MATH 108 - ELEMENTARY AND
INTERMEDIATE ALGEBRA - Determining
When a Limit does not Exist Question Bank
Question 1
Solution: To determine if the limit exists as xapproaches 1, we will evaluate
the limit from the left and the limit from the right separately.
1. Limit from the left:
lim
x→1−
x2−1
x−1= lim
x→1−
(x−1)(x+ 1)
x−1
= lim
x→1−
(x+ 1) = 1 + 1 = 2
2. Limit from the right:
lim
x→1+
x2−1
x−1= lim
x→1+
(x−1)(x+ 1)
x−1
= lim
x→1+(x+ 1) = 1 + 1 = 2
Since the limit from the left and the limit from the right both equal 2, the
overall limit exists and is equal to 2.
Therefore, limx→1
x2−1
x−1= 2.Question 1: Determine the limit, if it
exists, or state that the limit does not exist for the function f(x) =
x2−1
x−1as xapproaches 1.
Solution: To determine if the limit exists as xapproaches 1, we will
evaluate the limit from the left and the limit from the right separately.
1. Limit from the left:
lim
x→1−
x2−1
x−1= lim
x→1−
(x−1)(x+ 1)
x−1
= lim
x→1−
(x+ 1) = 1 + 1 = 2
2. Limit from the right:
lim
x→1+
x2−1
x−1= lim
x→1+
(x−1)(x+ 1)
x−1
1
= lim
x→1+(x+ 1) = 1 + 1 = 2
Since the limit from the left and the limit from the right both
equal 2, the overall limit exists and is equal to 2.
Therefore, limx→1
x2−1
x−1= 2.
Question 2
Solution: To determine if the limit exists, we can try to simplify
the expression and see what happens as xapproaches 2.
1. Direct substitution: Plugging in x= 2 into the expression, we
get: 22−4
2−2=0
0. Since the denominator is 0, this is an indeterminate
form, indicating that further simplification is needed.
2. Factor the numerator: Rewriting the expression, we have:
limx→2(x−2)(x+2)
x−2.
3. Cancel out the common factor: Cancelling out the common
factor of (x−2), we get: limx→2x+ 2 = 2 + 2 = 4.
Therefore, the limit exists and is equal to 4 for the given expres-
sion.Question 2: Determine if the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine if the limit exists, we can try to simplify
the expression and see what happens as xapproaches 2.
1. Direct substitution: Plugging in x= 2 into the expression, we
get: 22−4
2−2=0
0. Since the denominator is 0, this is an indeterminate
form, indicating that further simplification is needed.
2. Factor the numerator: Rewriting the expression, we have:
limx→2(x−2)(x+2)
x−2.
3. Cancel out the common factor: Cancelling out the common
factor of (x−2), we get: limx→2x+ 2 = 2 + 2 = 4.
Therefore, the limit exists and is equal to 4 for the given expres-
sion.
Question 3
f(x) = x2−9
x−3
Step-by-step Solution: To determine the limit as xapproaches 3
for the given function, we need to simplify the expression first. We’ll
start by factoring the numerator:
f(x) = (x+ 3)(x−3)
x−3
Next, we can simplify the expression by canceling out the common
factor of x−3in the numerator and denominator:
2
f(x) = x+ 3
Now, we can simply evaluate the function at x= 3 to find the limit:
f(3) = 3 + 3 = 6
Therefore, the limit as xapproaches 3 for the function f(x) = x2−9
x−3
is 6.Question 3: Determine the limit, if it exists, as xapproaches 3
for the function:
f(x) = x2−9
x−3
Step-by-step Solution: To determine the limit as xapproaches 3
for the given function, we need to simplify the expression first. We’ll
start by factoring the numerator:
f(x) = (x+ 3)(x−3)
x−3
Next, we can simplify the expression by canceling out the common
factor of x−3in the numerator and denominator:
f(x) = x+ 3
Now, we can simply evaluate the function at x= 3 to find the limit:
f(3) = 3 + 3 = 6
Therefore, the limit as xapproaches 3 for the function f(x) = x2−9
x−3
is 6.
Question 4
Step-by-step Solution: To determine if the limit exists, we need to
evaluate the limit as xapproaches 2 from both sides and check if the
left-hand limit is equal to the right-hand limit.
1. Let’s first simplify the given expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
= lim
x→2(x+ 2)
2. Now, substitute x= 2 into the expression:
= 2 + 2 = 4
3. Since the limit approaches the same value from both sides as x
approaches 2, the limit exists.
3
Therefore, the limit of x2−4
x−2as xapproaches 2 is 4.Question 4:
Determine whether the following limit exists:
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine if the limit exists, we need to
evaluate the limit as xapproaches 2 from both sides and check if the
left-hand limit is equal to the right-hand limit.
1. Let’s first simplify the given expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
= lim
x→2(x+ 2)
2. Now, substitute x= 2 into the expression:
= 2 + 2 = 4
3. Since the limit approaches the same value from both sides as x
approaches 2, the limit exists.
Therefore, the limit of x2−4
x−2as xapproaches 2 is 4.
Question 5
Step-by-step solution: 1. As xapproaches 2 from the left side
(x < 2), the expression becomes x−2
2−x=−1. 2. As xapproaches 2
from the right side (x > 2), the expression becomes x−2
x−2= 1. 3. Since
the limits from the left and right sides are different, the overall limit
limx→2
x−2
|x−2|does not exist.Question 5: Determine whether the limit
exists or not: limx→2
x−2
|x−2|.
Step-by-step solution: 1. As xapproaches 2 from the left side
(x < 2), the expression becomes x−2
2−x=−1. 2. As xapproaches 2
from the right side (x > 2), the expression becomes x−2
x−2= 1. 3. Since
the limits from the left and right sides are different, the overall limit
limx→2
x−2
|x−2|does not exist.
Question 6
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine the limit of the given ex-
pression as xapproaches 2, we can first simplify the expression by
factoring the numerator:
4
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Next, we can cancel out the common factor of x−2:
= lim
x→2(x+ 2)
Now, we can evaluate the limit by direct substitution:
= 2 + 2 = 4
Therefore, the value of the limit as xapproaches 2 is 4. Hence, the
limit exists and is equal to 4.Question 6: Determine the value of the
limit, if it exists, or state that the limit does not exist. If the limit
does not exist, explain why.
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine the limit of the given ex-
pression as xapproaches 2, we can first simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Next, we can cancel out the common factor of x−2:
= lim
x→2(x+ 2)
Now, we can evaluate the limit by direct substitution:
= 2 + 2 = 4
Therefore, the value of the limit as xapproaches 2 is 4. Hence,
the limit exists and is equal to 4.
Question 7
lim
x→3
x2−9
x−3
Solution:
To determine if the limit exists at x= 3, we will first attempt to
directly substitute x= 3 into the expression.
Substitute x= 3:
lim
x→3
(3)2−9
3−3= lim
x→3
0
0
5
Since we obtained an indeterminate form (0
0), we need to further
simplify the expression to determine the limit.
Factor the numerator:
lim
x→3
(x+ 3)(x−3)
x−3
Simplify the expression:
= lim
x→3(x+ 3)
Now, we can directly substitute x= 3 into the simplified expression
to find the limit:
= 3 + 3 = 6
Therefore, the limit of the function as xapproaches 3 exists and
is equal to 6.Question 7: Determine whether the limit exists or not,
and justify your answer. If it does not exist, explain why.
lim
x→3
x2−9
x−3
Solution:
To determine if the limit exists at x= 3, we will first attempt to
directly substitute x= 3 into the expression.
Substitute x= 3:
lim
x→3
(3)2−9
3−3= lim
x→3
0
0
Since we obtained an indeterminate form (0
0), we need to further
simplify the expression to determine the limit.
Factor the numerator:
lim
x→3
(x+ 3)(x−3)
x−3
Simplify the expression:
= lim
x→3(x+ 3)
Now, we can directly substitute x= 3 into the simplified expression
to find the limit:
= 3 + 3 = 6
Therefore, the limit of the function as xapproaches 3 exists and
is equal to 6.
6
Question 8
Determine whether the limit exists or not:
lim
x→4
x2−16
x−4
Solution:
To determine whether the limit exists at x= 4, we can try to
simplify the expression or analyze the function behavior around the
point x= 4.
