Math 115 Spring 11
Written Homework 10 Solutions
1. For following limits, state what indeterminate form the limits are in and evaluate the
limits.
3x2 − 4x − 4
x→2
2x2 − 8
(a) lim
0
. Algebraically, we hope to be able to factor the
0
numerator and denominator and convert the argument of the limit to a function defined at
Solution: This is in indeterminate form
x = 2.
3x2 − 4x − 4
x→2
2x2 − 8
lim
=
=
→
=
=
(3x + 2)(x − 2)
x→2 2(x + 2)(x − 2)
3x + 2
since, by properties of limits, x 6= 2
lim
x→2 2(x + 2)
3(2) + 2
2(2 + 2)
8
8
1
lim
8t3 + t
t→∞ (2t − 1)(2t2 + 1)
(b) lim
Solution: This is the limit of a rational function. The associated indeterminate form is
8t3 + t
8t3 + t
=
lim
t→∞ 4t3 − 2t2 + 2t − 1
t→∞ (2t − 1)(2t2 + 1)
8t3 + t
1/t3
= lim 3
·
t→∞ 4t − 2t2 + 2t − 1 1/t3
8 + 1/t2
= lim
t→∞ 4 − 2/t + 2/t2 − 1/t3
8+0
=
4−0+0−0
= 2
lim
∞
.
∞
(1 + h)3 − 1
h→0
h
(c) lim
Solution: This is indeterminate form
0
.
0
(1 + h)3 − 1
(1 + 3h + 3h2 + h3 ) − 1
= lim
h→0
h→0
h
h
3h + 3h2 + h3
= lim
h→0
h
= lim (3 + 3h + h2 )
lim
h→0
= 3+0+0
= 3
√
(d) lim
h→0
4+h−2
h
Solution: This is indeterminate form
0
.
0
√
√
√
4+h−2
4+h−2
4+h+2
√
= lim
lim
h→0
h→0
h
h
4+h+2
(4 + h) − 4
= lim
h→0
h
h
= lim √
h→0 h 4 + h + 2
1
= lim √
h→0 1 4 + h + 2
1
= √
1 4+0+2
1
=
4
(e)
lim (x2 −
x→∞
√
x4 − x2 − 1)
Solution: We talked of limits in the form ∞ − ∞ while discussing the long-run behavior of
polynomials. We will re-address this as yet another indeterminate form ∞ − ∞. As with all
indeterminate forms, we try an algebraic solution before attempting a sequential one. Here,
√
as in (d), we multiply by the conjugate of x2 − x4 − x2 − 1.
2
lim (x −
x→∞
√
x4
−
x2
− 1) =
=
=
=
=
=
=
2 √ 4
√
x + x − x2 − 1
4
2
√
lim (x − x − x − 1)
x→∞
x2 + x4 − x2 − 1
√
√
(x2 − x4 − x2 − 1)(x2 + x4 − x2 − 1)
√
lim
x→∞
x2 + x4 − x2 − 1
x4 − (x4 − x2 − 1)
√
lim
x→∞ x2 +
x4 − x2 − 1
x2 + 1
√
lim
x→∞ x2 +
x4 − x2 − 1
1
x2 + 1
x2
√
lim
4
2
x→∞ x2 +
x − x − 1 x12
1 + 1/x2
√
lim
x→∞ 1 + 1
x4 − x2 − 1
x2
1 + 1/x2
q
lim
x→∞
1 + x14 (x4 − x2 − 1)
2
1 + 1/x2
p
x→∞ 1 +
1 − 1/x2 − 1/x4
1+0
√
=
1+ 1−0−0
1
=
2
=
lim
2. Explain why the following are not in indeterminate form. Evaluate the limit. Fully
justify your answer.
1
(a) lim 4
x→0 x
Solution: This limit is in the form
1
. We need to use sequences.
0
Notice that the x4 term in the denominator will turn any input value x into a positive
number. As a result, the limit from both sides will be the same. Rather than using a limit
from both sides, use the sequence generated by f (n) = (−1)n /n. We know that the limit of
this sequence is 0, but alternates sign with each term.
