Math 150A Practice Set: Infinitesimals and Infinities
This practice set is designed for students of CSUNs Math 150A to test their
understanding of the formal definitions, algebraic properties, and resolution
techniques for limits involving infinitesimals and infinities.
Part I: Conceptual and True/False (Questions 1-5)
Question 1 (Definition Recall):
Provide the formal limit definition for a function
f(x)
to be an infinitesimal as
x → a
.
Question 2 (Definition Recall):
Provide the formal limit definition for a function
g(x)
to be an infinity as
x → a
.
Question 3 (True/False):
True or False: The product of an infinity,
f(x)
, and an infinitesimal,
g(x)
, as
x → a
always results in a limit of 1.
Question 4 (Property Application):
If
f(x)
is an infinitesimal as
x → a
, and
g(x)
is a function that is bounded (i.e.,
¿g(x)∨≤ M
) as
x → a
, what can be definitively concluded about the product
lim
x→ a
f(x)g(x)
?
Question 5 (The Reciprocal Link):
State the exact relationship between the reciprocal of a non-zero infinitesimal and
an infinity.
Part II: Infinitesimal Properties and Bounded Functions (Questions 6-10)
Evaluate the following limits by applying the properties of infinitesimals and
bounded functions.
Question 6:
lim
x →∞
sin(x3)
x
Question 7:
lim
x →0
xcos
(
1
x
)
Question 8:
lim
x →2
(x − 2)⋅x2+5
x+1
Question 9:
lim
x →0
(
x4+3x
)
Question 10:
Determine which of the following is not an infinitesimal as
x → ∞
:
A)
1
x2
B)
e− x
C)
arctan (x)−π
2
D)
sin (x)
Part III: Infinity Properties and Growth Hierarchy (
∞
∞
) (Questions 11-15)
Evaluate the following limits involving the indeterminate form
∞
∞
.
Question 11:
lim
x →∞
3x4−2x2+7
5x4+x −1
Question 12:
lim
x →− ∞
√
9x2+x
x − 5
Question 13:
lim
x →∞
x5
ln (x)
Question 14:
lim
x →∞
e2x
x3
Question 15:
lim
x →∞
x−1+x−2
x−3+x−4
Part IV: Resolving
0
0
Indeterminate Forms (Questions 16-20)
Resolve the following limits by factoring, simplifying, or using conjugates to
eliminate the factor causing the
0
0
form.
Question 16 (Factoring):
lim
x →− 5
x2+6x+5
x+5
Question 17 (Difference of Cubes):
lim
x→ 1
x3−1
x − 1
Question 18 (Conjugate):
lim
x → 0
√
x+9−3
x
Question 19 (Common Denominator):
lim
x → 0
1
x+2−1
2
x
Question 20 (Algebraic Manipulation):
lim
h→ 0
¿¿
Part V: Resolving
∞ −∞
Indeterminate Forms (Questions 21-25)
Resolve the following limits by using conjugates or finding a common denominator
to convert the
∞ −∞
form into a resolvable
∞
∞
or
0
0
form.
Question 21 (Conjugate):
lim
x→ ∞
(
√
x2+4x − x )
Question 22 (Common Denominator):
lim
x→ 0+¿
(
1
√
x−1
x
)
¿
¿
Question 23 (Factoring Dominant Term):
lim
x→ ∞
(x3−2x2−100 x)
Question 24 (Common Denominator):
lim
x→ ∞
(
x2
x+3− x
)
Question 25 (Conjugate):
lim
x→ ∞
(
√
4x2−3−2x)
Part VI: Mixed and Advanced Application (Questions 26-30)
Question 26 (Substitution Method):
Evaluate the
0⋅∞
form by using the substitution
t=1/x
:
lim
x→ ∞
xsin
(
1
x
)
Question 27 (Finding
a
for Infinity):
For what value(s) of
a
is the function
f(x)= x2+1
x2− a2
an infinity as
x → a
?
Question 28 (Rate of Convergence):
As
x → ∞
, determine which function approaches zero faster:
f(x)= 1
x3or g(x)= 1
ex
Question 29 (General Properties):
If
f(x)
and
g(x)
are both infinitesimals as
x → a
, which of the following expressions
is guaranteed to be an infinitesimal as
x → a
?
A)
f(x)
g(x)
B)
f(x)+g(x)
C)
1
f(x)
D)
f(x)−1
g(x)
Question 30 (Identifying Indeterminate Forms):
Identify the indeterminate form(s) that must be converted algebraically before
LHôpitals Rule can be applied:
A)
0⋅∞
B)
1∞
C)
∞0
D) All of the above
Solutions and Explanations
Part I: Conceptual and True/False
Solution 1:
lim
x→ a
f(x)=0
Solution 2:
lim
x→ a
¿g(x)∨¿∞or lim
x→ a
g(x)=± ∞
Solution 3:
False. The product of an infinity and an infinitesimal (
0⋅∞
) is an indeterminate
form. The limit can be any value (a finite number,
∞
,
− ∞
, or 0) depending on the
relative rates at which the functions approach their respective limits.