Given expression: x2−16
x−4
Factorizing x2−16 using the difference of squares formula gives:
x2−16 = (x−4)(x+ 4)
Substitute this back into the expression:
lim
x→4
(x−4)(x+ 4)
x−4
Now, simplify the expression by canceling out the common factor
of x−4:
lim
x→4(x+ 4)
Now, substitute x= 4 into the simplified expression:
4 + 4 = 8
Therefore, the limit exists and is equal to 8 at x= 4.Question 8:
Determine whether the limit exists or not:
lim
x→4
x2−16
x−4
Solution:
To determine whether the limit exists at x= 4, we can try to
simplify the expression or analyze the function behavior around the
point x= 4.
Given expression: x2−16
x−4
Factorizing x2−16 using the difference of squares formula gives:
x2−16 = (x−4)(x+ 4)
Substitute this back into the expression:
lim
x→4
(x−4)(x+ 4)
x−4
7
Now, simplify the expression by canceling out the common factor
of x−4:
lim
x→4(x+ 4)
Now, substitute x= 4 into the simplified expression:
4 + 4 = 8
Therefore, the limit exists and is equal to 8 at x= 4.
Question 9
Step-by-step solution: 1. Let’s first simplify the given expression:
x2−4
x−2=(x+ 2)(x−2)
x−2= (x+ 2)
2. Now, we want to show that the limit as x approaches 2 of (x +
2) does not exist.
3. Assume the limit exists and is equal to L:
lim
x→2(x+ 2) = L
4. We need to show that for any value of L, we can find two
different paths approaching 2 that give different limits.
5. Let’s consider two sequences that approach 2: an= 2 + 1
nand
bn= 2 −
1
n, where n is a positive integer.
6. As n approaches infinity, an= 2+and bn= 2−.
7. The value of the function (x + 2) for these sequences is:
f(an) = 2 + 1
n+ 2 = 4 + 1
n
f(bn)=2−
1
n+ 2 = 4 −
1
n
8. As n approaches infinity, f(an) = 4+and f(bn) = 4−, which are
different.
9. Therefore, since the function gives different values as it ap-
proaches 2 from the right and left sides, the limit does not exist.Question
9: Use the definition of limit to show that the limit does not exist:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first simplify the given expression:
x2−4
x−2=(x+ 2)(x−2)
x−2= (x+ 2)
8
2. Now, we want to show that the limit as x approaches 2 of (x +
2) does not exist.
3. Assume the limit exists and is equal to L:
lim
x→2(x+ 2) = L
4. We need to show that for any value of L, we can find two
different paths approaching 2 that give different limits.
5. Let’s consider two sequences that approach 2: an= 2 + 1
nand
bn= 2 −
1
n, where n is a positive integer.
6. As n approaches infinity, an= 2+and bn= 2−.
7. The value of the function (x + 2) for these sequences is:
f(an) = 2 + 1
n+ 2 = 4 + 1
n
f(bn)=2−
1
n+ 2 = 4 −
1
n
8. As n approaches infinity, f(an) = 4+and f(bn) = 4−, which are
different.
9. Therefore, since the function gives different values as it ap-
proaches 2 from the right and left sides, the limit does not exist.
Question 10
Step-by-step Solution: 1. Direct Substitution Method: Let us first
attempt to evaluate the limit by direct substitution:
lim
x→3
x2−9
x−3= lim
x→3
(3)2−9
3−3= lim
x→3
0
0
2. Factorization: Since we obtained the indeterminate form 0/0,
we can try to factorize the expression:
x2−9
x−3=(x−3)(x+ 3)
x−3=x+ 3
3. Apply the Limit:
lim
x→3(x+ 3) = 3 + 3 = 6
4. Conclusion: Since we were able to simplify the expression to
a non-indeterminate form and evaluate the limit to a finite number,
the limit limx→3
x2−9
x−3exists and is equal to 6.Question 10: Determine
whether the limit limx→3
x2−9
x−3exists.
9
Step-by-step Solution: 1. Direct Substitution Method: Let us first
attempt to evaluate the limit by direct substitution:
lim
x→3
x2−9
x−3= lim
x→3
(3)2−9
3−3= lim
x→3
0
0
2. Factorization: Since we obtained the indeterminate form 0/0,
we can try to factorize the expression:
x2−9
x−3=(x−3)(x+ 3)
x−3=x+ 3
3. Apply the Limit:
lim
x→3(x+ 3) = 3 + 3 = 6
4. Conclusion: Since we were able to simplify the expression to
a non-indeterminate form and evaluate the limit to a finite number,
the limit limx→3
x2−9
x−3exists and is equal to 6.
Question 11
lim
x→2
x2−4
x−2
Step-by-step solution: To determine if the limit exists, we first
simplify the expression by factoring the numerator:
lim
x→2
(x−2)(x+ 2)
x−2
Next, we cancel out the common factor of (x-2):
lim
x→2x+ 2
Now, we can evaluate the expression at x = 2:
2 + 2 = 4
Therefore, the limit of the function as x approaches 2 exists and is
equal to 4.Question 11: Determine whether the limit exists or does
not exist:
lim
x→2
x2−4
x−2
Step-by-step solution: To determine if the limit exists, we first
simplify the expression by factoring the numerator:
10
lim
x→2
(x−2)(x+ 2)
x−2
Next, we cancel out the common factor of (x-2):
lim
x→2x+ 2
Now, we can evaluate the expression at x = 2:
2 + 2 = 4
Therefore, the limit of the function as x approaches 2 exists and
is equal to 4.
Question 12
Question 12: Determine whether the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine whether the limit exists, we will evaluate
the limit from the left side and the right side of x= 2.
Left-side limit: limx→2−
x2−4
x−2= limx→2−
(x−2)(x+2)
x−2= limx→2−(x+ 2) =
4.
Right-side limit: limx→2+
x2−4
x−2= limx→2+(x−2)(x+2)
x−2= limx→2+(x+2) =
4.
Since the left-side and right-side limits are equal, the limit exists
and is equal to 4.
Therefore, limx→2
x2−4
x−2= 4.Sure, here is the question and its solu-
tion in LateX code:
Question 12: Determine whether the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine whether the limit exists, we will evaluate
the limit from the left side and the right side of x= 2.
Left-side limit: limx→2−
x2−4
x−2= limx→2−
(x−2)(x+2)
x−2= limx→2−(x+ 2) =
4.
Right-side limit: limx→2+
x2−4
x−2= limx→2+(x−2)(x+2)
x−2= limx→2+(x+2) =
4.
Since the left-side and right-side limits are equal, the limit exists
and is equal to 4.
Therefore, limx→2
x2−4
x−2= 4.
Question 13
Step-by-step solution: To determine whether the limit exists or
not, we first try to simplify the expression by factoring out the com-
mon factor in the numerator:
11
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
Next, cancel out the common factor (x - 3) from the numerator
and denominator:
= lim
x→3(x+ 3)
Now, substitute x = 3 into the expression:
= 3 + 3
= 6
Therefore, the limit as x approaches 3 for the given expression is
6, which exists.Question 13: Determine whether the limit exists or
not:
lim
x→3
x2−9
x−3
Step-by-step solution: To determine whether the limit exists or
not, we first try to simplify the expression by factoring out the com-
mon factor in the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
Next, cancel out the common factor (x - 3) from the numerator
and denominator:
= lim
x→3(x+ 3)
Now, substitute x = 3 into the expression:
= 3 + 3
= 6
Therefore, the limit as x approaches 3 for the given expression is
6, which exists.
12
Question 14
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first try to directly substitute x= 2
into the expression:
lim
x→2
x2−4
x−2=22−4
2−2=0
0
2. The expression becomes 0
0, which is an indeterminate form.
3. To further analyze this limit, we can simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
4. Simplifying, we get:
lim
x→2(x+ 2) = 4
5. Therefore, the limit of the given expression as xapproaches 2
exists and is equal to 4.
Thus, the limit exists and is equal to 4.Question 14: Determine
whether the limit exists or not:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first try to directly substitute x= 2
into the expression:
lim
x→2
x2−4
x−2=22−4
2−2=0
0
2. The expression becomes 0
0, which is an indeterminate form.
3. To further analyze this limit, we can simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
4. Simplifying, we get:
lim
x→2(x+ 2) = 4
5. Therefore, the limit of the given expression as xapproaches 2
exists and is equal to 4.
Thus, the limit exists and is equal to 4.
13
Question 15
Solution: To determine whether the limit exists or not, we can
analyze the behavior of the function as xapproaches 2 from both the
left and the right sides.