1
n→∞ ((−1)n /n)4
1
= lim
n→∞ ((−1)4n /n4 )
1
= lim
n→∞ (1/n4 )
= lim n4
1
=
x→0 x4
lim
lim
n→∞
= ∞
(b)
lim (−x2 −
x→∞
√
x4 − x3 − 1)
Solution: As x goes to infinity, −x2 also goes to −∞. For large values of x, x4 − x3 −
√
1 > 0. Hence, as x goes to infinity, − x4 − x3 − 1 also goes to −∞. Hence, lim (−x2 −
x→∞
√
4
3
x − x − 1) = −∞.
3. Using limits, determine the horizontal asymptotes of the graphs of the following functions, if any. If the graph does not have any horizontal asymptotes, you must clearly show
why.
(a) f (x) :=
3x2 − 4x − 4
2x2 − 8
3x2 − 4x − 4
. Note that this limit is in indeterminate form ∞
.
∞
x→∞
2x2 − 8
2
1/x
Since the degree of the polynomial in the denominator is 2, we multiply the quotient by 1/x
2.
Solution: Evaluate lim
3x2 − 4x − 4
lim
=
x→∞
2x2 − 8
3x2 − 4x − 4 1/x2
lim
x→∞
2x2 − 8
1/x2
2
3 − 4/x − 4/x
= lim
x→∞
2 − 8/x2
3−0−0
since 1/x2 and 1/x → 0 as x → ∞
=
2−0
= 3/2
3
p(x)
A horizontal asymptote occurs at y = . Recall that for rational functions, if lim
x→∞ q(x)
2
p(x)
is finite, lim
approaches the same limit.
x→−∞ q(x)
(b) g(x) :=
3x3 − 4x − 4
2x2 − 8
3x3 − 4x − 4
.
. Note that this limit is in indeterminate form ∞
∞
x→∞
2x2 − 8
2
1/x
Since the degree of the polynomial in the denominator is 2, we multiply the quotient by 1/x
2.
Solution: Evaluate lim
3x3 − 4x − 4 1/x2
lim
x→∞
2x2 − 8
1/x2
3x − 4/x − 4/x2
= lim
x→∞
2 − 8/x
3x
= lim
since 1/x2 and 1/x → 0 as x → ∞
x→∞ 2
= ∞ since 3x → ∞ as x → ∞
3x3 − 4x − 4
lim
=
x→∞
2x2 − 8
3x
3x3 − 4x − 4
= lim
= −∞.
2
x→∞
x→∞ 2
2x − 8
By the same arithmetic, lim
The graph of this function has no horizontal asymptote.
(c) h(x) :=
3x3 − 4x − 4
2x5 − 8
3x3 − 4x − 4
. Note that this limit is in indeterminate form ∞
.
∞
x→∞
2x5 − 8
5
1/x
Since the degree of the polynomial in the denominator is 5, we multiply the quotient by 1/x
5.
Solution: Evaluate lim
3x3 − 4x − 4 1/x5
lim
x→∞
2x5 − 8
1/x5
3/x2 − 4/x4 − 4/x5
= lim
x→∞
2 − 8/x5
0−0−0
since for p > 0, 1/xp → 0 as x → ∞
=
2−0
= 0
3x3 − 4x − 4
=
lim
x→∞
2x5 − 8
A horizontal asymptote occurs at y = 0.
4.
(a) How many horizontal asymptotes can the graph of a rational function have? Explain.
p(x)
, we know from lecture it will have a
q(x)
horizontal asymptote if the degree of q is greater than or equal to the degree of p. When
Solution: Given a rational function R(x) :=
degree of q is greater than p, the limit to both +∞ and −∞ of R(x) is 0. There is only
one horizontal asymptote. When the degree of q is equal to the degree of p, the limit is
dependent on the leading coefficents of the two polynomials. If the leading term of p(x) is
cn xn and the leading term of q(x) is dn xn ,
cn x n
cn
cn
=
.
=
lim
x→±∞ dn xn
x→±∞ dn
dn
lim R(x) = lim
x→±∞
Again, there is only one horizontal asymptote.
If a rational function has a horizontal asymptote, the same asymptote is approached as
x goes to both +∞ and −∞.
(b) How many horizontal asymptotes can the graph of an arbitrary function have? Can
a function intersect a horizontal asymptote? Explain.
