MATH 108 - ELEMENTARY AND
INTERMEDIATE ALGEBRA - Determining
Limits Algebraically Question Bank
Question 1
Determine the limit algebraically:
lim
x→3
x2−9
x−3
Step-by-step Solution:
To determine the limit algebraically, we first simplify the expression by fac-
toring the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x−3)(x+ 3)
x−3
Now, we can cancel out the common factor of (x−3) in the numerator and
denominator:
= lim
x→3(x+ 3)
Finally, we can substitute x= 3 into the simplified expression to find the
limit:
= 3 + 3 = 6
Therefore, the limit of x2−9
x−3as xapproaches 3 is 6.Question 1:
Determine the limit algebraically:
lim
x→3
x2−9
x−3
Step-by-step Solution:
To determine the limit algebraically, we first simplify the expres-
sion by factoring the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x−3)(x+ 3)
x−3
1
Now, we can cancel out the common factor of (x−3) in the numer-
ator and denominator:
= lim
x→3(x+ 3)
Finally, we can substitute x= 3 into the simplified expression to
find the limit:
= 3 + 3 = 6
Therefore, the limit of x2−9
x−3as xapproaches 3 is 6.
Question 2
Solution:
We can simplify the given expression by factoring the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+3)(x−3)
x−3
Next, cancel out the common factor of (x−3):
= lim
x→3x+ 3
Now, we can directly substitute x= 3:
= 3 + 3 = 6
Therefore, the limit is 6.Question 2: Determine the limit alge-
braically: lim
x→3
x2−9
x−3.
Solution:
We can simplify the given expression by factoring the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+3)(x−3)
x−3
Next, cancel out the common factor of (x−3):
= lim
x→3x+ 3
Now, we can directly substitute x= 3:
= 3 + 3 = 6
Therefore, the limit is 6.
Question 3
Step 1: Plug in x= 2 directly into the expression:
22−4
2−2=0
0
Step 2: Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
2
Step 3: Simplify the expression by canceling out the common fac-
tor of (x−2):
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
Step 4: Plug in x= 2 into the simplified expression:
lim
x→2(x+ 2) = 2 + 2 = 4
Answer: The limit of x2−4
x−2as xapproaches 2 is 4.Question 3: De-
termine the limit algebraically:
lim
x→2
x2−4
x−2
Step 1: Plug in x= 2 directly into the expression:
22−4
2−2=0
0
Step 2: Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
Step 3: Simplify the expression by canceling out the common fac-
tor of (x−2):
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
Step 4: Plug in x= 2 into the simplified expression:
lim
x→2(x+ 2) = 2 + 2 = 4
Answer: The limit of x2−4
x−2as xapproaches 2 is 4.
Question 4
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
lim
x→2
x2−4
x−2=4−4
0
3
2. Simplify the expression: Since the denominator is 0, we cannot
directly evaluate the limit. To simplify, factor the numerator as the
difference of squares:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
3. Evaluate the limit after cancellation:
lim
x→2
x2−4
x−2= lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the expression as xapproaches 2 is 4.Ques-
tion 4: Determine the limit algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
lim
x→2
x2−4
x−2=4−4
0
2. Simplify the expression: Since the denominator is 0, we cannot
directly evaluate the limit. To simplify, factor the numerator as the
difference of squares:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
3. Evaluate the limit after cancellation:
lim
x→2
x2−4
x−2= lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the expression as xapproaches 2 is 4.
Question 5
Find the limit algebraically:
lim
x→2
x2−4
x−2
4
Step-by-step solution:
1. Substitute x= 2 into the expression to see if it is indeterminate:
22−4
2−2=4−4
0=0
0
Since we obtained an indeterminate form, we proceed with alge-
braic simplification.
2. Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
3. Cancel out the common factor of (x−2):
lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the given expression as xapproaches 2 is
4.Question 5:
Find the limit algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution:
1. Substitute x= 2 into the expression to see if it is indeterminate:
22−4
2−2=4−4
0=0
0
Since we obtained an indeterminate form, we proceed with alge-
braic simplification.
2. Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
3. Cancel out the common factor of (x−2):
lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the given expression as xapproaches 2 is 4.