Left-hand limit: Let’s find the limit as xapproaches 2 from the
left side, denoted by limx→2−
x2−4
x−2.
lim
x→2−
x2−4
x−2= lim
x→2−
(x−2)(x+ 2)
x−2
= lim
x→2−
(x+ 2)
= 4
Right-hand limit: Similarly, let’s find the limit as xapproaches 2
from the right side, denoted by limx→2+
x2−4
x−2.
lim
x→2+
x2−4
x−2= lim
x→2+
(x−2)(x+ 2)
x−2
= lim
x→2+(x+ 2)
= 4
Since the left-hand limit and the right-hand limit both approach
the same value of 4 as xapproaches 2, we can conclude that limx→2
x2−4
x−2
exists and is equal to 4.Question 15: Determine whether the limit
limx→2
x2−4
x−2exists. If it does not exist, explain why.
Solution: To determine whether the limit exists or not, we can
analyze the behavior of the function as xapproaches 2 from both the
left and the right sides.
Left-hand limit: Let’s find the limit as xapproaches 2 from the
left side, denoted by limx→2−
x2−4
x−2.
lim
x→2−
x2−4
x−2= lim
x→2−
(x−2)(x+ 2)
x−2
= lim
x→2−
(x+ 2)
= 4
Right-hand limit: Similarly, let’s find the limit as xapproaches 2
from the right side, denoted by limx→2+
x2−4
x−2.
lim
x→2+
x2−4
x−2= lim
x→2+
(x−2)(x+ 2)
x−2
= lim
x→2+(x+ 2)
= 4
14
Since the left-hand limit and the right-hand limit both approach
the same value of 4 as xapproaches 2, we can conclude that limx→2
x2−4
x−2
exists and is equal to 4.
Question 16
Step-by-step solution: 1. Let’s start by simplifying the expression
using the difference of squares formula: (x2−4) = (x+ 2)(x−2). 2.
This gives us limx→2(x+2)(x−2)
x−2. 3. Cancel out the common factor of
(x−2) in the numerator and denominator to get limx→2(x+2). 4. Now,
we can directly substitute x= 2 into the expression to find the limit:
limx→2(x+ 2) = 2 + 2 = 4.
Therefore, the limit limx→2
x2−4
x−2exists and is equal to 4.Question
16: Determine whether the limit limx→2
x2−4
x−2exists. If it does not
exist, explain why.
Step-by-step solution: 1. Let’s start by simplifying the expression
using the difference of squares formula: (x2−4) = (x+ 2)(x−2). 2.
This gives us limx→2(x+2)(x−2)
x−2. 3. Cancel out the common factor of
(x−2) in the numerator and denominator to get limx→2(x+2). 4. Now,
we can directly substitute x= 2 into the expression to find the limit:
limx→2(x+ 2) = 2 + 2 = 4.
Therefore, the limit limx→2
x2−4
x−2exists and is equal to 4.
Question 17
lim
x→−2
x2−4
x+ 2
Step-by-step Solution: To determine if the limit exists at x=−2,
we can first try to evaluate the expression at x=−2.
Substitute x=−2into the expression:
(−2)2−4
−2+2 =4−4
0=0
0
Notice that we get an indeterminate form of 0
0when attempting
to evaluate the expression.
To further investigate if the limit exists, we can simplify the ex-
pression:
lim
x→−2
x2−4
x+ 2 = lim
x→−2
(x−2)(x+ 2)
x+ 2 = lim
x→−2(x−2)
Now, as xapproaches −2, the expression simplifies to:
15
lim
x→−2(x−2) = −2−2 = −4
Therefore, the limit of the given function as xapproaches −2exists
and the value is −4.Question 17: Determine whether the following
limit exists:
lim
x→−2
x2−4
x+ 2
Step-by-step Solution: To determine if the limit exists at x=−2,
we can first try to evaluate the expression at x=−2.
Substitute x=−2into the expression:
(−2)2−4
−2+2 =4−4
0=0
0
Notice that we get an indeterminate form of 0
0when attempting
to evaluate the expression.
To further investigate if the limit exists, we can simplify the ex-
pression:
lim
x→−2
x2−4
x+ 2 = lim
x→−2
(x−2)(x+ 2)
x+ 2 = lim
x→−2(x−2)
Now, as xapproaches −2, the expression simplifies to:
lim
x→−2(x−2) = −2−2 = −4
Therefore, the limit of the given function as xapproaches −2exists
and the value is −4.
Question 18
Step-by-step Solution: 1. Direct Substitution: If we substitute
x= 2 directly into the expression, we get an indeterminate form (0
0).
2. Factorization:
x2−4
x−2=(x+ 2)(x−2)
x−2
Simplifying, we get:
f(x) = x+ 2
3. Limit Calculation: Now, as xapproaches 2, the expression
f(x) = x+ 2 approaches 4. Therefore, the limit as xapproaches 2 of
x2−4
x−2is 4.
16
Conclusion: Since the limit exists and is equal to 4, we can con-
clude that the limit exists for the given function.Question 18: Deter-
mine whether the limit exists or not:
lim
x→2
x2−4
x−2
Step-by-step Solution: 1. Direct Substitution: If we substitute
x= 2 directly into the expression, we get an indeterminate form (0
0).
2. Factorization:
x2−4
x−2=(x+ 2)(x−2)
x−2
Simplifying, we get:
f(x) = x+ 2
3. Limit Calculation: Now, as xapproaches 2, the expression
f(x) = x+ 2 approaches 4. Therefore, the limit as xapproaches 2 of
x2−4
x−2is 4.
Conclusion: Since the limit exists and is equal to 4, we can con-
clude that the limit exists for the given function.
Question 19
Solution: To determine whether the limit exists or not, we will
analyze the behavior as xapproaches 0.
We know that limx→0sin(x)
x= 1.
The limit of sin(x)
xas xapproaches 0 is a well-known limit and
evaluates to 1. Therefore, the limit exists and equals 1.Question 19:
Determine whether the limit limx→0sin(x)
xexists. If it does not exist,
explain why.
Solution: To determine whether the limit exists or not, we will
analyze the behavior as xapproaches 0.
We know that limx→0sin(x)
x= 1.
The limit of sin(x)
xas xapproaches 0 is a well-known limit and
evaluates to 1. Therefore, the limit exists and equals 1.
Question 20
Solution: To determine whether the limit exists, we will simplify
the expression and see if we can find the limit.
Given function: f(x) = x−2
x2−4
1. Substitute x= 2 into the expression: f(2) = 2−2
22−4=0
0
2. Factor the denominator: x2−4=(x−2)(x+ 2)
17
= lim
x→1+(x+ 1) = 1 + 1 = 2
Since the limit from the left and the limit from the right both
equal 2, the overall limit exists and is equal to 2.
Therefore, limx→1
x2−1
x−1= 2.
Question 2
Solution: To determine if the limit exists, we can try to simplify
the expression and see what happens as xapproaches 2.
1. Direct substitution: Plugging in x= 2 into the expression, we
get: 22−4
2−2=0
0. Since the denominator is 0, this is an indeterminate
form, indicating that further simplification is needed.
2. Factor the numerator: Rewriting the expression, we have:
limx→2(x−2)(x+2)
x−2.
3. Cancel out the common factor: Cancelling out the common
factor of (x−2), we get: limx→2x+ 2 = 2 + 2 = 4.
Therefore, the limit exists and is equal to 4 for the given expres-
sion.Question 2: Determine if the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine if the limit exists, we can try to simplify
the expression and see what happens as xapproaches 2.
1. Direct substitution: Plugging in x= 2 into the expression, we
get: 22−4
2−2=0
0. Since the denominator is 0, this is an indeterminate
form, indicating that further simplification is needed.
2. Factor the numerator: Rewriting the expression, we have:
limx→2(x−2)(x+2)
x−2.
3. Cancel out the common factor: Cancelling out the common
factor of (x−2), we get: limx→2x+ 2 = 2 + 2 = 4.
Therefore, the limit exists and is equal to 4 for the given expres-
sion.
Question 3
f(x) = x2−9
x−3
Step-by-step Solution: To determine the limit as xapproaches 3
for the given function, we need to simplify the expression first. We’ll
start by factoring the numerator:
f(x) = (x+ 3)(x−3)
x−3
Next, we can simplify the expression by canceling out the common
factor of x−3in the numerator and denominator:
2
f(x) = x+ 3
Now, we can simply evaluate the function at x= 3 to find the limit:
f(3) = 3 + 3 = 6
Therefore, the limit as xapproaches 3 for the function f(x) = x2−9
x−3
is 6.Question 3: Determine the limit, if it exists, as xapproaches 3
for the function:
f(x) = x2−9
x−3
Step-by-step Solution: To determine the limit as xapproaches 3
for the given function, we need to simplify the expression first. We’ll
start by factoring the numerator:
f(x) = (x+ 3)(x−3)
x−3
Next, we can simplify the expression by canceling out the common
factor of x−3in the numerator and denominator:
f(x) = x+ 3
Now, we can simply evaluate the function at x= 3 to find the limit:
f(3) = 3 + 3 = 6
Therefore, the limit as xapproaches 3 for the function f(x) = x2−9
x−3
is 6.