Solution: In general, there is no reason that lim f (x) and lim f (x) have to both exist and
x→∞
x→−∞
be the same. The graph of an arbitrary function can have 0, one or (at most) two horizontal
asymptotes.
A function can intersect its horizontal asymptote. For example, the graph of the last
example in lecture on rational functions did this. Horizontal asymptotes are information
about what happens in the long run: as x → ∞ and as x → −∞. The say nothing about
what happens in the middle of the graph.
5. Evaluate the following limits.
2x2 + x − 6
(a) lim
x→0
x2 + 2x
2x2 + x − 6
is of the form
x→0
x2 + 2x
sequences and check the limit from both sides.
−6
.
0
Solution: The limit lim
As a result, we will need to use
1. from the right of 0
Use the sequence O+ (n) := n1 . By the continuity of the rational function g(x) :=
2x2 + x − 6
, we expect lim+ g(x) = lim g(O+ (n)).
n→∞
x→0
x2 + 2x
lim+
x→0
2x2 + x − 6
=
x2 + 2x
2(1/n)2 + (1/n) − 6
n→∞
(1/n)2 + 2(1/n)
2
+ n1 − 6
n2
= lim
1
n→∞
+ n2
n2
=
=
lim
lim
2
n2
+
1
n2
n→∞
lim
2
n
n→∞
1
n
−6 n
·
n
+ n2
+1−
1
+2
n
6
n
− n6
n→∞ 2
= −∞
=
lim
2. from the left of 0
Use the sequence O− (n) := − n1 . By the continuity of the rational function g(x) :=
2x2 + x − 6
, we expect lim− g(x) = lim g(O− (n)).
n→∞
x→0
x2 + 2x
lim+
x→0
2x2 + x − 6
=
x2 + 2x
lim
2(−1/n)2 + (−1/n) − 6
n→∞
(−1/n)2 + 2(−1/n)
2
− n1 − 6
n2
= lim
1
n→∞
− n2
n2
=
lim
=
2
n2
1
n2
n→∞
lim
n→∞
−
2
n
lim
−6 n
·
n
− n2
−1−
1
−2
n
− n6
n→∞ −2
= ∞
=
1
n
6
n
Since the limit from the right of zero and the limit from the left of zero don’t grow
2x2 + x − 6
in the same direction, we CAN NOT write lim
= something. Hence, we say
x→0
x2 + 2x
2x2 + x − 6
lim
does not exist.
x→0
x2 + 2x
x4 − 16
(b) lim 4
x→0 x − 4x2
x4 − 16
is of the form
x→0 x4 − 4x2
sequences and check the limit from both sides.
Solution: The limit lim
−16
.
0
As a result, we will need to use
1. from the right of 0
Use the sequence O+ (n) := n1 . By the continuity of the rational function r(x) :=
x4 − 16
, we expect lim+ r(x) = lim r(O+ (n)).
n→∞
x→0
x4 − 4x2
lim+
x→0
(1/n)4 − 16
n→∞ (1/n)4 − 4(1/n)2
1
− 16
n4
= lim 1
n→∞ 4 − 42
n
n
x4 − 16
=
x4 − 4x2
=
=
lim
1
− 16 n2
n4
· 2
n→∞ 14 − 42
n
n
n
1
2
2 − 16n
lim n 1
n→∞
−4
n2
2
lim
−16n
n→∞
−4
= ∞
=
lim
2. from the left of 0
Use the sequence O− (n) := − n1 . By the continuity of the rational function r(x) :=
x4 − 16
, we expect lim− r(x) = lim r(O− (n)).
n→∞
x→0
x4 − 4x2
4
(−1/n)4 − 16
x − 16
= lim
lim
n→∞ (−1/n)4 − 4(−1/n)2
x→0− x4 − 4x2
1
4 − 16
= lim n1
n→∞ 4 − 42
n
n
=
=
1
− 16 n2
n4
· 2
lim 1
n→∞ 4 − 42
n
n
n
1
2
2 − 16n
lim n 1
n→∞
−4
n2
2
−16n
n→∞
−4
= ∞
=
lim
Since the limit from the right of zero and the limit from the left of zero grow in the same
direction, we write
2x2 + x − 6
= ∞.
x→0
x2 + 2x
lim
x4 − 16
x→2 x4 − 4x2
(c) lim
x4 − 16
is of the form
x→2 x4 − 4x2
Solution: The limit lim
−16
.
