Download Math 3 Additional Problems Suggested Book Problems: 1.3: 2, 3, 27

Math 3 Additional Problems
Suggested Book Problems:
1.3:
1.4:
1.5:
1.6:
1.7:
2,
1,
2,
1,
1,
3,
5
4,
2,
2,
27, 35, 40, 52, 65
5, 12, 17, 40a
11, 15, 18, 49
19
Other Suggested Problems:
(1) Let f (x) =
√
x + 1. What is the domain of f ? Evaluate lim f (x).
x→1
Solution:
We must have
√ domain is [−1, ∞). Since f is continuous on its
à x + 1ó 0, so the
domain, lim x + 1 = 1 + 1 = 2.
x→1
(2) Let f (x) =
x2 −4
x−2
What is the domain of f ? Evaluate lim f (x).
x→2
Solution:
We must have x − 2 6= 0, so the domain of f is (−∞, 2) ∪ (2, ∞) (the union of
the two intervals). Alternatively, we could write {x : x 6= 2}. Both answers are
acceptable.
(3) Let f (x) =
ator.]
x2 −3x+2
.
(x−1)(x−5)
At what points does the limit exist? [Hint: factor the numer-
Solution:
Factoring, we get f (x) = (x−1)(x−2)
. Since f is a rational function, it is continuous
(x−1)(x−5)
at the points in the domain. Points we have to check (those that are not in the
domain) are x = 1 and x = 5. At points near x = 1, we have f (x) = x−2
, so
x−5
1−2
lim f (x) =
= 1/4. At x = 5, f has a vertical asymptote. If x > 5 f (x) =
x→1
1−5
x−2
> 0, since x − 2 > 0 and x − 5 > 0. If 2 < x < 5, we have x − 2 > 0 and x − 5 < 0,
x−5
x−2
so x−5 < 0. Therefore, the limit does not exist because lim+ = ∞ and lim− = −∞.
x→5
1
x→5
√
(4) Evaluate lim
x→1
x2 + 3 − 4
.
x−2
Solution:
√
√
√
The numerator is a root function, so lim x2 + 3 − 4 = 1 + 3 − 4 = 4 − 4 =
x→1
2 − 4 = −2. The denominator is a polynomial so lim x − 2 = 1 − 2 = −1. Thus,
x→1
√
x2 + 3 − 4
−2
lim
=
= 2.
x→1
x−2
−1
(5) Evaluate limπ
x→ 3
sin(x)
. Which limit law are you using?
cos(x)
Solution:
Both sin(x) and
cos(x) are continuous at π3. We can first evaluate separately
π π
and limπ cos(x) = cos
. Because cos π3 6= 0, we can use
limπ sin(x) = sin
x→ 3
x→ 3
3
3 √
sin π3
sin(x)
the quotient limit law. Thus, limπ
=
=
3.
x→ 3 cos(x)
cos π3
(6) State the formal definition of a limit: what does it mean if lim f (x) = L? Can you
x→a
explain intuitively what this means?
Solution:
We say that lim f (x) = L if for every ε > 0, there is a number δ > 0 such that if
x→a
|x − a| < δ, then |f (x) − L| < ε. This means that in order for the limit L to exist at
x = a, it must be the case that as x gets closer and closer to a, f (x) gets closer and
closer to L. In fact, we can find values of x close to a such that f (x) is arbitrarily
close to L (as close as we want it!)
(7) Prove that lim 5x + 2 = 12 using the ε − δ definition of a limit. What value of δ did
x→2
you use? Why?
Solution:
Step 1: Since f (x) = 5x + 2 is linear, we can check f (2) = 10 + 2 = 12. This is
our candidate for L.
Step 2: Fix an arbitrary ε > 0.
2
Step 3: We need to find the values of x such that |f (x) − L| < ε.
|5x + 2 − 12| < ε ⇒ |5x − 10| < ε
⇒ −ε < 5x − 10 < ε
⇒ −ε + 10 < 5x < ε + 10
ε + 10
ε + 10
⇒−
<x<
5
5
Step 4: To find delta, we must solve for |x − a|, which is |x − 2|. We have
< x < ε+10
, so − ε+10
− 2 < x − 2 < ε+10
− 2. Finally, −2 = −10
, so this
5
5
5
5
ε
ε
ε
becomes − 5 < x − 2 < 5 . Thus, δ = 5 .