Question 4
Step-by-step Solution: To determine if the limit exists, we need to
evaluate the limit as xapproaches 2 from both sides and check if the
left-hand limit is equal to the right-hand limit.
1. Let’s first simplify the given expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
= lim
x→2(x+ 2)
2. Now, substitute x= 2 into the expression:
= 2 + 2 = 4
3. Since the limit approaches the same value from both sides as x
approaches 2, the limit exists.
3
Therefore, the limit of x2−4
x−2as xapproaches 2 is 4.Question 4:
Determine whether the following limit exists:
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine if the limit exists, we need to
evaluate the limit as xapproaches 2 from both sides and check if the
left-hand limit is equal to the right-hand limit.
1. Let’s first simplify the given expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
= lim
x→2(x+ 2)
2. Now, substitute x= 2 into the expression:
= 2 + 2 = 4
3. Since the limit approaches the same value from both sides as x
approaches 2, the limit exists.
Therefore, the limit of x2−4
x−2as xapproaches 2 is 4.
Question 5
Step-by-step solution: 1. As xapproaches 2 from the left side
(x < 2), the expression becomes x−2
2−x=−1. 2. As xapproaches 2
from the right side (x > 2), the expression becomes x−2
x−2= 1. 3. Since
the limits from the left and right sides are different, the overall limit
limx→2
x−2
|x−2|does not exist.Question 5: Determine whether the limit
exists or not: limx→2
x−2
|x−2|.
Step-by-step solution: 1. As xapproaches 2 from the left side
(x < 2), the expression becomes x−2
2−x=−1. 2. As xapproaches 2
from the right side (x > 2), the expression becomes x−2
x−2= 1. 3. Since
the limits from the left and right sides are different, the overall limit
limx→2
x−2
|x−2|does not exist.
Question 6
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine the limit of the given ex-
pression as xapproaches 2, we can first simplify the expression by
factoring the numerator:
4
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Next, we can cancel out the common factor of x−2:
= lim
x→2(x+ 2)
Now, we can evaluate the limit by direct substitution:
= 2 + 2 = 4
Therefore, the value of the limit as xapproaches 2 is 4. Hence, the
limit exists and is equal to 4.Question 6: Determine the value of the
limit, if it exists, or state that the limit does not exist. If the limit
does not exist, explain why.
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine the limit of the given ex-
pression as xapproaches 2, we can first simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Next, we can cancel out the common factor of x−2:
= lim
x→2(x+ 2)
Now, we can evaluate the limit by direct substitution:
= 2 + 2 = 4
Therefore, the value of the limit as xapproaches 2 is 4. Hence,
the limit exists and is equal to 4.
Question 7
lim
x→3
x2−9
x−3
Solution:
To determine if the limit exists at x= 3, we will first attempt to
directly substitute x= 3 into the expression.
Substitute x= 3:
lim
x→3
(3)2−9
3−3= lim
x→3
0
0
5
Since we obtained an indeterminate form ( 0
0), we need to further
simplify the expression to determine the limit.
Factor the numerator:
lim
x→3
(x+ 3)(x−3)
x−3
Simplify the expression:
= lim
x→3(x+ 3)
Now, we can directly substitute x= 3 into the simplified expression
to find the limit:
= 3 + 3 = 6
Therefore, the limit of the function as xapproaches 3 exists and
is equal to 6.Question 7: Determine whether the limit exists or not,
and justify your answer. If it does not exist, explain why.
lim
x→3
x2−9
x−3
Solution:
To determine if the limit exists at x= 3, we will first attempt to
directly substitute x= 3 into the expression.
Substitute x= 3:
lim
x→3
(3)2−9
3−3= lim
x→3
0
0
Since we obtained an indeterminate form ( 0
0), we need to further
simplify the expression to determine the limit.
Factor the numerator:
lim
x→3
(x+ 3)(x−3)
x−3
Simplify the expression:
= lim
x→3(x+ 3)
Now, we can directly substitute x= 3 into the simplified expression
to find the limit:
= 3 + 3 = 6
Therefore, the limit of the function as xapproaches 3 exists and
is equal to 6.
6
Question 8
Determine whether the limit exists or not:
lim
x→4
x2−16
x−4
Solution:
To determine whether the limit exists at x= 4, we can try to
simplify the expression or analyze the function behavior around the
point x= 4.
Given expression: x2−16
x−4
Factorizing x2−16 using the difference of squares formula gives:
x2−16 = (x−4)(x+ 4)
Substitute this back into the expression:
lim
x→4
(x−4)(x+ 4)
x−4
Now, simplify the expression by canceling out the common factor
of x−4:
lim
x→4(x+ 4)
Now, substitute x= 4 into the simplified expression:
4 + 4 = 8
Therefore, the limit exists and is equal to 8 at x= 4.Question 8:
Determine whether the limit exists or not:
lim
x→4
x2−16
x−4
Solution:
To determine whether the limit exists at x= 4, we can try to
simplify the expression or analyze the function behavior around the
point x= 4.
Given expression: x2−16
x−4
Factorizing x2−16 using the difference of squares formula gives:
x2−16 = (x−4)(x+ 4)
Substitute this back into the expression:
lim
x→4
(x−4)(x+ 4)
x−4
7
Now, simplify the expression by canceling out the common factor
of x−4:
lim
x→4(x+ 4)
Now, substitute x= 4 into the simplified expression:
4 + 4 = 8
Therefore, the limit exists and is equal to 8 at x= 4.
Question 9
Step-by-step solution: 1. Let’s first simplify the given expression:
x2−4
x−2=(x+ 2)(x−2)
x−2= (x+ 2)
2. Now, we want to show that the limit as x approaches 2 of (x +
2) does not exist.
3. Assume the limit exists and is equal to L:
lim
x→2(x+ 2) = L
4. We need to show that for any value of L, we can find two
different paths approaching 2 that give different limits.
5. Let’s consider two sequences that approach 2: an= 2 + 1
nand
bn= 2 −
1
n, where n is a positive integer.
6. As n approaches infinity, an= 2+and bn= 2−.
7. The value of the function (x + 2) for these sequences is:
f(an) = 2 + 1
n+ 2 = 4 + 1
n
f(bn)=2−
1
n+ 2 = 4 −
1
n
8. As n approaches infinity, f(an) = 4+and f(bn) = 4−, which are
different.
9. Therefore, since the function gives different values as it ap-
proaches 2 from the right and left sides, the limit does not exist.Question
9: Use the definition of limit to show that the limit does not exist:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first simplify the given expression:
x2−4
x−2=(x+ 2)(x−2)
x−2= (x+ 2)
8
2. Now, we want to show that the limit as x approaches 2 of (x +
2) does not exist.
3. Assume the limit exists and is equal to L:
lim
x→2(x+ 2) = L
4. We need to show that for any value of L, we can find two
different paths approaching 2 that give different limits.
5. Let’s consider two sequences that approach 2: an= 2 + 1
nand
bn= 2 −
1
n, where n is a positive integer.
6. As n approaches infinity, an= 2+and bn= 2−.
7. The value of the function (x + 2) for these sequences is:
f(an) = 2 + 1
n+ 2 = 4 + 1
n
f(bn)=2−
1
n+ 2 = 4 −
1
n
8. As n approaches infinity, f(an) = 4+and f(bn) = 4−, which are
different.
9. Therefore, since the function gives different values as it ap-
proaches 2 from the right and left sides, the limit does not exist.
Question 10
Step-by-step Solution: 1. Direct Substitution Method: Let us first
attempt to evaluate the limit by direct substitution:
lim
x→3
x2−9
x−3= lim
x→3
(3)2−9
3−3= lim
x→3
0
0
2. Factorization: Since we obtained the indeterminate form 0/0,
we can try to factorize the expression:
x2−9
x−3=(x−3)(x+ 3)
x−3=x+ 3
3. Apply the Limit:
lim
x→3(x+ 3) = 3 + 3 = 6
4. Conclusion: Since we were able to simplify the expression to
a non-indeterminate form and evaluate the limit to a finite number,
the limit limx→3
x2−9
x−3exists and is equal to 6.Question 10: Determine
whether the limit limx→3
x2−9
x−3exists.