Question 6
Question 6: Find the limit of the function as x approaches 3:
lim
x→3
x2−9
x−3
Solution:
5
To find the limit of the function as x approaches 3, we can sim-
plify the expression by factoring the numerator and canceling out the
common factor in the denominator.
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
= lim
x→3(x+ 3)
Now, we can directly substitute x = 3 into the expression:
= 3 + 3 = 6
Therefore, the limit of the function as x approaches 3 is 6.
This is the solution to question 6.Certainly! Here is a question on
determining limits algebraically:
Question 6: Find the limit of the function as x approaches 3:
lim
x→3
x2−9
x−3
Solution:
To find the limit of the function as x approaches 3, we can sim-
plify the expression by factoring the numerator and canceling out the
common factor in the denominator.
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
= lim
x→3(x+ 3)
Now, we can directly substitute x = 3 into the expression:
= 3 + 3 = 6
Therefore, the limit of the function as x approaches 3 is 6.
This is the solution to question 6.
Question 7
Solution: To find the limit, we can simplify the expression first by
factoring the numerator: lim
x→3
x2−9
x−3= lim
x→3
(x−3)(x+3)
x−3
Next, we can cancel out the common factor of (x−3) from the
numerator and the denominator: = lim
x→3(x+ 3)
Finally, evaluate the limit by substituting x= 3 into the simplified
expression: = 3 + 3 = 6
6
Therefore, lim
x→3
x2−9
x−3= 6. “‘“‘latex Question 7: Find the limit alge-
braically: lim
x→3
x2−9
x−3.
Solution: To find the limit, we can simplify the expression first by
factoring the numerator: lim
x→3
x2−9
x−3= lim
x→3
(x−3)(x+3)
x−3
Next, we can cancel out the common factor of (x−3) from the
numerator and the denominator: = lim
x→3(x+ 3)
Finally, evaluate the limit by substituting x= 3 into the simplified
expression: = 3 + 3 = 6
Therefore, lim
x→3
x2−9
x−3= 6. “‘
Question 8
Solution: To determine the limit algebraically, we first simplify
the expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
= lim
x→2(x+ 2)
Since this expression is now in a form that we can directly substi-
tute the value of x into, we can evaluate the limit:
= 2 + 2
= 4
Therefore,
lim
x→2
x2−4
x−2= 4
.Question 8: Determine the limit algebraically:
lim
x→2
x2−4
x−2
Solution: To determine the limit algebraically, we first simplify
the expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
7
= lim
x→2(x+ 2)
Since this expression is now in a form that we can directly substi-
tute the value of x into, we can evaluate the limit:
= 2 + 2
= 4
Therefore,
lim
x→2
x2−4
x−2= 4
.
Question 9
Question 9: Determine the limit algebraically:
lim
x→0
3x2−2x
4x
Step-by-step solution: We can simplify the expression by factoring
out an xfrom the numerator:
lim
x→0
3x2−2x
4x= lim
x→0
x(3x−2)
4x
Now we can cancel out the xterms:
lim
x→0
3x−2
4
Next, we substitute x= 0 into the expression:
3(0) −2
4=−1
2
Therefore, the limit is −1
2.Sure, here is the question and step-by-
step solution formatted in LateX code:
Question 9: Determine the limit algebraically:
lim
x→0
3x2−2x
4x
Step-by-step solution: We can simplify the expression by factoring
out an xfrom the numerator:
lim
x→0
3x2−2x
4x= lim
x→0
x(3x−2)
4x
8
Now we can cancel out the xterms:
lim
x→0
3x−2
4
Next, we substitute x= 0 into the expression:
3(0) −2
4=−1
2
Therefore, the limit is −1
2.
Question 10
Find the limit of the function algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution:
1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
2. Simplify the expression:
=4−4
0=0
0
3. Factor out the numerator:
=(x−2)(x+ 2)
x−2
4. Cancel out the common factor of x−2:
=x+ 2
5. Substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, the limit of the function as xapproaches 2 is 4.Question
10:
Find the limit of the function algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution:
9
1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
2. Simplify the expression:
=4−4
0=0
0
3. Factor out the numerator:
=(x−2)(x+ 2)
x−2
4. Cancel out the common factor of x−2:
=x+ 2
5. Substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, the limit of the function as xapproaches 2 is 4.
Question 11
Question 11: Find the limit of the function f(x) = 3x2−4x+1
2x2−x−3as x
approaches 1.