− ε+10
5
Step 5: We must verify that if |x − 2| < δ, then |f (x) − 12| < ε:
ε
5
⇒ |5x − 10| < ε
|x − 2| < δ ⇒ |x − 2| <
⇒ |5x + 2 − 12| < ε
⇒ |f (x) − L| < ε
(8) Suppose that f , g and h are functions, and that lim f (x) = 2, lim g(x) = 1 and
x→2
x→2
lim h(x) = 0. Determine the following limits if they exist and state which limit laws
x→2
you are using. If the limit does not exist, explain why:
a. lim (f + 3g)(x)
x→2
b. lim f g(x)
x→2
f
c. lim (x)
x→2 h
f
d. lim
+ h (x)
x→2
g
Solution:
a. Using the limit laws for addition and multiplication by a constant, we have
lim f (x) = 2 and lim 3g(x) = 3(1) = 3, so lim (f + 3g)(x) = 2 + 3 = 5.
x→2
x→2
x→2
b. Using the limit laws for multiplication of two functions, we have lim f (x) = 2
x→2
and lim g(x) = 1, so lim f g(x) = 2(1) = 2.
x→2
x→2
3
c. Note that the denominator is h(x) and lim h(x) = 0, so f has a vertical asympx→2
f
tote at x = 2. We cannot say if lim (x) exists, because it depends on whether or
x→2 h
not the right and left limits are the same (they will both be either ∞ or −∞).
f
d. Using the limit laws for addition and quotients of functions, we have lim
+ h (x) =
x→2
g
2
+ 0 = 2.
1
(9) Draw the graph of a function f satisfying all of the following properties:
i. lim+ f (x) = 1
x→2
ii. lim− f (x) = 3
x→2
iii. f (2) = 0
iv. lim f (x) = 0
x→4
v. f has a vertical asymptote at x = 0
Solution:
There are many different ways to draw such a function. Here is one:
1
(10) Evaluate lim x sin
. What theorem are you using?
x→0
x
2
Solution:
Note that −1 ≤ sin x1 ≤ 1. Since
x2 is always positive, we can multiply all sides
by x2 to obtain −x2 < x2 sin x1 < x2 . Since lim −x2 = 0 and lim x2 = 0, we have
x→0
x→0
1
2
lim x sin
= 0 by the Squeeze Theorem.
x→0
x
4
(11) Let f (x) = sin(x) + 2 and g(x) = x cos
1
x
. Evaluate lim [f (x) − g(x)]. What
x→π/2
rules/theorems are you using?
Solution:
π + 2, since sin(x) is continuous. −g(x) is also continuous
2
π
at π2 , since it is the product of a linear function
and a trig function, and 2 is in its
π
2
domain. Therefore limx→ π2 −g(x) = − 2 cos π .
First, limπ f (x) = sin
x→ 2
(12) Let f (x) =
x2 +4x−5
.
x(x−1)(x−3)
Where does f have vertical asymptotes, if any?
Solution:
A vertical asymptote occurs when the denominator is 0. Thus, the points we must
check are x = 0, x = 1 and x = 3. Factoring the numerator, we have f (x) =
(x+5)(x−1)
x+5
. At points near 0, we can cancel so we get f (x) = x(x−3)
. No more
x(x−1)(x−3)
canceling can be done, so we have a vertical asymptote at x = 0.
x−5
1−5
−4
(x − 5)(x − 1)
= lim
=
=
= 2.
We have lim f (x) = lim
x→1 x(x − 3)
x→1
x→1 x(x − 1)(x − 3)
1(1 − 3)
−2
Therefore, there is no vertical asymptote at x = 1.
Finally, using the same reasoning as for x = 0, at points near x = 3, we have
x+5
f (x) = x(x−3)
. No more canceling can be done, so we have a vertical asymptote at
x = 3.
5