0
As a result, we hope an algebraic
solution will suffice to answer this question. Begin by factoring r(x) :=
r(x) :=
=
=
x4 − 16
.
x4 − 4x2
x4 − 16
x4 − 4x2
(x2 + 4)(x2 − 4)
x2 (x2 − 4)
x2 + 4
when x is not − 2 or 2
x2
Hence,
x4 − 16
(x2 + 4)(x2 − 4)
=
lim
x→2 x4 − 4x2
x→2
x2 (x2 − 4)
(x2 + 4)
= lim
x→2
x2
since, by properties of limits, x infinitesimally close to 2, but not equal to 2
((2)2 + 4)
=
(2)2
(x2 + 4)
since r̂(x) :=
continuous at x = 2
x2
= 2
lim
(d)
(x + 1)(x − 2)2
x→−∞
x2 − 4
lim
Solution: Note that this is the long-run behavior of the rational function d(x) :=
x3 − 3x2 + 4
∞
. The indeterminate form here is
.
2
x −4
∞
(x + 1)(x − 2)2
=
x→−∞
x2 − 4
x3 − 3x2 + 4
x→−∞
x2 − 4
3
x − 3x2 + 4
·
= lim
x→−∞
x2 − 4
lim
(x + 1)(x − 2)2
=
x2 − 4
lim
1
x2
1
x2
x − 3 + x42
x→−∞
1 − x42
x
= lim
x→−∞ 1
= −∞
=
lim
6. For each rational function given, use the end behavior, intercepts, asymptotes, and continuity to draw a rough sketch of the graph of the function.
2x2 + x − 6
(a) g(x) :=
x2 + 2x
Solution:
1. domain
Required that x2 + 2x 6= 0. Hence, x 6= 0 or x 6= −2. The domain of g is (∞, −2) ∪
(−2, 0) ∪ (0, ∞).
2. end behavior
Need to examine the following limits:
x → ∞, x → −∞, x → −2, and x → 0.
We have already done many of these.
(a) In lecture, we have shown that lim g(x) =
x→−2
7
. This tells us that there will be a
2
hole in the graph at (−2, 7/2).
2x2 + x − 6
2x2 + x − 6
=
−∞
and
lim
= ∞.
x→0
x→0−
x2 + 2x
x2 + 2x
Thus, the graph has a vertical asymptote at x = 0.
(b) In Prob 5a, we showed lim+
(c) For the limit to infinity,
2x2 + x − 6
=
x→∞
x2 + 2x
lim
2x2 + x − 6 1/x2
·
x→∞
x2 + 2x
1/x2
2 + x1 − x62
= lim
x→∞
1 + x2
2
= lim
x→∞ 1
= 2
lim
Hence there is a horizontal asymptote at y = 2. (Recall rational functions that
have a horizontal asymptote have the same long-run behavior at both ends: Prob
4.)
3. intercepts
(a) y-intercept
Since g(0) is undefined, there is no y intercept.
(b) x-intercept(s)
We need the roots of the numerator 2x2 + x − 6 that are in the domain of g. Since
2x2 + x − 6 = (2x − 3)(x + 2) = 0 has the solutions x = −2 and x = 3/2, the
potential intercepts occur at x = −2 and x = 3/2. But, x = −2 is not in the
domain of g. Hence, the only x-intercept is at (3/2, 0).
4. graph
We begin by graphing the hole, the x-intercept, the behavior as x approaches the
vertical asymptote at x = 0, and the horizontal asymptote at y = 2. Then we use
continuity to complete the graph.
(b) r(x) :=
x4 − 16
x4 − 4x2
Solution:
1. domain
Required that x4 − 4x2 6= 0. Since x4 − 4x2 = x2 (x + 2)(x − 2), this requires x 6= 0,
x 6= 2 or x 6= −2. The domain of g is (∞, −2) ∪ (−2, 0) ∪ (0, 2) ∪ (2, ∞).
