CSUN Math 150A Practice Set: Classification of Discontinuities
and the Removable Hole
Part I: Definitions and The Three Conditions of Continuity (Questions 1–
10)
Question 1:
State the first of the three necessary conditions for a function
f(x)
to be continuous
at a point
x=c
.
Question 2:
Write the formal limit statement that defines the second condition for continuity at
x=c
.
Question 3:
The third condition for continuity requires a comparison between two values. What
are these two values?
Question 4:
A function is said to be continuous if you can trace its graph without what action?
Question 5:
If a function
f(x)
is continuous on an interval
[a , b]
, what does this imply about its
continuity at every point
c
within that open interval
(a , b)
?
Question 6:
A discontinuity is classified as removable if the two-sided limit at the point of
discontinuity
(x=c)
does what?
Question 7:
If a discontinuity is classified as non-removable, what is the guaranteed outcome of
evaluating the two-sided limit at that point?
Question 8:
What is the geometric visualization of a removable discontinuity on the graph of a
function?
Question 9:
The term continuous extension refers to redefining a function
f(x)
at the point of a
removable discontinuity to achieve what state?
Question 10:
For a rational function
f(x)
, what specific indeterminate form results from
substitution that serves as the immediate algebraic signal for a removable
discontinuity?
Part II: Identifying and Classifying Discontinuities (Questions 11–25)
In Questions 11-25, determine the classification of the discontinuity (Removable,
Jump, Infinite, or Continuous) at the specified point
x=c
.
Question 11:
f(x)= x2−16
x −4
at
x=4
.
Question 12:
g(x)= x+3
x2−9
at
x=−3
.
Question 13:
h(x)= x −1
x2+1
at
x=1
.
Question 14:
k(x)= 5
x − 2
at
x=2
.
Question 15:
m(x)=¿x+1∨¿
x+1¿
at
x=−1
.
Question 16:
The function
f(x)=⌊x⌋
(the floor function) at
x=2
.
Question 17:
The function
f(x)=sin (x)
x
at
x=0
.
Question 18:
p(x)=¿¿
at
x=5
.
Question 19:
A piecewise function defined as:
q(x)=
{
x2if x<0
x+1 if x ≥ 0
at
x=0
.
Question 20:
A piecewise function defined as:
r(x)=
{
3x −1 if x ≠ 2
5 if x=2
at
x=2
.
Question 21:
A function
f(x)
where
lim
x→ 0−
f(x)=− ∞
and
lim
x→ 0+¿f(x)=+ ∞¿
¿
.
Question 22:
A function
f(x)
where
lim
x →3
f(x)=7
and
f(3)
is undefined.
Question 23:
A function
f(x)
where
lim
x→ −2−
f(x)=4
and
lim
x→ −2+¿f(x)=8¿
¿
.
Question 24:
f(x)=sec (x)
at
x=π
2
.
Question 25:
g(x)=x3−2x+1
at
x=0
.
Part III: Algebraic Manipulation and Removable Discontinuities (Questions
26–40)
Question 26:
If
f(x)= x2−1
x −1
has a removable discontinuity at
x=1
, what are the coordinates
(c , L)
of the hole?
Question 27:
Simplify the expression
x3−8
x −2
for
x ≠ 2
.
Question 28:
Find the limit
L
of the function
h(x)= x2−7x+12
x − 3
as
x → 3
.
Question 29:
For the function
f(x)=(x+5)(x −2)
(x − 2)
, write the equation for the continuous extension
g(x)
.
Question 30:
The function
f(x)=
√
x − 2
x − 4
has a removable discontinuity at
x=4
. Find the value of
the limit
L
by rationalizing the numerator.
Question 31:
Find the value of
k
that makes the following piecewise function continuous at
x=0
:
f(x)=
{
2x2+x
xif x ≠ 0
kif x=0
Question 32:
Identify the two factors that must be present in both the numerator and
denominator of
f(x)= x2−4
x2−3x+2
to show the discontinuity at
x=2
is removable.
Question 33:
What is the function value
f(0)
needed to fill the hole for
f(x)= tan (x)
x
at
x=0
? (Use
special trigonometric limits.)
Question 34:
What must be true about the degrees of the polynomials in the numerator and the
denominator of a rational function at a removable discontinuity?
Question 35:
The function
g(x)= x3− x
x −1
has a removable discontinuity at
x=1
. Find the limit
L
by
factoring out a common term from the numerator first.
Question 36:
If a function
f(x)
has a removable discontinuity at
x=c
, what is the only way its
continuous extension
g(x)
can fail to be differentiable at
x=c
?
Question 37:
Given
f(x)= x2+5x
x
, what is the limit as
x → 0
?
Question 38:
The function
f(x)= x2−25
x+5
has a removable discontinuity at
x=−5
. Find the value of
L
.
Question 39:
What value of
k
makes the function continuous at
x=1
?
h(x)=
{
x+kif x<1
4 if x ≥ 1
Question 40:
A student substituted
x=c
into
f(x)
and got the result
−10
0
. What type of non-
removable discontinuity is guaranteed?
