Review for Math 150A Exam 1 Answer Key
(1.2) In problems 1-4, use the graph at right to evaluate each. If the quantity does not exist, explain why.
1. a. 𝑓(0) = 𝐷𝑁𝐸 (undefined) 𝑓(𝑥)
b. 𝑓(2) = 2
c. lim+ 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→0
d. lim 𝑓(𝑥) = 1
𝑥→2
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = 0 because 𝑓(0) = 𝐷𝑁𝐸
OR because lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸).
𝑥→0
Discontinuous at 𝑥 = 2 because lim 𝑓(𝑥) ≠ 𝑓(2).
𝑥→2
2. a. 𝑓(−4) = 2 𝑓(𝑥)
b. 𝑓(−1) = 𝐷𝑁𝐸 (undefined)
c. lim − 𝑓(𝑥) = −∞ (𝐷𝑁𝐸)
𝑥→−4
d. lim 𝑓(𝑥) = 1
𝑥→−1
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = −4 because lim 𝑓(𝑥) = 𝐷𝑁𝐸.
𝑥→−4
Discontinuous at 𝑥 = −1 because 𝑓(−1) = 𝐷𝑁𝐸.
3. a. 𝑓(−3) = 𝐷𝑁𝐸 (undefined) 𝑓(𝑥)
b. 𝑓(1) = −1
c. lim 𝑓(𝑥) = 𝐷𝑁𝐸 (left & right limits do not match)
𝑥→1
d. lim+ 𝑓(𝑥) = 3
𝑥→3
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = −3 because lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸).
𝑥→−3
Discontinuous at 𝑥 = 1 because lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥), therefore lim 𝑓(𝑥) = 𝐷𝑁𝐸.
𝑥→1 𝑥→1 𝑥→1
Discontinuous at 𝑥 = 3 because lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥), therefore lim 𝑓(𝑥) = 𝐷𝑁𝐸.
𝑥→3 𝑥→3 𝑥→3
Math 150A, Exam 1 Review with solutions, Page 1 of 17, Cypress College
,4. a. 𝑓(−2) = −2 𝑓(𝑥)
b. 𝑓(1) = 𝐷𝑁𝐸 (undefined)
c. lim 𝑓(𝑥) = 1
𝑥→1
d. lim 𝑓(𝑥) = 𝐷𝑁𝐸 (left & right limits do not match)
𝑥→−2
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = −2 because lim 𝑓(𝑥) ≠ lim + 𝑓(𝑥), therefore lim 𝑓(𝑥) = 𝐷𝑁𝐸
𝑥→−2− 𝑥→−2 𝑥→−2
Discontinuous at 𝑥 = 1 because 𝑓(1) = 𝐷𝑁𝐸.
f is defined for all values except the one given. Sketch the graph of a function that satisfies all given
conditions:
5. 𝑓 is defined for all values of 𝑥 ≠ 4 and the other value given.
𝑓(−1) = 𝐷𝑁𝐸
lim 𝑓(𝑥) = −2
𝑥→−1
lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→4
*Each graph you see for problems 5-8 will represent one of many
possible solutions. If you ever have a question as to whether your graph satisfies all conditions, please ask me
before the exam and I will take a look.
6. 𝑓 is defined for all values of 𝑥 ≠ −2
𝑓(4) = 0
lim 𝑓(𝑥) = 𝐷𝑁𝐸
𝑥→4
lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→−2+
lim 𝑓(𝑥) = −∞ (𝐷𝑁𝐸)
𝑥→−2−
7. 𝑓 is defined for all values of 𝑥 ≠ 3
𝑓(−4) = 2
lim 𝑓(𝑥) = 3
𝑥→−4
lim 𝑓(𝑥) = 𝐷𝑁𝐸
𝑥→3
Math 150A, Exam 1 Review with solutions, Page 2 of 17, Cypress College
, 8. 𝑓 is defined for all values of 𝑥 ≠ ± 3
𝑓(0) = 0
𝑓(−5) = 1
lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→−3+
(1.2) Write the proof:
9. lim (−2𝑥 + 5) = −9
𝑥→7
𝜀
For any 𝜀 > 0, there exists a 𝛿 = 2 such that,
if |𝑥 − 7| < 𝛿, then
|(−2𝑥 + 5) − (−9)| = | − 2𝑥 + 14|
= | − 2| ∙ |𝑥 − 7|
= 2|𝑥 − 7|
< 2𝛿
𝜀
< 2 (2)
<𝜀
1
10. lim (2 𝑥 + 6) = 5
𝑥→−2
For any 𝜀 > 0, there exists a 𝛿 = 2𝜀 such that,
if |𝑥 − (−2)| = |𝑥 + 2| < 𝛿, then
1 1
|(2 𝑥 + 6) − (5)| = |2 𝑥 + 1|
1
= |2| ∙ |𝑥 + 2|
1
= 2 |𝑥 + 2|
1
< 2𝛿
1
< 2 (2𝜀)
<𝜀
Math 150A, Exam 1 Review with solutions, Page 3 of 17, Cypress College
(1.2) In problems 1-4, use the graph at right to evaluate each. If the quantity does not exist, explain why.
