
=0
3. Compute: limx→4
(a)
x+4
x2 −3x−4
=
1. Try plugging in:
limx→4
x+4
x2 −3x−4
(4)+4
(4)2 −3(4)−4
=
8
0
=
No Good - Division by Zero
2. Try factoring and canceling
No Good - factoring and canceling only works when Step #1 yields 00 .
3. Analyze one-sided limits
limx→4−
x+4
= limx→4−
x2 −3x−4
( 85 )
x+4
(x+1)(x−4)
=
8
(5)(−ε)
=
(−ε)
x+4
(x+1)(x−4)
=
8
(5)(+ε)
=
(+ε)
= −∞
x → 4−
⇒ x<4
⇒ x−4<0
limx→4−
x+4
x2 −3x−4
= limx→4−
( 85 )
= +∞
x → 4+
⇒ x>4
⇒ x−4>0
Since the one-sided limits are not equal, limx→4
4. Compute: limx→3
(a)
√
√
x+7− 10
x−3
x+4
x2 −3x−4
Does Not Exist.
=
1. Try plugging in:
√
√
10
limx→3 x+7−
x−3
=
√
√
(3)+7− 10
(3)−3
=
0
0
No Good - Division by Zero
2. Try factoring and canceling
√
√
10
limx→3 x+7−
x−3
limx→3
=
√
√
10
limx→3 x+7−
x−3
(x+7)−10
√
√
(x−3)[ x+7+ 10]
1
k√
√ l
(3)+7+ 10
i.e., limx→3
=
= limx→3
√ 1√
10+ 10
√
√
x+7− 10
x−3
=
=
√1
2 10
√1
2 10
2
x−3 √
√
(x−3)[ x+7+ 10]
=
=
·
√
√
√x+7+√10
x+7+ 10
√
10
20
√
10
20
=
√
√
2
2
x+7) −( 10)
limx→3 (x−3) √x+7+√10
[
]
= limx→3
(
1 √
√
x+7+ 10]
[
=
=
5. Compute: limx→−∞
limx→∞
2x3 +3x+2
3x3 +8x2 +2
i.e., limx→∞
6.
2x3 +3x+2
3x3 +8x2 +2
= limx→∞
2x3 +3x+2
3x3 +8x2 +2
=
=
2x3
3x3
= limx→∞
2
3
=
2
3
2
3
x=
f (x) =
x=
f (x) =
2.5
2.9
2.99
2.999
2.9999
−15.1
−227.8
−1212.3
−21156.3
−834561.9
3.5
3.1
3.01
3.001
3.0001
−15.1
−227.8
−1212.3
−21156.3
−834561.9
Based on the information in the table above, do the following:
(a) limx→3− f (x) = −∞
(b) limx→3+ f (x) = −∞
(c) Graph f (x)
x=3
3
7. Determine whether f (x) is continuous at the point x = 3.
⎧ x2 −2x−3
for x ≤ 3
⎨ x−3
f (x) =
⎩
2x − 2
for x > 3
f (x) is continuous at the point x = 3 exactly when limx→3 f (x) = f (3) .
We will check this by computing limx→3 f (x) .
Since f (x) is defined differently for x < 3 than it is for x > 3, we must compute the
one-sided limits in order to compute limx→3 f (x) .
limx→3− f (x) = limx→3−
x2 −2x−3
x−3
= limx→3−
(x+1)(x−3)
(x−3)
= limx→3− (x + 1) = ((3) + 1) = 4
limx→3+ f (x) = limx→3+ (2x − 2) = (2 (3) − 2) = 4
Since the one-sided limits are equal, limx→3 f (x) exists and is equal to the common
value of the one-sided lmits.
i.e., limx→3 f (x) = 4
However, f (3) =
(3)2 −2(3)−3
(3)−3
= 00 , which is undefined.
Since f (3) is undefined, limx→3 f (x) 6= f (3) .
Since limx→3 f (x) 6= f (3) , f (x) is NOT continuous at the point, x = 3.
4
8. f (x) =
x+1
.
x2 −4
Find the asymptotes and graph
Verticals
Find x-values that cause division by zero.
⇒ x2 − 4 = 0
⇒ (x + 2) (x − 2) = 0
⇒ x = −2 and x = 2
possible vertical asymptotes
Compute the one-sided limits
x = −2
limx→−2−
x+1
x2 −4
= limx→−2−
x+1
(x+2)(x−2)
=
−1
(−ε)(−4)
=
−1
(ε)(4)
=
(− 14 )
ε
x → −2−
⇒ x < −2
⇒ x+2<0
limx→−2+
x+1
x2 −4
= −∞
Infinite limits tell us
that x = −2 IS a
vertical asymptote
= limx→−2+
x+1
(x+2)(x−2)
=
−1
(ε)(−4)
=
1
(ε)(4)
=
( 14 )
ε
= +∞
x → −2+
⇒ x > −2
⇒ x+2>0
x=2
limx→2−
x+1
x2 −4
= limx→2−
x+1
(x+2)(x−2)
=
3
(4)(−ε)
=
( 34 )
−ε
= −∞
x → 2−
⇒ x<2
⇒ x−2<0
limx→2+
x+1
x2 −4
= limx→2+
Infinite limits tell us
that x = 2 IS a
vertical asymptote
x+1
(x+2)(x−2)
=
3
(4)(ε)
x → 2+
⇒ x>2
⇒ x−2>0
5
=
( 34 )
ε
= +∞
Horizontals
Compute limits as x → −∞ and x → +∞.
limx→−∞
x+1
x2 −4
= limx→−∞
x
x2
= limx→−∞
1
x
=0
Limits are finite and
constant. Hence, y = 0 IS
a horizontal asymptote
limx→+∞
x+1
x2 −4
= limx→+∞
Graph f (x) =
x+1
x2
= limx→+∞
1
x
=0
x+1
x2 −4
x = !2
x=2
Extra (10 pts - WOW!)
9. Compute: limx→−∞
limx→−∞
√ x−1
x2 −2x
√ x−1
x2 −2x
=
= limx→−∞
√x
x2
x
= limx→−∞ |x|
= limx→−∞
x
−x
= limx→−∞ (−1) = −1
i.e. limx→−∞ √xx−1
2 −2x = −1
Remark: Since x → −∞, we could assume that x < 0. Hence, we were justified in
making the substitution −x for |x| .
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