2. end behavior
Need to examine the following limits:
x → ∞, x → −∞, x → −2, x → 2, and x → 0.
Again, we have already done many of these.
x4 − 16
= ∞. Thus, the graph has a vertical asymp(a) In Prob 5b, we showed lim 4
x→0 x − 4x2
tote at x = 0.
x4 − 16
= 2. This tells us that there will be a hole
x→2 x4 − 4x2
(b) In Prob 5c, we showed lim
in the graph at (2, 2).
x4 − 16
(c) We have not yet checked the lim 4
case. That said, the algebra in this
x→−2 x − 4x2
x4 − 16
computation is identical to our solution to Prob 5c above. Since lim 4
is
x→−2 x − 4x2
in indeterminate for
0
x4 − 16
x2 + 4
and 4
=
on r’s domain, we get
0
x − 4x2
x2
x4 − 16
x2 + 4
=
lim
= 2.
x→−2 x4 − 4x2
x→−2
x2
lim
This tells us that there is also a hole in the graph at (−2, 2).
(d) For the limit to infinity,
x4 − 16 1/x4
·
x→∞ x4 − 4x2 1/x4
1 − 16/x4
= lim
x→∞ 1 − 4/x2
1
= lim
x→∞ 1
= 1
x4 − 16
=
x→∞ x4 − 4x2
lim
lim
Hence there is a horizontal asymptote at y = 1. (Recall rational functions that
have a horizontal asymptote have the same long-run behavior at both ends: Prob
4.)
3. intercepts
(a) y-intercept
Since r(0) is undefined, there is no y intercept.
(b) x-intercept(s)
We need the roots of the numerator x4 − 16 that are in the domain of r. Since
x4 − 16 = (x2 + 4)(x − 2)(x + 2) = 0 has the solutions x = −2 and x = 2, the
potential intercepts occur at x = −2 and x = 2. But, neither of these values are
in the domain of r. Hence, there are no x-intercepts either.
4. graph
We begin by graphing the holes, the horizontal asymptote at y = 1, and the behavior
as x approaches the vertical asymptote at x = 0. Then we use continuity to complete
the graph.
(c) p(x) :=
x3 − 3x + 2
x+2
Solution:
1. domain
Required that x + 2 6= 0. The domain of g is (∞, −2) ∪ (−2, ∞).
2. end behavior
Need to examine the following limits:
x → ∞, x → −∞, and x → −2.
x3 − 3x + 2
case. Note that this limit is in the indeterminate form
(a) The lim
x→−2
x+2
0
. As a result, we factor the numerator knowing that x = −2 is a root of the
0
polynomial x3 − 3x + 2.
x2 − 2x + 1
x+2
x3
− 3x + 2
− x3 − 2x2
− 2x2 − 3x
2x2 + 4x
x+2
−x−2
0
Then,
x3 − 3x + 2
(x + 2)(x2 − 2x + 1)
= lim
= lim (x2 −2x+1) = (−2)2 −2(−2)+1 = 9.
x→−2
x→−2
x→−2
x+2
x+2
lim
This tells us that there will be a hole in the graph at (−2, 9).
(b) For the limit to infinity,
x3 − 3x + 2
=
x→∞
x+2
lim
x3 − 3x + 2 1/x
·
x→∞
x+2
1/x
2
x − 3 + 2/x
= lim
x→∞
1 + 2/2
2
x
= lim
x→∞ 1
= ∞
lim
Hence there no horizontal asymptote for the graph of this rational function. Now,
when there is no horizontal asymptote, we are not guaranteed that the end behavior will be the same on both sides. We need to check the limit to −∞.
x3 − 3x + 2 1/x
·
x→−∞
x+2
1/x
2
x − 3 + 2/x
= lim
x→−∞
1 + 2/2
2
x
= lim
x→−∞ 1
= ∞
x3 − 3x + 2
=
x→−∞
x+2
lim
lim
Hence, the values of the function grow unbounded on both infinite ends of its
domain.
3. intercepts
(a) y-intercept
03 − 3(0) + 2
= 1. Hence the graph of p has the y-intercept (0, 1).