1. a. 𝑓(0) = 𝐷𝑁𝐸 (undefined) 𝑓(𝑥)
b. 𝑓(2) = 2
c. lim+ 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→0
d. lim 𝑓(𝑥) = 1
𝑥→2
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = 0 because 𝑓(0) = 𝐷𝑁𝐸
OR because lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸).
𝑥→0
Discontinuous at 𝑥 = 2 because lim 𝑓(𝑥) ≠ 𝑓(2).
𝑥→2
2. a. 𝑓(−4) = 2 𝑓(𝑥)
b. 𝑓(−1) = 𝐷𝑁𝐸 (undefined)
c. lim − 𝑓(𝑥) = −∞ (𝐷𝑁𝐸)
𝑥→−4
d. lim 𝑓(𝑥) = 1
𝑥→−1
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = −4 because lim 𝑓(𝑥) = 𝐷𝑁𝐸.
𝑥→−4
Discontinuous at 𝑥 = −1 because 𝑓(−1) = 𝐷𝑁𝐸.
3. a. 𝑓(−3) = 𝐷𝑁𝐸 (undefined) 𝑓(𝑥)
b. 𝑓(1) = −1
c. lim 𝑓(𝑥) = 𝐷𝑁𝐸 (left & right limits do not match)
𝑥→1
d. lim+ 𝑓(𝑥) = 3
𝑥→3
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = −3 because lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸).
𝑥→−3
Discontinuous at 𝑥 = 1 because lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥), therefore lim 𝑓(𝑥) = 𝐷𝑁𝐸.
𝑥→1 𝑥→1 𝑥→1
Discontinuous at 𝑥 = 3 because lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥), therefore lim 𝑓(𝑥) = 𝐷𝑁𝐸.
𝑥→3 𝑥→3 𝑥→3
Math 150A, Exam 1 Review with solutions, Page 1 of 17, Cypress College
,4. a. 𝑓(−2) = −2 𝑓(𝑥)
b. 𝑓(1) = 𝐷𝑁𝐸 (undefined)
c. lim 𝑓(𝑥) = 1
𝑥→1
d. lim 𝑓(𝑥) = 𝐷𝑁𝐸 (left & right limits do not match)
𝑥→−2
e. Find all points of discontinuity. Explain which part of the definition
of continuity fails at each point.
Discontinuous at 𝑥 = −2 because lim 𝑓(𝑥) ≠ lim + 𝑓(𝑥), therefore lim 𝑓(𝑥) = 𝐷𝑁𝐸
𝑥→−2− 𝑥→−2 𝑥→−2
Discontinuous at 𝑥 = 1 because 𝑓(1) = 𝐷𝑁𝐸.
f is defined for all values except the one given. Sketch the graph of a function that satisfies all given
conditions:
5. 𝑓 is defined for all values of 𝑥 ≠ 4 and the other value given.
𝑓(−1) = 𝐷𝑁𝐸
lim 𝑓(𝑥) = −2
𝑥→−1
lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→4
*Each graph you see for problems 5-8 will represent one of many
possible solutions. If you ever have a question as to whether your graph satisfies all conditions, please ask me
before the exam and I will take a look.
6. 𝑓 is defined for all values of 𝑥 ≠ −2
𝑓(4) = 0
lim 𝑓(𝑥) = 𝐷𝑁𝐸
𝑥→4
lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→−2+
lim 𝑓(𝑥) = −∞ (𝐷𝑁𝐸)
𝑥→−2−
7. 𝑓 is defined for all values of 𝑥 ≠ 3
𝑓(−4) = 2
lim 𝑓(𝑥) = 3
𝑥→−4
lim 𝑓(𝑥) = 𝐷𝑁𝐸
𝑥→3
Math 150A, Exam 1 Review with solutions, Page 2 of 17, Cypress College
, 8. 𝑓 is defined for all values of 𝑥 ≠ ± 3
𝑓(0) = 0
𝑓(−5) = 1
lim 𝑓(𝑥) = ∞ (𝐷𝑁𝐸)
𝑥→−3+
(1.2) Write the proof:
9. lim (−2𝑥 + 5) = −9
𝑥→7
𝜀
For any 𝜀 > 0, there exists a 𝛿 = 2 such that,
if |𝑥 − 7| < 𝛿, then
|(−2𝑥 + 5) − (−9)| = | − 2𝑥 + 14|
= | − 2| ∙ |𝑥 − 7|
= 2|𝑥 − 7|
< 2𝛿
𝜀
< 2 (2)
<𝜀
1
10. lim (2 𝑥 + 6) = 5
𝑥→−2
For any 𝜀 > 0, there exists a 𝛿 = 2𝜀 such that,
if |𝑥 − (−2)| = |𝑥 + 2| < 𝛿, then
1 1
|(2 𝑥 + 6) − (5)| = |2 𝑥 + 1|
1
= |2| ∙ |𝑥 + 2|
1
= 2 |𝑥 + 2|
1
< 2𝛿
1
< 2 (2𝜀)
<𝜀
Math 150A, Exam 1 Review with solutions, Page 3 of 17, Cypress College