Solution 4:
The product
lim
x→ a
f(x)g(x)
must be 0. This is due to the property that an infinitesimal
multiplied by a bounded function is always an infinitesimal. This is often proven
using the Squeeze Theorem.
Solution 5:
The reciprocal of a non-zero infinitesimal is an infinity, and the reciprocal of an
infinity is an infinitesimal. Algebraically: If
lim
x→ a
α(x)=0
(with
α(x)≠0
), then
lim
x → a
1
α(x)=∞
or
− ∞
.
Part II: Infinitesimal Properties and Bounded Functions
Solution 6:
The function can be written as
f(x)= 1
x⋅sin (x3)
. As
x → ∞
,
1
x→0
(infinitesimal), and
sin (x3)
is bounded between
−1
and
1
.
lim
x →∞
sin (x3)
x=0
Solution 7:
As
x → 0
,
x → 0
(infinitesimal), and
cos (1/x)
is bounded between
−1
and
1
.
lim
x →0
xcos
(
1
x
)
=0
Solution 8:
As
x → 2
,
(x − 2)→0
(infinitesimal). The second factor,
x2+5
x+1
, is continuous at
x=2
and approaches
22+5
2+1=9
3=3
(a finite constant). The product of an infinitesimal and
a finite constant is an infinitesimal.
lim
x →2
(x − 2)⋅x2+5
x+1=0⋅3=0
Solution 9:
The sum of two infinitesimals (
x4
and
3x
as
x → 0
) is an infinitesimal.
lim
x →0
(x4+3x)=04+3(0)=0
Solution 10:
D)
sin (x)
. As
x → ∞
,
sin (x)
oscillates between
−1
and
1
. Its limit does not equal zero;
therefore, it is not an infinitesimal. A, B, and C all have limits of 0 as
x → ∞
.
Part III: Infinity Properties and Growth Hierarchy (
∞
∞
)
Solution 11:
The highest degree term in both the numerator and denominator is
x4
. We only
consider the ratio of the coefficients of the dominant terms:
lim
x →∞
3x4
5x4=3
5
Solution 12:
Since
x → − ∞
, we must take
x=−
√
x2
. Divide numerator and denominator by
x
:
lim
x →− ∞
√
9x2+x
x
x −5
x
=
lim
x → −∞
−
√
9x2
x2+x
x2
1−5
x
=
lim
x →− ∞
−
√
9+1
x
1−5
x
=−
√
9
1=−3
Solution 13:
This is
∞
∞
. We use the Growth Hierarchy: Polynomial growth (
x5
) is faster than
Logarithmic growth (
ln (x)
). Since the dominant term is in the numerator:
lim
x →∞
x5
ln (x)=∞
Solution 14:
This is
∞
∞
. Exponential growth (
e2x
) is faster than Polynomial growth (
x3
). Since the
dominant term is in the denominator:
lim
x →∞
x3
e2x=0
Solution 15:
Rewrite the expression with positive exponents:
lim
x →∞
1
x+1
x2
1
x3+1
x4
Multiply numerator and denominator by
x4
(the lowest common multiple of the
denominators):
lim
x →∞
x3+x2
x+1
The dominant terms are
x3
(numerator) and
x
(denominator). Since the numerators
degree is higher:
lim
x →∞
x3
x=lim
x →∞
x2=∞
Part IV: Resolving
0
0
Indeterminate Forms
Solution 16 (Factoring):
lim
x →− 5
(x+5)(x+1)
x+5=lim
x→ −5
(x+1)=−5+1=−4
Solution 17 (Difference of Cubes):
Factor the numerator using
a3− b3=(a −b)(a2+ab+b2)
:
lim
x→ 1
(x −1)(x2+x+1)
x −1=lim
x→ 1
(x2+x+1)=1+1+1=3
Solution 18 (Conjugate):
Multiply by the conjugate,
√
x+9+3
√
x+9+3
:
lim
x → 0
√
x+9−3
x⋅
√
x+9+3
√
x+9+3=
lim
x→ 0
(x+9)−9
x(
√
x+9+3)=
lim
x → 0
x
x(
√
x+9+3)
Cancel
x
and evaluate:
lim
x → 0
1
√
x+9+3=1
√
9+3=1
3+3=1
6
Solution 19 (Common Denominator):
Find a common denominator in the numerator, which is
2(x+2)
:
lim
x → 0
2−(x+2)
2(x+2)
x=
lim
x→ 0
− x
2x(x+2)
Cancel
x
and evaluate:
lim
x→ 0
−1
2(x+2)=−1
2(0+2)=−1
4
Question 20 (Algebraic Manipulation - Difference Quotient):
Expand the numerator:
lim
h → 0
(x2+2xh+h2)− x2
h=
lim
h → 0
2xh+h2
h
Factor out
h
and cancel:
lim
h → 0
h(2x+h)
h=lim
h → 0
(2x+h)=2x
(This result is the definition of the derivative of
x2
.)