Solution: To find the limit as xapproaches 1, we can directly
substitute x= 1 into the function:
f(1) = 3(1)2−4(1) + 1
2(1)2−1−3
f(1) = 3−4+1
2−1−3
f(1) = 0
−2
f(1) = 0
Therefore, the limit of the function as xapproaches 1 is 0.Sure!
Here is a question on determining limits algebraically:
Question 11: Find the limit of the function f(x) = 3x2−4x+1
2x2−x−3as x
approaches 1.
Solution: To find the limit as xapproaches 1, we can directly
substitute x= 1 into the function:
10
f(1) = 3(1)2−4(1) + 1
2(1)2−1−3
f(1) = 3−4+1
2−1−3
f(1) = 0
−2
f(1) = 0
Therefore, the limit of the function as xapproaches 1 is 0.
Question 12
Question 12: Find the limit of the function
f(x) = x2−4
x−2
as xapproaches 2.
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can try to simplify the expression by factoring the
numerator:
f(x) = x2−4
x−2=(x+ 2)(x−2)
x−2
Now, we see that x−2appears in both the numerator and denom-
inator, so we can cancel it to simplify the expression:
f(x) = x+ 2
Now we can find the limit as xapproaches 2 by directly substitut-
ing the value 2 into the simplified expression:
f(2) = 2 + 2 = 4
Therefore, the limit of the function f(x)as xapproaches 2 is 4.
You can use this solution to practice calculating limits algebraically.Certainly!
Here is a question on determining limits algebraically:
Question 12: Find the limit of the function
f(x) = x2−4
x−2
as xapproaches 2.
11
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can try to simplify the expression by factoring the
numerator:
f(x) = x2−4
x−2=(x+ 2)(x−2)
x−2
Now, we see that x−2appears in both the numerator and denom-
inator, so we can cancel it to simplify the expression:
f(x) = x+ 2
Now we can find the limit as xapproaches 2 by directly substitut-
ing the value 2 into the simplified expression:
f(2) = 2 + 2 = 4
Therefore, the limit of the function f(x)as xapproaches 2 is 4.
You can use this solution to practice calculating limits algebraically.
Question 13
Question 13: Find the limit:
lim
x→2
x2−4
x−2.
Solution: To find the limit, we can simplify the expression using
factorization:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2.
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
lim
x→2(x+ 2) = 2 + 2 = 4.
Therefore,
lim
x→2
x2−4
x−2= 4.
The limit of the expression as x approaches 2 is 4.Sure, here is a
question along with the step-by-step solution on determining limits
algebraically:
Question 13: Find the limit:
lim
x→2
x2−4
x−2.
12
Solution: To find the limit, we can simplify the expression using
factorization:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2.
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
lim
x→2(x+ 2) = 2 + 2 = 4.
Therefore,
lim
x→2
x2−4
x−2= 4.
The limit of the expression as x approaches 2 is 4.
Question 14
Question 14: Find the limit algebraically: limx→2
x2−4
x−2
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can simplify the expression by factoring the numer-
ator:
limx→2
x2−4
x−2= limx→2(x+2)(x−2)
x−2
Now, we can cancel out the common factor of (x−2):
= limx→2(x+ 2)
Now, substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, limx→2
x2−4
x−2= 4.Sure, here is a question with step-by-
step solutions on Determining Limits Algebraically:
Question 14: Find the limit algebraically: limx→2
x2−4
x−2
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can simplify the expression by factoring the numer-
ator:
limx→2
x2−4
x−2= limx→2(x+2)(x−2)
x−2
Now, we can cancel out the common factor of (x−2):
= limx→2(x+ 2)
Now, substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, limx→2
x2−4
x−2= 4.