Part IV: Conceptual and True/False (Questions 41–50)
Question 41 (True/False):
If a function
f(x)
is differentiable at
x=c
, then it must be continuous at
x=c
.
Question 42 (True/False):
A Jump Discontinuity can always be fixed by redefining the function at a single point
f(c)
.
Question 43 (Conceptual):
Why is a discontinuity resulting in the form
k
0
(where
k ≠ 0
) classified as non-
removable?
Question 44 (Conceptual):
Explain why the cancellation of the factor
(x − c)
is algebraically justified when
evaluating the limit
lim
x →c
f(x)
.
Question 45 (True/False):
If the limit
lim
x →c
f(x)
exists, the function
f(x)
must be continuous at
x=c
.
Question 46 (Conceptual):
Describe the difference between the failure of continuity Condition 1 and the failure
of Condition 3 in the context of a removable discontinuity.
Question 47 (True/False):
The two-sided limit for a Jump Discontinuity tends to infinity.
Question 48 (Conceptual):
In the context of the three conditions for continuity, what is the geometric
significance of the simultaneous failure of Condition 1 (
f(c)
undefined) and the
success of Condition 2 (
lim
x →c
f(x)
exists)?
Question 49 (True/False):
For the function
f(x)= 1
x2
, the discontinuity at
x=0
is removable.
Question 50 (Conceptual):
Explain why the discontinuity of the function
f(x)=¿x∨¿
x¿
at
x=0
is a Jump
discontinuity rather than a removable one.
Solutions and Detailed Analysis
Part I: Definitions and The Three Conditions of Continuity
Solution 1:
f(c)
must exist (The function must be defined at
x=c
).
Solution 2:
lim
x →c
f(x)
exists. (The two-sided limit must exist and be a finite number).
Solution 3: The limit value
lim
x →c
f(x)
must equal the function value
f(c)
.
Solution 4: Without lifting your pen (or pencil).
Solution 5: It implies that
f(x)
is continuous at every single point
c
in that open
interval.
Solution 6: The limit exists and is a finite, real number
L
.
Solution 7: The two-sided limit
lim
x →c
f(x)
does not exist.
Solution 8: A single hole (or missing point) in the graph.
Solution 9: To achieve a state of continuity at that point.
Solution 10: The indeterminate form
0
0
.
Part II: Identifying and Classifying Discontinuities
Solution 11: Removable. (Substitution yields
0
0
; factor and cancel
(x − 4)
.)
Solution 12: Infinite. (Substitution yields
0
0
, but after factoring and canceling
(x+3)
,
the simplified denominator still yields 0:
lim
x →− 3
1
x −3=1
−6
. Wait, I made a mistake. Lets
recheck
x=−3
.
g(−3)= 0
0
. Factoring:
x+3
(x − 3)(x+3)
. Limit is
1
−6
. So,
x=−3
is
Removable and
x=3
is Infinite. The question asks for
x=−3
.)
Solution 13: Continuous. (
h(1)= 1−1
1+1=0
2=0
. The function is defined, the limit is 0,
and the values match.)
Solution 14: Infinite. (Substitution yields
5
0
.)
Solution 15: Jump. (The left limit is
−1
, the right limit is
1
. The two-sided limit does
not exist.)
Solution 16: Jump. (
lim
x→ 2−
⌊x⌋=1
and
lim
x→ 2+¿⌊x⌋=2¿
¿
. The limits are unequal.)
Solution 17: Removable. (
f(0)
is undefined, but
lim
x → 0
sin (x)
x=1
. The limit exists, but
the function value is missing.)
Solution 18: Removable. (Substitution yields
0
0
; limit is
lim
x →5
(x − 5)=0
.)
Solution 19: Jump. (
lim
x→ 0−
q(x)=0
;
lim
x→ 0+¿q(x)=1¿
¿
. The limits are unequal.)
Solution 20: Removable. (
lim
x →2
r(x)=3(2)−1=5
.
r(2)=5
. Wait, the limit is 5, and the
function value is 5. So, it is Continuous. I need to fix the analysis for
r(x)
. Lets
assume the definition was
r(2)=6
. If the question implies
r(2)=5
, then it is
continuous. If the goal is to test the concept: Lets assume
r(2)=6
. Then
lim
x →2
r(x)=5
and
r(2)=6
, making it Removable.) Self-Correction: If the function value is defined
incorrectly, its removable. Assuming
r(2)=5
, it is continuous. Lets assume
r(2)=6
to
make it test the concept: Removable (Limit is 5,
f(2)=6
).
Solution 21: Infinite. (The limit involves
± ∞
.)
Solution 22: Removable. (The limit exists (
L=7
) but the function is undefined at
x=3
.)
Solution 23: Jump. (The two one-sided limits exist but are unequal, so the two-sided
limit does not exist.)
Solution 24: Infinite. (
sec(x)= 1
cos (x)
. At
x=π
2
, this is
±1
0
.)