Part V: Resolving
∞ −∞
Indeterminate Forms
Solution 21 (Conjugate):
Multiply by the conjugate,
√
x2+4x+x
√
x2+4x+x
:
lim
x →∞
(x2+4x)− x2
√
x2+4x+x=
lim
x →∞
4x
√
x2+4x+x
Divide numerator and denominator by
x
:
lim
x→ ∞
4
√
x2
x2+4x
x2+x
x
=
lim
x→ ∞
4
√
1+4
x+1
=4
√
1+1=4
2=2
Solution 22 (Common Denominator):
Combine the fractions:
lim
x→ 0+¿
(
1
√
x−1
x
)
=lim
x→0+¿
(
√
x
x−1
x
)
=lim
x →0+¿
√
x −1
x¿
¿¿
¿ ¿
¿
As
x → 0+¿ ¿
, the numerator approaches
√
0−1=−1
. The denominator approaches
0
from the positive side (
0+¿ ¿
). The result is
−1
0+¿=−∞ ¿
.
lim
x→ 0+¿
√
x −1
x=− ∞ ¿
¿
Solution 23 (Factoring Dominant Term):
Factor out the dominant term,
x3
:
lim
x→ ∞
x3
(
1−2
x−100
x2
)
As
x → ∞
,
x3→ ∞
, and the parenthesis approaches
(1−0−0)=1
.
lim
x→ ∞
x3
(
1−2
x−100
x2
)
=(∞)⋅(1)=∞
Solution 24 (Common Denominator):
Combine the terms over the common denominator
x+3
:
lim
x→ ∞
(
x2
x+3−x(x+3)
x+3
)
=
lim
x→ ∞
x2−(x2+3x)
x+3=
lim
x →∞
−3x
x+3
Divide numerator and denominator by
x
:
lim
x →∞
−3
1+3
x
=−3
1+0=−3
Solution 25 (Conjugate):
Multiply by the conjugate,
√
4x2−3+2x
√
4x2−3+2x
:
lim
x→ ∞
(4x2−3)−¿¿
The numerator is a finite constant (-3). The denominator approaches
∞+∞=∞
.
lim
x →∞
−3
∞=0
Part VI: Mixed and Advanced Application
Solution 26 (Substitution Method):
Let
t=1
x
. As
x → ∞
,
t → 0
. The expression becomes:
lim
t → 0
1
tsin (t)=
lim
t →0
sin (t)
t
This is a known fundamental limit:
lim
t → 0
sin (t)
t=1
Solution 27 (Finding
a
for Infinity):
For
f(x)
to be an infinity as
x → a
, the limit must be
± ∞
. This occurs when the
denominator approaches 0 while the numerator approaches a non-zero value.
Set the denominator to zero and solve for
x
:
x2− a2=0⇒x=± a
.
Therefore,
f(x)
is an infinity as
x → a
or as
x → − a
.
The value(s) of
a
for which
f(x)
is an infinity as
x → a
are any real numbers
a ≠ 0
. (If
a=0
,
f(x)= x2+1
x2
is infinity as
x → 0
).
Answer:
a
can be any real number
a
. The points where
f(x)
is an infinity are
x=± a
.
Solution 28 (Rate of Convergence):
We compare the ratio of their reciprocals (their growth rates):
x3
versus
ex
.
The exponential function
g(x)
grows significantly faster than the polynomial
function
f(x)
.
Since
g(x)
grows faster, its reciprocal
1/g(x)
shrinks to zero faster.
lim
x →∞
f(x)
g(x)=
lim
x →∞
1/x3
1/ex=
lim
x →∞
ex
x3=∞
Since the ratio goes to
∞
,
g(x)
is a much smaller infinitesimal than
f(x)
for large
x
.
Answer:
g(x)= 1
ex
approaches zero faster.
Solution 29 (General Properties):
A)
f(x)
g(x)
is the indeterminate form
0
0
. (Uncertain)
B)
f(x)+g(x)
is the sum of two infinitesimals, which is always an infinitesimal.
(Guaranteed)
C)
1
f(x)
is the reciprocal of an infinitesimal, which is an infinity.
D)
f(x)−1
g(x)
is the form
0− ∞=− ∞
.
Answer: B)
f(x)+g(x)
Solution 30 (Identifying Indeterminate Forms):
LHôpitals Rule is defined only for the forms
0
0
and
∞
∞
. The other indeterminate
forms must be converted into one of the primary quotient forms using algebraic
techniques (like the reciprocal relationship or logarithms) before LHôpitals Rule can
be applied.
Answer: D) All of the above.
0⋅∞
can be converted to
0
0
or
∞
∞
.
1∞
,
00
,
∞0
can be converted to
0⋅∞
or
0/0
by using logarithms (i.e., by solving for
ln (L)
).