0+2
(b) x-intercept(s)
Since p(0) =
We need the roots of the numerator x3 − 3x + 2 that are in the domain of p. Since
x3 − 3x + 2 = (x + 2)(x2 − 2x + 1) = (x + 2)(x − 1)(x − 1) = 0 has the solutions
x = −2 and x = 1, the potential intercepts occur at x = −2 and x = 1. But only
x = 1 is in the domain of p. Hence, the x-intercept is (1, 0).
4. graph
We begin by graphing the hole, the intercepts, and the behavior at the infinite ends of
the domain. Then we use continuity to complete the graph.
7.
(a) If lim− f (x) = 5 and lim+ f (x) = 5, what do you know about lim f (x)? What do you
x→1
x→1
x→1
know about f (1)? Explain.
Solution: Since the limit from the right equals the limit from the left, we know that
lim f (x) = 5. These limits tell us NOTHING about how the function behaves AT x = 1.
x→1
Only how the function behaves NEAR x = 1. Hence, we know nothing about the value f (1),
if it even exists.
(b) If lim+ f (x) = 8, but lim f (x) does not exist, what do you know about lim− f (x)?
x→2
x→2
x→2
Explain.
Solution: We know that the limit from the left is not the same as the limit from the right.
Hence, lim− f (x) 6= 8.
x→2
(c) If lim f (x) = ∞, lim f (x) = 3, and lim+ f (x) = ∞, what do you know about any
x→−∞
x→∞
x→1
horizontal and vertical asymptotes of the graph of y = f (x)? Explain.
Solution: By the definition of a horizontal asymptote, since lim f (x) = 3, the graph of
x→∞
y = f (x) has a horizontal asymptote at y = 3.
By the definition of a vertical asymptote, since lim+ f (x) = ∞, the graph of y = f (x) has a
x→1
vertical asymptote at x = 1.
8. Sketch the graph of a function with all of the following limits and values:
lim f (x) = 1, lim f (x) DNE, f (5) = 4, f (0) = 1, lim f (x) = 0, lim f (x) = ∞
x→5−
x→5
x→∞
x→−∞
Solution: There are many possible correct graphs that exhibit all of these behaviors. Only
two possibilities are given here.
9. Use the following graph of y = g(x) to answer the questions below:
(a) State the domain and range of g in interval notation.
Solution: There is a point on the graph corresponding to every x value except x = 2. Hence,
the domain of g is (−∞, 2) ∪ (2, 5]. The range is (−∞, −1) ∪ (1, 3]
(b) What is g(−2)? What is g(2)? What is g(4)?
Solution: Here, looking for the y-values corresponding to the points where x = −2, x = 2
and x = 4. We see that g(−2) = 2, g(2) is undefined (as x = 2 not in the domain of g), and
g(4) = 3.
(c) For what values of x does (i) g(x) = 2?, (ii) g(x) = −2?, (iii) g(x) = 1?
Solution: We start by looking as the set of points on the graph of g that intersect the
horizontal lines, y = 2, y = −2 and y = 1, respectively.
For y = 2 we see that y = 2 intersects the graph at the points (1, 2), (3, 2) and the line
interval from (−2, 2) to (0, 2). Hence, the solutions to the equation g(x) = 2 are the set
{x ∈ R| − 2 ≤ x ≤ 0, x = 1 or x = 3}.
For y = −2, there is only one point of intersection with the graph; (−3, −2). Hence,
g(x) = −2 when x = −3.
For y = 1, there is no point of intersection with the graph. Hence, there are no solutions
to the equation g(x) = 1.
(d) Determine all of the following limits from the graph of y = g(x):
• lim g(x)
x→−∞
•
•
lim g(x)
x→−2−
lim g(x)
x→−2+
• lim g(x)
x→0
• lim− g(x)
x→4
• lim+ g(x)
x→4
• lim g(x)
x→2
Solution:
• lim g(x) = −1
x→−∞
•
•
lim g(x) = −∞
x→−2−
lim g(x) = 2
x→−2+
• lim g(x) does not exist (since the limit from the right of x = 0 is 1 and the limit from
x→0
the left of x = 0 is 2)
• lim− g(x) = 1
x→4
• lim+ g(x) = −2
x→4
• lim g(x) = 3
x→2