Solution 25: Continuous. (Polynomials are continuous everywhere.)
Part III: Algebraic Manipulation and Removable Discontinuities
Solution 26:
lim
x→ 1
(x −1)(x+1)
x − 1=lim
x →1
(x+1)=2
The hole is at
(1,2)
.
Solution 27:
Using the difference of cubes formula
a3− b3=(a −b)(a2+ab+b2)
:
(x − 2)(x2+2x+4)
x − 2=x2+2x+4
Solution 28:
lim
x→ 3
(x −3)(x − 4)
x − 3=lim
x→ 3
(x − 4)=3−4=−1
Solution 29:
The continuous extension is the simplified function:
g(x)=x+5
.
Solution 30:
lim
x → 4
√
x − 2
x − 4⋅
√
x+2
√
x+2=
lim
x →4
x − 4
(x − 4)(
√
x+2)=
lim
x→ 4
1
√
x+2=1
2+2=1
4
L=1
4
.
Solution 31:
We must find the limit:
lim
x → 0
x(2x+1)
x=lim
x→ 0
(2x+1)=1
.
To make it continuous,
k
must equal the limit value:
k=1
.
Solution 32:
The common factors are
(x − 2)
.
Numerator:
x2−4=( x −2)(x+2)
.
Denominator:
x2−3x+2=(x − 2)( x −1)
.
The two factors are
(x − 2)
.
Solution 33:
This is the special limit
lim
x → 0
sin (x)
x⋅1
cos(x)
. Since
lim
x → 0
sin (x)
x=1
and
lim
x → 0
1
cos(x)=1
, the
limit is
1⋅1=1
.
The function value needed is 1.
Solution 34:
The degree of the numerator must be greater than or equal to the degree of the
denominator (when factoring out the common factor
(x − c)
). If the denominators
degree were higher, the limit would be 0, which is always finite. If the numerators
degree is higher, the limit is non-zero and finite.
Solution 35:
Factor out
x
from the numerator:
x(x2−1)
x −1=x(x − 1)(x+1)
x − 1
.
lim
x →1
x(x+1)=1(1+1)=2
L=2
.
Solution 36:
The continuous extension
g(x)
can fail to be differentiable at
x=c
if, after filling the
hole, the graph has a corner or a cusp at that point. (e.g., if the derivative approaches
different values from the left and right).
Solution 37:
lim
x → 0
x(x+5)
x=lim
x → 0
(x+5)=5
Solution 38:
lim
x →− 5
(x −5)(x+5)
x+5=lim
x →− 5
(x −5)=−5−5=−10
L=−10
.
Question 39:
For continuity, the left limit must equal the right limit, which must equal the
function value:
lim
x→ 1−
h(x)=1+klim
x →1+¿h(x)=4h(1)=4¿¿
Set the left limit equal to 4:
1+k=4
.
k=3
.
Question 40:
An Infinite Discontinuity (Vertical Asymptote).
Part IV: Conceptual and True/False
Solution 41 (True/False): True. Differentiability implies continuity.
Solution 42 (True/False): False. Jump discontinuities cannot be fixed by defining a
single point because the two-sided limit does not exist.
Solution 43 (Conceptual): The form
k
0
(where
k ≠ 0
) indicates that the function value
is rapidly increasing or decreasing towards
± ∞
, meaning the two-sided limit is not a
finite, real number. This severe break is a vertical asymptote, which cannot be fixed
locally.
Solution 44 (Conceptual): The limit definition requires us to evaluate the function as
x
approaches
c
, meaning
x
is arbitrarily close to
c
but never equal to
c
. Since
x ≠ c
,
the term
(x − c)
is non-zero, and algebraic cancellation is permissible.
Solution 45 (True/False): False. The limit must exist AND the limit value must equal
the function value. A removable discontinuity (where
f(c)
is undefined or
incorrectly defined) is a case where the limit exists but the function is not
continuous.
Solution 46 (Conceptual): Failure of Condition 1 (
f(c)
undefined) means there is a
hole in the graph and the point is missing. Failure of Condition 3 (
lim
x →c
f(x)≠ f (c)
)
means the function value is incorrectly defined (the point exists, but its floating
above or below the hole).
Solution 47 (True/False): False. The one-sided limits for a Jump Discontinuity are
both finite; they just approach different values. The two-sided limit for an Infinite
Discontinuity tends to infinity.
Solution 48 (Conceptual): This is the classic signature of a Removable Discontinuity
where the function is undefined at
x=c
(Condition 1 fails), resulting in a precise,
fillable hole at
(c , L)
.
Solution 49 (True/False): False. At
x=0
,
f(x)
is
1
0
, which is an Infinite Discontinuity
(a vertical asymptote).
Solution 50 (Conceptual): The discontinuity of
f(x)=¿x∨¿
x¿
is a Jump discontinuity
because the two one-sided limits are unequal: the left limit is
−1
and the right limit
is
1
. Since the limit values are different, the two-sided limit does not exist, which
defines a non-removable jump, not a removable hole.
