Math 100 - Homework Set 1 (Limits and Continuity)
due date: Thursday February the 4th at 8am (in class or by e-mail)
Basic Skills required to work through the problems:
• factoring quadratic and cubic polynomials;
• manipulating (factoring, simplifying, razionalizing, etc.) expressions with ratios
and square roots of polynomials;
• interpreting and manipulating expressions containing absolute values;
• factoring and simplifying expressions containing ratios and square roots of polynomials;
• determining the domain of a function;
Learning Goals: After completing this set, you should be able to:
• master the basic skills listed above;
• compute the limit of a function at a point using limit laws and algebraic manipulation techniques, or determine that such limit does not exists.
• compute the limit of a function at a point using limit laws, algebraic manipulation
techniques, or the squeeze theorem, or determine that such limit does not exist;
• evaluate the limit of a function at infinity using limit laws and algebraic manipulation techniques, or determine that such limit does not exist;
• determine if a function has horizontal or vertical asymptotes, and find their equation;
• identify whether a function is discontinuous and, if possible, find conditions for
which the discontinuity can be eliminated;
• apply the Intermediate Value Theorem, use it to construct simple proofs about a
given statement.
1. A warm up. Write out factorizations of the following expressions:
(a) x2 − 25
Solution: x2 − 25 = (x − 5)(x + 5).
(b) x2 − x − 2
Solution: x2 − x − 2 = (x − 2)(x + 1).
(c) x3 − 8
Solution: x3 − 8 = (x − 2)(x2 + 2x + 4).
2. More warm up. Simplify:
x2 − 25
(a)
x+5
Solution: x2 − 25 = (x − 5)(x + 5) so, cancelling the factor x + 5, (x2 −
25)/(x + 5) = x − 5 for x 6= −5.
(b)
x2 − x − 2
x−2
Solution: x2 − x − 2 = (x − 2)(x + 1) so, cancelling the factor x − 2,
(x2 − x − 2)/(x − 2) = x + 1 for x 6= 2.
(c)
x2 − 4
x3 − 8
Solution: x3 − 8 = (x − 2)(x2 + 2x + 4) so,
(x − 2)(x + 2)
x+2
x2 − 4
=
=
x3 − 8
(x − 2)(x2 + 2x + 4)
x2 + 2x + 4
for x 6= 2.
3. Evaluate the following limits.
x2 − 25
(a) lim
x→−5 x + 5
Solution: Using the result of question 2(a) above,
x2 − 25
= lim (x − 5) = −5 − 5 = −10.
x→−5 x + 5
x→−5
lim
(b) lim
x→−1
x2 − 25
x+5
Solution: Here, direct substitution works since we have a rational function
that is defined at the limiting point; the answer is
(−1)2 − 25
−24
=
= −6.
−1 + 5
4
x2 − x − 2
(c) lim
x→2
x−2
Page 2
Solution: Using the result of question 2(b), the limit is lim (x + 1) =
x→2
2 + 1 = 3.
(d) lim
x→2
x2 − 4
x3 − 8
Solution: Using the result of question 2(c), the limit is
lim
x→2 x2
x+2
,
+ 2x + 4
and this can be evaluated using direct substitution and equals
2+2
4
1
=
= .
4+4+4
12
3
4. Evaluate the following limit:
1
+ 12
x
x→−2 x3 + 8
lim
Solution:
1
+ 21
x
lim
x→−2 x3 + 8
=
2+x
2x
lim
x→−2 x3 + 8
2+x
x→−2 2x(x3 + 8)
2+x
= lim
x→−2 2x(x + 2)(x2 − 2x + 4)
1
= lim
x→−2 2x(x2 − 2x + 4)
1
1
=
=−
−48
48
= lim
5. Evaluate the following limit:
lim
x→0
|2x − 2| − |2x + 2|
x
Page 3
Solution: There are a few ways to deal with absolute values but you can always
fall back on the basic definitions
(
(
2x − 2 if 2x − 2 ≥ 0
2x + 2
if 2x + 2 ≥ 0
|2x − 2| =
|2x + 2| =
2 − 2x if 2x − 2 < 0
−2 − 2x if 2x + 2 < 0
Now for x near 0 we have 2x − 2 < 0 and 2x + 1 ≥ 0 and so for x near 0
|2x − 2| − |2x + 2| = −(2x − 2) − (2x + 2) = −4x
Then
lim
x→0
−4x
|2x − 2| − |2x + 2|
= lim
x→0 x
x
−4
= −4.
= lim
x→0 1
6. (a) What is the domain of the function g(t) =
√1
t 1+t
−
1
t
?