9
Step-by-step Solution: 1. Direct Substitution Method: Let us first
attempt to evaluate the limit by direct substitution:
lim
x→3
x2−9
x−3= lim
x→3
(3)2−9
3−3= lim
x→3
0
0
2. Factorization: Since we obtained the indeterminate form 0/0,
we can try to factorize the expression:
x2−9
x−3=(x−3)(x+ 3)
x−3=x+ 3
3. Apply the Limit:
lim
x→3(x+ 3) = 3 + 3 = 6
4. Conclusion: Since we were able to simplify the expression to
a non-indeterminate form and evaluate the limit to a finite number,
the limit limx→3
x2−9
x−3exists and is equal to 6.
Question 11
lim
x→2
x2−4
x−2
Step-by-step solution: To determine if the limit exists, we first
simplify the expression by factoring the numerator:
lim
x→2
(x−2)(x+ 2)
x−2
Next, we cancel out the common factor of (x-2):
lim
x→2x+ 2
Now, we can evaluate the expression at x = 2:
2 + 2 = 4
Therefore, the limit of the function as x approaches 2 exists and is
equal to 4.Question 11: Determine whether the limit exists or does
not exist:
lim
x→2
x2−4
x−2
Step-by-step solution: To determine if the limit exists, we first
simplify the expression by factoring the numerator:
10
lim
x→2
(x−2)(x+ 2)
x−2
Next, we cancel out the common factor of (x-2):
lim
x→2x+ 2
Now, we can evaluate the expression at x = 2:
2 + 2 = 4
Therefore, the limit of the function as x approaches 2 exists and
is equal to 4.
Question 12
Question 12: Determine whether the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine whether the limit exists, we will evaluate
the limit from the left side and the right side of x= 2.
Left-side limit: limx→2−
x2−4
x−2= limx→2−
(x−2)(x+2)
x−2= limx→2−(x+ 2) =
4.
Right-side limit: limx→2+
x2−4
x−2= limx→2+(x−2)(x+2)
x−2= limx→2+(x+2) =
4.
Since the left-side and right-side limits are equal, the limit exists
and is equal to 4.
Therefore, limx→2
x2−4
x−2= 4.Sure, here is the question and its solu-
tion in LateX code:
Question 12: Determine whether the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine whether the limit exists, we will evaluate
the limit from the left side and the right side of x= 2.
Left-side limit: limx→2−
x2−4
x−2= limx→2−
(x−2)(x+2)
x−2= limx→2−(x+ 2) =
4.
Right-side limit: limx→2+
x2−4
x−2= limx→2+(x−2)(x+2)
x−2= limx→2+(x+2) =
4.
Since the left-side and right-side limits are equal, the limit exists
and is equal to 4.
Therefore, limx→2
x2−4
x−2= 4.
Question 13
Step-by-step solution: To determine whether the limit exists or
not, we first try to simplify the expression by factoring out the com-
mon factor in the numerator:
11
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
Next, cancel out the common factor (x - 3) from the numerator
and denominator:
= lim
x→3(x+ 3)
Now, substitute x = 3 into the expression:
= 3 + 3
= 6
Therefore, the limit as x approaches 3 for the given expression is
6, which exists.Question 13: Determine whether the limit exists or
not:
lim
x→3
x2−9
x−3
Step-by-step solution: To determine whether the limit exists or
not, we first try to simplify the expression by factoring out the com-
mon factor in the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
Next, cancel out the common factor (x - 3) from the numerator
and denominator:
= lim
x→3(x+ 3)
Now, substitute x = 3 into the expression:
= 3 + 3
= 6
Therefore, the limit as x approaches 3 for the given expression is
6, which exists.
12
Question 14
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first try to directly substitute x= 2
into the expression:
lim
x→2
x2−4
x−2=22−4
2−2=0
0
2. The expression becomes 0
0, which is an indeterminate form.
3. To further analyze this limit, we can simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
4. Simplifying, we get:
lim
x→2(x+ 2) = 4
5. Therefore, the limit of the given expression as xapproaches 2
exists and is equal to 4.
Thus, the limit exists and is equal to 4.Question 14: Determine
whether the limit exists or not:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first try to directly substitute x= 2
into the expression:
lim
x→2
x2−4
x−2=22−4
2−2=0
0
2. The expression becomes 0
0, which is an indeterminate form.
3. To further analyze this limit, we can simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
4. Simplifying, we get:
lim
x→2(x+ 2) = 4
5. Therefore, the limit of the given expression as xapproaches 2
exists and is equal to 4.
Thus, the limit exists and is equal to 4.
13
Question 15
Solution: To determine whether the limit exists or not, we can
analyze the behavior of the function as xapproaches 2 from both the
left and the right sides.
Left-hand limit: Let’s find the limit as xapproaches 2 from the
left side, denoted by limx→2−
x2−4
x−2.
lim
x→2−
x2−4
x−2= lim
x→2−
(x−2)(x+ 2)
x−2
= lim
x→2−
(x+ 2)
= 4
Right-hand limit: Similarly, let’s find the limit as xapproaches 2
from the right side, denoted by limx→2+
x2−4
x−2.
lim
x→2+
x2−4
x−2= lim
x→2+
(x−2)(x+ 2)
x−2
= lim
x→2+(x+ 2)
= 4
Since the left-hand limit and the right-hand limit both approach
the same value of 4 as xapproaches 2, we can conclude that limx→2
x2−4
x−2
exists and is equal to 4.Question 15: Determine whether the limit
limx→2
x2−4
x−2exists. If it does not exist, explain why.
Solution: To determine whether the limit exists or not, we can
analyze the behavior of the function as xapproaches 2 from both the
left and the right sides.
Left-hand limit: Let’s find the limit as xapproaches 2 from the
left side, denoted by limx→2−
x2−4
x−2.
lim
x→2−
x2−4
x−2= lim
x→2−
(x−2)(x+ 2)
x−2
= lim
x→2−
(x+ 2)
= 4
Right-hand limit: Similarly, let’s find the limit as xapproaches 2
from the right side, denoted by limx→2+
x2−4
x−2.
lim
x→2+
x2−4
x−2= lim
x→2+
(x−2)(x+ 2)
x−2
= lim
x→2+(x+ 2)
= 4
14
Since the left-hand limit and the right-hand limit both approach
the same value of 4 as xapproaches 2, we can conclude that limx→2
x2−4
x−2
exists and is equal to 4.
Question 16
Step-by-step solution: 1. Let’s start by simplifying the expression
using the difference of squares formula: (x2−4) = (x+ 2)(x−2). 2.
This gives us limx→2(x+2)(x−2)
x−2. 3. Cancel out the common factor of
(x−2) in the numerator and denominator to get limx→2(x+2). 4. Now,
we can directly substitute x= 2 into the expression to find the limit:
limx→2(x+ 2) = 2 + 2 = 4.
Therefore, the limit limx→2
x2−4
x−2exists and is equal to 4.Question
16: Determine whether the limit limx→2
x2−4
x−2exists. If it does not
exist, explain why.
Step-by-step solution: 1. Let’s start by simplifying the expression
using the difference of squares formula: (x2−4) = (x+ 2)(x−2). 2.
This gives us limx→2(x+2)(x−2)
x−2. 3. Cancel out the common factor of
(x−2) in the numerator and denominator to get limx→2(x+2). 4. Now,
we can directly substitute x= 2 into the expression to find the limit:
limx→2(x+ 2) = 2 + 2 = 4.
Therefore, the limit limx→2
x2−4
x−2exists and is equal to 4.
Question 17
lim
x→−2
x2−4
x+ 2
Step-by-step Solution: To determine if the limit exists at x=−2,
we can first try to evaluate the expression at x=−2.
Substitute x=−2into the expression:
(−2)2−4
−2+2 =4−4
0=0
0
Notice that we get an indeterminate form of 0
0when attempting
to evaluate the expression.
To further investigate if the limit exists, we can simplify the ex-
pression:
lim
x→−2
x2−4
x+ 2 = lim
x→−2
(x−2)(x+ 2)
x+ 2 = lim
x→−2(x−2)
Now, as xapproaches −2, the expression simplifies to:
15
lim
x→−2(x−2) = −2−2 = −4
Therefore, the limit of the given function as xapproaches −2exists
and the value is −4.Question 17: Determine whether the following
limit exists:
lim
x→−2
x2−4
x+ 2
Step-by-step Solution: To determine if the limit exists at x=−2,
we can first try to evaluate the expression at x=−2.
Substitute x=−2into the expression:
(−2)2−4
−2+2 =4−4
0=0
0
Notice that we get an indeterminate form of 0
0when attempting
to evaluate the expression.
To further investigate if the limit exists, we can simplify the ex-
pression:
lim
x→−2
x2−4
x+ 2 = lim
x→−2
(x−2)(x+ 2)
x+ 2 = lim
x→−2(x−2)
Now, as xapproaches −2, the expression simplifies to:
lim
x→−2(x−2) = −2−2 = −4
Therefore, the limit of the given function as xapproaches −2exists
and the value is −4.
Question 18
Step-by-step Solution: 1. Direct Substitution: If we substitute
x= 2 directly into the expression, we get an indeterminate form ( 0
0).
2. Factorization:
x2−4
x−2=(x+ 2)(x−2)
x−2
Simplifying, we get:
f(x) = x+ 2
3. Limit Calculation: Now, as xapproaches 2, the expression
f(x) = x+ 2 approaches 4. Therefore, the limit as xapproaches 2 of
x2−4
x−2is 4.
16
Conclusion: Since the limit exists and is equal to 4, we can con-
clude that the limit exists for the given function.Question 18: Deter-
mine whether the limit exists or not:
lim
x→2
x2−4
x−2
Step-by-step Solution: 1. Direct Substitution: If we substitute
x= 2 directly into the expression, we get an indeterminate form ( 0
0).
2. Factorization:
x2−4
x−2=(x+ 2)(x−2)
x−2
Simplifying, we get:
f(x) = x+ 2
3. Limit Calculation: Now, as xapproaches 2, the expression
f(x) = x+ 2 approaches 4. Therefore, the limit as xapproaches 2 of
x2−4
x−2is 4.
Conclusion: Since the limit exists and is equal to 4, we can con-
clude that the limit exists for the given function.
Question 19
Solution: To determine whether the limit exists or not, we will
analyze the behavior as xapproaches 0.
We know that limx→0sin(x)
x= 1.
The limit of sin(x)
xas xapproaches 0 is a well-known limit and
evaluates to 1. Therefore, the limit exists and equals 1.Question 19:
Determine whether the limit limx→0sin(x)
xexists. If it does not exist,
explain why.
Solution: To determine whether the limit exists or not, we will
analyze the behavior as xapproaches 0.
We know that limx→0sin(x)
x= 1.
The limit of sin(x)
xas xapproaches 0 is a well-known limit and
evaluates to 1. Therefore, the limit exists and equals 1.
Question 20
Solution: To determine whether the limit exists, we will simplify
the expression and see if we can find the limit.
Given function: f(x) = x−2
x2−4
1. Substitute x= 2 into the expression: f(2) = 2−2
22−4=0
0
2. Factor the denominator: x2−4=(x−2)(x+ 2)
17
= lim
x→1+(x+ 1) = 1 + 1 = 2
Since the limit from the left and the limit from the right both
equal 2, the overall limit exists and is equal to 2.
Therefore, limx→1
x2−1
x−1= 2.
Question 2
Solution: To determine if the limit exists, we can try to simplify
the expression and see what happens as xapproaches 2.
1. Direct substitution: Plugging in x= 2 into the expression, we
get: 22−4
2−2=0
0. Since the denominator is 0, this is an indeterminate
form, indicating that further simplification is needed.
2. Factor the numerator: Rewriting the expression, we have:
limx→2(x−2)(x+2)
x−2.
3. Cancel out the common factor: Cancelling out the common
factor of (x−2), we get: limx→2x+ 2 = 2 + 2 = 4.
Therefore, the limit exists and is equal to 4 for the given expres-
sion.Question 2: Determine if the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine if the limit exists, we can try to simplify
the expression and see what happens as xapproaches 2.
1. Direct substitution: Plugging in x= 2 into the expression, we
get: 22−4
2−2=0
0. Since the denominator is 0, this is an indeterminate
form, indicating that further simplification is needed.
2. Factor the numerator: Rewriting the expression, we have:
limx→2(x−2)(x+2)
x−2.
3. Cancel out the common factor: Cancelling out the common
factor of (x−2), we get: limx→2x+ 2 = 2 + 2 = 4.
Therefore, the limit exists and is equal to 4 for the given expres-
sion.
Question 3
f(x) = x2−9
x−3
Step-by-step Solution: To determine the limit as xapproaches 3
for the given function, we need to simplify the expression first. We’ll
start by factoring the numerator:
f(x) = (x+ 3)(x−3)
x−3
Next, we can simplify the expression by canceling out the common
factor of x−3in the numerator and denominator:
2
f(x) = x+ 3
Now, we can simply evaluate the function at x= 3 to find the limit:
f(3) = 3 + 3 = 6
Therefore, the limit as xapproaches 3 for the function f(x) = x2−9
x−3
is 6.Question 3: Determine the limit, if it exists, as xapproaches 3
for the function:
f(x) = x2−9
x−3
Step-by-step Solution: To determine the limit as xapproaches 3
for the given function, we need to simplify the expression first. We’ll
start by factoring the numerator:
f(x) = (x+ 3)(x−3)
x−3
Next, we can simplify the expression by canceling out the common
factor of x−3in the numerator and denominator:
f(x) = x+ 3
Now, we can simply evaluate the function at x= 3 to find the limit:
f(3) = 3 + 3 = 6
Therefore, the limit as xapproaches 3 for the function f(x) = x2−9
x−3
is 6.
Question 4
Step-by-step Solution: To determine if the limit exists, we need to
evaluate the limit as xapproaches 2 from both sides and check if the
left-hand limit is equal to the right-hand limit.
1. Let’s first simplify the given expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
= lim
x→2(x+ 2)
2. Now, substitute x= 2 into the expression:
= 2 + 2 = 4
3. Since the limit approaches the same value from both sides as x
approaches 2, the limit exists.
3
Therefore, the limit of x2−4
x−2as xapproaches 2 is 4.Question 4:
Determine whether the following limit exists:
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine if the limit exists, we need to
evaluate the limit as xapproaches 2 from both sides and check if the
left-hand limit is equal to the right-hand limit.
1. Let’s first simplify the given expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
= lim
x→2(x+ 2)
2. Now, substitute x= 2 into the expression:
= 2 + 2 = 4
3. Since the limit approaches the same value from both sides as x
approaches 2, the limit exists.
Therefore, the limit of x2−4
x−2as xapproaches 2 is 4.
Question 5
Step-by-step solution: 1. As xapproaches 2 from the left side
(x < 2), the expression becomes x−2
2−x=−1. 2. As xapproaches 2
from the right side (x > 2), the expression becomes x−2
x−2= 1. 3. Since
the limits from the left and right sides are different, the overall limit
limx→2
x−2
|x−2|does not exist.Question 5: Determine whether the limit
exists or not: limx→2
x−2
|x−2|.
Step-by-step solution: 1. As xapproaches 2 from the left side
(x < 2), the expression becomes x−2
2−x=−1. 2. As xapproaches 2
from the right side (x > 2), the expression becomes x−2
x−2= 1. 3. Since
the limits from the left and right sides are different, the overall limit
limx→2
x−2
|x−2|does not exist.
Question 6
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine the limit of the given ex-
pression as xapproaches 2, we can first simplify the expression by
factoring the numerator:
4
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Next, we can cancel out the common factor of x−2:
= lim
x→2(x+ 2)
Now, we can evaluate the limit by direct substitution:
= 2 + 2 = 4
Therefore, the value of the limit as xapproaches 2 is 4. Hence, the
limit exists and is equal to 4.Question 6: Determine the value of the
limit, if it exists, or state that the limit does not exist. If the limit
does not exist, explain why.
lim
x→2
x2−4
x−2
Step-by-step Solution: To determine the limit of the given ex-
pression as xapproaches 2, we can first simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Next, we can cancel out the common factor of x−2:
= lim
x→2(x+ 2)
Now, we can evaluate the limit by direct substitution:
= 2 + 2 = 4
Therefore, the value of the limit as xapproaches 2 is 4. Hence,
the limit exists and is equal to 4.
Question 7
lim
x→3
x2−9
x−3
Solution:
To determine if the limit exists at x= 3, we will first attempt to
directly substitute x= 3 into the expression.
Substitute x= 3:
lim
x→3
(3)2−9
3−3= lim
x→3
0
0
5
Since we obtained an indeterminate form ( 0
0), we need to further
simplify the expression to determine the limit.
Factor the numerator:
lim
x→3
(x+ 3)(x−3)
x−3
Simplify the expression:
= lim
x→3(x+ 3)
Now, we can directly substitute x= 3 into the simplified expression
to find the limit:
= 3 + 3 = 6
Therefore, the limit of the function as xapproaches 3 exists and
is equal to 6.Question 7: Determine whether the limit exists or not,
and justify your answer. If it does not exist, explain why.
lim
x→3
x2−9
x−3
Solution:
To determine if the limit exists at x= 3, we will first attempt to
directly substitute x= 3 into the expression.