Solution: t 6= 0 and 1 + t > 0 → t > −1, so t ∈ (−1, 0) ∪ (0, +∞)
(b) Evaluate the following limit:
lim
t→0
1
1
√
−
t 1+t t
Solution:
√
1
1
1− 1+t
lim √
− = lim √
t→0 t 1 + t
t→0 t 1 + t
t
√
√
(1 − 1 + t)(1 + 1 + t)
√
√
= lim
t→0 (t 1 + t)(1 +
1 + t)
1 − (1 + t)
√
= lim √
t→0 t 1 + t(1 +
1 + t)
−t
√
= lim √
t→0 t 1 + t(1 +
1 + t)
−1
√
= lim √
t→0
1 + t(1 + 1 + t)
−1
√
=√
1 + 0(1 + 1 + 0)
−1
1
=
=− .
2
2
Page 4
7. Evaluate the following limit:
√
25 + h − 4
lim
h→0
h
Solution:
Multiplying and dividing by the rationalizing factor
√
25 + h + 4 yields
9+h
(25 + h) − 16
√
= lim √
,
h→0 h( 25 + h + 4)
h→0 h( 25 + h + 4)
lim
lim+
9+h
√
= +∞,
h( 25 + h + 4)
lim−
9+h
√
= −∞,
h( 25 + h + 4)
h→0
while
h→0
so the limit doesn’t exist.
8. True or False. If false, give a counter-example.
(a) If f (−1) = −1 and f (1) = 1, then f (0) = 0.
Solution: This statement is false. There are many functions f that satisfy
f (−1) = −1, f (1) = 1, but f (0) 6= 0. One such counterexample is the
function that takes value −1 for all x < 0 and value 1 for all x ≥ 0; for
this function, f (0) = 1 6= 0.
(b) If f (−1) = −1 and f (1) = 1, then there is a point c such that −1 < c < 1 and
f (c) = 0.
Solution: This statement is also false, and the same counterexample as
in part (i) works.
(c) If f (−1) = −1 and f (1) = 1, and f (x) is continuous then there is a point c
such that −1 < c < 1 and f (c) = 0.
Solution: This statement is true, and is a direct consequence of the Intermediate Value Theorem.
9. Use the Intermediate Value Theorem to show that the function f (x) = x3 + 2x2 −
4x − 1 has at least three zeros in the interval [−4, 4].
Page 5
Solution: We compute a table of values:
x
f (x)
−4 −3 −2 −1 0
1 2
−17 2
7
4 −1 −2 7
3
32
4
79
From the table we see that by the intermediate value theorem, there must be a
root in (−4, −3). This is because the function is continuous (it is a polynomial)
and is negative at x = −4 and positive at x = −3 and so for some value p with
−4 < p < −3, we will have f (p) = 0. Also there is a root in (−1, 0) and a root
in (1, 2). There are other clever strategies that would avoid computing so many
values. You only need the values for −4, −3, 0, 2 for example.
10. Consider the following function, where a is an unspecified parameter:
(
f (x) =
(x − 3)2 + a, x ≥ 1;
1
+ 2a − 3, x < 1.
2x
Then f (x) could be discontinuous for, at most, two values of x. Identify these
x-values. Choose a so that one of these discontinuities disappears.
Solution: f (x) is discontinuous at x = 0 and could be discontinuous for x = 1.
In order to be continuous we must have
lim−
x→1
1
+ 2a + 3 = lim+ (x − 3)2 + a
x→1
2x
1
Now limx→1− 2x
+ 2a − 3 = 21 + 2a − 3 = − 52 + 2a and limx→1+ (x − 3)2 + a =
(1 − 3)2 + a = 4 + a. Thus we can make f (x) continuous at x = 1 by choosing
a so that − 25 + 2a = 4 + a, and so for a = 13/2.
4 − 3x
.
x+7
(a) Find the following limits: (i) lim f (x), (ii) lim f (x), (iii) lim f (x).
11. Let f (x) =
x→∞
Solution: (i) limx→∞
4−3x
x+7
= limx→∞
x→−∞
4−3x
x+7
1
x
1
x
= limx→∞
x→−7
4
−3
x
1+ x7
=
−3
1
= −3
(ii) limx→−∞ f (x) = −3 by same argument.
(iii) limx→−7 4−3x
does not exist. In particular, limx→−7+ 4−3x
= ∞ since
x+7
x+7
+
+
limx→−7 (4 − 3x) = 25 > 0 and limx→−7 (x + 7) = 0 and x + 7 > 0 for
= −∞ and hence the limit as x → −7
x → −7+ . Similarly limx→−7− 4−3x
x+7
does not exist.
(b) Specify any horizontal or vertical asymptotes (if any).
Page 6
Solution: Horizontal asymptote y = −3; vertical asymptote x = −7.
(c) Roughly sketch the graph of f (x) using the information you found in parts (a)
and (b).
12. (Optional) Show by means of example that lim [f (x) + g(x)] may exist even though
x→a
neither lim f (x) nor lim g(x) exists.
x→a
x→a
Solution: There are many ways to construct such examples. The easiest way
is to let f (x) be a function that does not have a limit as x approaches a and
g(x) = −f (x), which will also not have a limit. But then f (x) + g(x) is the 0
function, which certainly does have a limit, namely 0! To be specific, we could
take f (x) = 1/x, g(x) = −1/x, and a = 0.
Page 7