Substitute x= 3:
lim
x→3
(3)2−9
3−3= lim
x→3
0
0
Since we obtained an indeterminate form ( 0
0), we need to further
simplify the expression to determine the limit.
Factor the numerator:
lim
x→3
(x+ 3)(x−3)
x−3
Simplify the expression:
= lim
x→3(x+ 3)
Now, we can directly substitute x= 3 into the simplified expression
to find the limit:
= 3 + 3 = 6
Therefore, the limit of the function as xapproaches 3 exists and
is equal to 6.
6
Question 8
Determine whether the limit exists or not:
lim
x→4
x2−16
x−4
Solution:
To determine whether the limit exists at x= 4, we can try to
simplify the expression or analyze the function behavior around the
point x= 4.
Given expression: x2−16
x−4
Factorizing x2−16 using the difference of squares formula gives:
x2−16 = (x−4)(x+ 4)
Substitute this back into the expression:
lim
x→4
(x−4)(x+ 4)
x−4
Now, simplify the expression by canceling out the common factor
of x−4:
lim
x→4(x+ 4)
Now, substitute x= 4 into the simplified expression:
4 + 4 = 8
Therefore, the limit exists and is equal to 8 at x= 4.Question 8:
Determine whether the limit exists or not:
lim
x→4
x2−16
x−4
Solution:
To determine whether the limit exists at x= 4, we can try to
simplify the expression or analyze the function behavior around the
point x= 4.
Given expression: x2−16
x−4
Factorizing x2−16 using the difference of squares formula gives:
x2−16 = (x−4)(x+ 4)
Substitute this back into the expression:
lim
x→4
(x−4)(x+ 4)
x−4
7
Now, simplify the expression by canceling out the common factor
of x−4:
lim
x→4(x+ 4)
Now, substitute x= 4 into the simplified expression:
4 + 4 = 8
Therefore, the limit exists and is equal to 8 at x= 4.
Question 9
Step-by-step solution: 1. Let’s first simplify the given expression:
x2−4
x−2=(x+ 2)(x−2)
x−2= (x+ 2)
2. Now, we want to show that the limit as x approaches 2 of (x +
2) does not exist.
3. Assume the limit exists and is equal to L:
lim
x→2(x+ 2) = L
4. We need to show that for any value of L, we can find two
different paths approaching 2 that give different limits.
5. Let’s consider two sequences that approach 2: an= 2 + 1
nand
bn= 2 −
1
n, where n is a positive integer.
6. As n approaches infinity, an= 2+and bn= 2−.
7. The value of the function (x + 2) for these sequences is:
f(an) = 2 + 1
n+ 2 = 4 + 1
n
f(bn)=2−
1
n+ 2 = 4 −
1
n
8. As n approaches infinity, f(an) = 4+and f(bn) = 4−, which are
different.
9. Therefore, since the function gives different values as it ap-
proaches 2 from the right and left sides, the limit does not exist.Question
9: Use the definition of limit to show that the limit does not exist:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first simplify the given expression:
x2−4
x−2=(x+ 2)(x−2)
x−2= (x+ 2)
8
2. Now, we want to show that the limit as x approaches 2 of (x +
2) does not exist.
3. Assume the limit exists and is equal to L:
lim
x→2(x+ 2) = L
4. We need to show that for any value of L, we can find two
different paths approaching 2 that give different limits.
5. Let’s consider two sequences that approach 2: an= 2 + 1
nand
bn= 2 −
1
n, where n is a positive integer.
6. As n approaches infinity, an= 2+and bn= 2−.
7. The value of the function (x + 2) for these sequences is:
f(an) = 2 + 1
n+ 2 = 4 + 1
n
f(bn)=2−
1
n+ 2 = 4 −
1
n
8. As n approaches infinity, f(an) = 4+and f(bn) = 4−, which are
different.
9. Therefore, since the function gives different values as it ap-
proaches 2 from the right and left sides, the limit does not exist.
Question 10
Step-by-step Solution: 1. Direct Substitution Method: Let us first
attempt to evaluate the limit by direct substitution:
lim
x→3
x2−9
x−3= lim
x→3
(3)2−9
3−3= lim
x→3
0
0
2. Factorization: Since we obtained the indeterminate form 0/0,
we can try to factorize the expression:
x2−9
x−3=(x−3)(x+ 3)
x−3=x+ 3
3. Apply the Limit:
lim
x→3(x+ 3) = 3 + 3 = 6
4. Conclusion: Since we were able to simplify the expression to
a non-indeterminate form and evaluate the limit to a finite number,
the limit limx→3
x2−9
x−3exists and is equal to 6.Question 10: Determine
whether the limit limx→3
x2−9
x−3exists.
9
Step-by-step Solution: 1. Direct Substitution Method: Let us first
attempt to evaluate the limit by direct substitution:
lim
x→3
x2−9
x−3= lim
x→3
(3)2−9
3−3= lim
x→3
0
0
2. Factorization: Since we obtained the indeterminate form 0/0,
we can try to factorize the expression:
x2−9
x−3=(x−3)(x+ 3)
x−3=x+ 3
3. Apply the Limit:
lim
x→3(x+ 3) = 3 + 3 = 6
4. Conclusion: Since we were able to simplify the expression to
a non-indeterminate form and evaluate the limit to a finite number,
the limit limx→3
x2−9
x−3exists and is equal to 6.
Question 11
lim
x→2
x2−4
x−2
Step-by-step solution: To determine if the limit exists, we first
simplify the expression by factoring the numerator:
lim
x→2
(x−2)(x+ 2)
x−2
Next, we cancel out the common factor of (x-2):
lim
x→2x+ 2
Now, we can evaluate the expression at x = 2:
2 + 2 = 4
Therefore, the limit of the function as x approaches 2 exists and is
equal to 4.Question 11: Determine whether the limit exists or does
not exist:
lim
x→2
x2−4
x−2
Step-by-step solution: To determine if the limit exists, we first
simplify the expression by factoring the numerator:
10
lim
x→2
(x−2)(x+ 2)
x−2
Next, we cancel out the common factor of (x-2):
lim
x→2x+ 2
Now, we can evaluate the expression at x = 2:
2 + 2 = 4
Therefore, the limit of the function as x approaches 2 exists and
is equal to 4.
Question 12
Question 12: Determine whether the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine whether the limit exists, we will evaluate
the limit from the left side and the right side of x= 2.
Left-side limit: limx→2−
x2−4
x−2= limx→2−
(x−2)(x+2)
x−2= limx→2−(x+ 2) =
4.
Right-side limit: limx→2+
x2−4
x−2= limx→2+(x−2)(x+2)
x−2= limx→2+(x+2) =
4.
Since the left-side and right-side limits are equal, the limit exists
and is equal to 4.
Therefore, limx→2
x2−4
x−2= 4.Sure, here is the question and its solu-
tion in LateX code:
Question 12: Determine whether the limit exists or not: limx→2
x2−4
x−2.
Solution: To determine whether the limit exists, we will evaluate
the limit from the left side and the right side of x= 2.
Left-side limit: limx→2−
x2−4
x−2= limx→2−
(x−2)(x+2)
x−2= limx→2−(x+ 2) =
4.
Right-side limit: limx→2+
x2−4
x−2= limx→2+(x−2)(x+2)
x−2= limx→2+(x+2) =
4.
Since the left-side and right-side limits are equal, the limit exists
and is equal to 4.
Therefore, limx→2
x2−4
x−2= 4.
Question 13
Step-by-step solution: To determine whether the limit exists or
not, we first try to simplify the expression by factoring out the com-
mon factor in the numerator:
11
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
Next, cancel out the common factor (x - 3) from the numerator
and denominator:
= lim
x→3(x+ 3)
Now, substitute x = 3 into the expression:
= 3 + 3
= 6
Therefore, the limit as x approaches 3 for the given expression is
6, which exists.Question 13: Determine whether the limit exists or
not:
lim
x→3
x2−9
x−3
Step-by-step solution: To determine whether the limit exists or
not, we first try to simplify the expression by factoring out the com-
mon factor in the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
Next, cancel out the common factor (x - 3) from the numerator
and denominator:
= lim
x→3(x+ 3)
Now, substitute x = 3 into the expression:
= 3 + 3
= 6
Therefore, the limit as x approaches 3 for the given expression is
6, which exists.
12
Question 14
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first try to directly substitute x= 2
into the expression:
lim
x→2
x2−4
x−2=22−4
2−2=0
0
2. The expression becomes 0
0, which is an indeterminate form.
3. To further analyze this limit, we can simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
4. Simplifying, we get:
lim
x→2(x+ 2) = 4
5. Therefore, the limit of the given expression as xapproaches 2
exists and is equal to 4.
Thus, the limit exists and is equal to 4.Question 14: Determine
whether the limit exists or not:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Let’s first try to directly substitute x= 2
into the expression:
lim
x→2
x2−4
x−2=22−4
2−2=0
0
2. The expression becomes 0
0, which is an indeterminate form.
3. To further analyze this limit, we can simplify the expression by
factoring the numerator:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
4. Simplifying, we get:
lim
x→2(x+ 2) = 4
5. Therefore, the limit of the given expression as xapproaches 2
exists and is equal to 4.
Thus, the limit exists and is equal to 4.
13
Question 15
Solution: To determine whether the limit exists or not, we can
analyze the behavior of the function as xapproaches 2 from both the
left and the right sides.
Left-hand limit: Let’s find the limit as xapproaches 2 from the
left side, denoted by limx→2−
x2−4
x−2.
lim
x→2−
x2−4
x−2= lim
x→2−
(x−2)(x+ 2)
x−2
= lim
x→2−
(x+ 2)
= 4
Right-hand limit: Similarly, let’s find the limit as xapproaches 2
from the right side, denoted by limx→2+
x2−4
x−2.
lim
x→2+
x2−4
x−2= lim
x→2+
(x−2)(x+ 2)
x−2
= lim
x→2+(x+ 2)
= 4
Since the left-hand limit and the right-hand limit both approach
the same value of 4 as xapproaches 2, we can conclude that limx→2
x2−4
x−2
exists and is equal to 4.Question 15: Determine whether the limit
limx→2
x2−4
x−2exists. If it does not exist, explain why.
Solution: To determine whether the limit exists or not, we can
analyze the behavior of the function as xapproaches 2 from both the
left and the right sides.
Left-hand limit: Let’s find the limit as xapproaches 2 from the
left side, denoted by limx→2−
x2−4
x−2.
lim
x→2−
x2−4
x−2= lim
x→2−
(x−2)(x+ 2)
x−2
= lim
x→2−
(x+ 2)
= 4
Right-hand limit: Similarly, let’s find the limit as xapproaches 2
from the right side, denoted by limx→2+
x2−4
x−2.
lim
x→2+
x2−4
x−2= lim
x→2+
(x−2)(x+ 2)
x−2
= lim
x→2+(x+ 2)
= 4
14
Since the left-hand limit and the right-hand limit both approach
the same value of 4 as xapproaches 2, we can conclude that limx→2
x2−4
x−2
exists and is equal to 4.
Question 16
Step-by-step solution: 1. Let’s start by simplifying the expression
using the difference of squares formula: (x2−4) = (x+ 2)(x−2). 2.
This gives us limx→2(x+2)(x−2)
x−2. 3. Cancel out the common factor of
(x−2) in the numerator and denominator to get limx→2(x+2). 4. Now,
we can directly substitute x= 2 into the expression to find the limit:
limx→2(x+ 2) = 2 + 2 = 4.
Therefore, the limit limx→2
x2−4
x−2exists and is equal to 4.Question
16: Determine whether the limit limx→2
x2−4
x−2exists. If it does not
exist, explain why.
Step-by-step solution: 1. Let’s start by simplifying the expression
using the difference of squares formula: (x2−4) = (x+ 2)(x−2). 2.
This gives us limx→2(x+2)(x−2)
x−2. 3. Cancel out the common factor of
(x−2) in the numerator and denominator to get limx→2(x+2). 4. Now,
we can directly substitute x= 2 into the expression to find the limit:
limx→2(x+ 2) = 2 + 2 = 4.
Therefore, the limit limx→2
x2−4
x−2exists and is equal to 4.
Question 17
lim
x→−2
x2−4
x+ 2
Step-by-step Solution: To determine if the limit exists at x=−2,
we can first try to evaluate the expression at x=−2.
Substitute x=−2into the expression:
(−2)2−4
−2+2 =4−4
0=0
0
Notice that we get an indeterminate form of 0
0when attempting
to evaluate the expression.
To further investigate if the limit exists, we can simplify the ex-
pression:
lim
x→−2
x2−4
x+ 2 = lim
x→−2
(x−2)(x+ 2)
x+ 2 = lim
x→−2(x−2)
Now, as xapproaches −2, the expression simplifies to:
15
lim
x→−2(x−2) = −2−2 = −4
Therefore, the limit of the given function as xapproaches −2exists
and the value is −4.Question 17: Determine whether the following
limit exists:
lim
x→−2
x2−4
x+ 2
Step-by-step Solution: To determine if the limit exists at x=−2,
we can first try to evaluate the expression at x=−2.
Substitute x=−2into the expression:
(−2)2−4
−2+2 =4−4
0=0
0
Notice that we get an indeterminate form of 0
0when attempting
to evaluate the expression.
To further investigate if the limit exists, we can simplify the ex-
pression:
lim
x→−2
x2−4
x+ 2 = lim
x→−2
(x−2)(x+ 2)
x+ 2 = lim
x→−2(x−2)
Now, as xapproaches −2, the expression simplifies to:
lim
x→−2(x−2) = −2−2 = −4
Therefore, the limit of the given function as xapproaches −2exists
and the value is −4.
Question 18
Step-by-step Solution: 1. Direct Substitution: If we substitute
x= 2 directly into the expression, we get an indeterminate form ( 0
0).
2. Factorization:
x2−4
x−2=(x+ 2)(x−2)
x−2
Simplifying, we get:
f(x) = x+ 2
3. Limit Calculation: Now, as xapproaches 2, the expression
f(x) = x+ 2 approaches 4. Therefore, the limit as xapproaches 2 of
x2−4
x−2is 4.
16
Conclusion: Since the limit exists and is equal to 4, we can con-
clude that the limit exists for the given function.Question 18: Deter-
mine whether the limit exists or not:
lim
x→2
x2−4
x−2
Step-by-step Solution: 1. Direct Substitution: If we substitute
x= 2 directly into the expression, we get an indeterminate form ( 0
0).
2. Factorization:
x2−4
x−2=(x+ 2)(x−2)
x−2
Simplifying, we get:
f(x) = x+ 2
3. Limit Calculation: Now, as xapproaches 2, the expression
f(x) = x+ 2 approaches 4. Therefore, the limit as xapproaches 2 of
x2−4
x−2is 4.
Conclusion: Since the limit exists and is equal to 4, we can con-
clude that the limit exists for the given function.
Question 19
Solution: To determine whether the limit exists or not, we will
analyze the behavior as xapproaches 0.
We know that limx→0sin(x)
x= 1.
The limit of sin(x)
xas xapproaches 0 is a well-known limit and
evaluates to 1. Therefore, the limit exists and equals 1.Question 19:
Determine whether the limit limx→0sin(x)
xexists. If it does not exist,
explain why.
Solution: To determine whether the limit exists or not, we will
analyze the behavior as xapproaches 0.
We know that limx→0sin(x)
x= 1.
The limit of sin(x)
xas xapproaches 0 is a well-known limit and
evaluates to 1. Therefore, the limit exists and equals 1.
Question 20
Solution: To determine whether the limit exists, we will simplify
the expression and see if we can find the limit.
Given function: f(x) = x−2
x2−4
1. Substitute x= 2 into the expression: f(2) = 2−2
22−4=0
0
2. Factor the denominator: x2−4=(x−2)(x+ 2)
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3. Rewrite the function: f(x) = x−2
(x−2)(x+2)
4. Simplify the expression: f(x) = 1
x+2
5. Now, substitute x= 2 into the simplified expression: f(2) =
1
2+2 =1
4
Since f(2) is finite and not equal to 0
0, the limit exists.
Therefore, limx→2
x−2
x2−4=1
4.Question 20: Determine whether the
limit exists or not: limx→2
x−2
x2−4.
Solution: To determine whether the limit exists, we will simplify
the expression and see if we can find the limit.
Given function: f(x) = x−2
x2−4
1. Substitute x= 2 into the expression: f(2) = 2−2
22−4=0
0
2. Factor the denominator: x2−4=(x−2)(x+ 2)
3. Rewrite the function: f(x) = x−2
(x−2)(x+2)
4. Simplify the expression: f(x) = 1
x+2
5. Now, substitute x= 2 into the simplified expression: f(2) =
1
2+2 =1
4
Since f(2) is finite and not equal to 0
0, the limit exists.
Therefore, limx→2
x−2
x2−4=1
4.
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