Question 15
lim
h→0
3h2+ 2h
h
13
Step-by-step solution: To find the limit limh→03h2+2h
h, we can sim-
plify the expression by factoring out an hfrom the numerator:
= lim
h→0
h(3h+ 2)
h
Next, cancel out the common factor of hin the numerator and
denominator:
= lim
h→0(3h+ 2)
Now, substitute h= 0 into the expression:
= 3(0) + 2
= 2
Therefore, the limit is equal to 2.Question 15: Determine the limit
algebraically:
lim
h→0
3h2+ 2h
h
Step-by-step solution: To find the limit limh→03h2+2h
h, we can sim-
plify the expression by factoring out an hfrom the numerator:
= lim
h→0
h(3h+ 2)
h
Next, cancel out the common factor of hin the numerator and
denominator:
= lim
h→0(3h+ 2)
Now, substitute h= 0 into the expression:
= 3(0) + 2
= 2
Therefore, the limit is equal to 2.
14
Question 16
Question 16: Find the limit algebraically. You may need to ratio-
nalize the denominator if there is a radical expression.
lim
x→4
√x−2
x−4
Step-by-step solution: 1. Substitute x= 4 into the expression:
√4−2
4−4=2−2
0=0
0
2. We have an indeterminate form of 0
0, so we need to simplify the
expression further. 3. Multiply the numerator and denominator by
the conjugate of the numerator to rationalize the denominator:
lim
x→4
√x−2
x−4= lim
x→4
√x−2
x−4×√x+ 2
√x+ 2 = lim
x→4
x−4
(x−4)(√x+ 2)
4. Simplify the expression:
= lim
x→4
1
√x+ 2
5. Substitute x= 4 into the simplified expression:
=1
√4+2 =1
2+2 =1
4
Therefore, the limit of √x−2
x−4as xapproaches 4 is 1
4.Sure, here is a ques-
tion and step-by-step solution on determining limits algebraically:
Question 16: Find the limit algebraically. You may need to ratio-
nalize the denominator if there is a radical expression.
lim
x→4
√x−2
x−4
Step-by-step solution: 1. Substitute x= 4 into the expression:
√4−2
4−4=2−2
0=0
0
2. We have an indeterminate form of 0
0, so we need to simplify the
expression further. 3. Multiply the numerator and denominator by
the conjugate of the numerator to rationalize the denominator:
lim
x→4
√x−2
x−4= lim
x→4
√x−2
x−4×√x+ 2
√x+ 2 = lim
x→4
x−4
(x−4)(√x+ 2)
15
4. Simplify the expression:
= lim
x→4
1
√x+ 2
5. Substitute x= 4 into the simplified expression:
=1
√4+2 =1
2+2 =1
4
Therefore, the limit of √x−2
x−4as xapproaches 4 is 1
4.
Question 17
Question 17: Find the limit of the following function algebraically:
lim
x→4
x2−16
x−4
Step-by-step solution: To find the limit of the function, we can try
simplifying the expression by factoring out the common factor in the
numerator:
lim
x→4
x2−16
x−4= lim
x→4
(x+ 4)(x−4)
x−4
Next, we can cancel out the common factor of x−4in the numer-
ator and denominator:
= lim
x→4x+ 4
Now, substitute x= 4 into the simplified expression:
= 4 + 4
Therefore, the limit of the given function as xapproaches 4 is
8.Sure, here is a question on Determining Limits Algebraically:
Question 17: Find the limit of the following function algebraically:
lim
x→4
x2−16
x−4
Step-by-step solution: To find the limit of the function, we can try
simplifying the expression by factoring out the common factor in the
numerator:
lim
x→4
x2−16
x−4= lim
x→4
(x+ 4)(x−4)
x−4
Next, we can cancel out the common factor of x−4in the numer-
ator and denominator:
= lim
x→4x+ 4
Now, substitute x= 4 into the simplified expression:
= 4 + 4
Therefore, the limit of the given function as xapproaches 4 is 8.
16
Question 18
Question 18: Find the limit of the function as x approaches 5:
lim
x→5
x2−25
x−5
Solution: To find the limit, we can simplify the expression by
factoring the numerator:
lim
x→5
x2−25
x−5= lim
x→5
(x−5)(x+ 5)
x−5
Next, we can cancel out the common factor of (x - 5) from the
numerator and denominator:
lim
x→5(x+ 5)
Now we can substitute x = 5 into the expression:
5 + 5 = 10
Therefore,
lim
x→5
x2−25
x−5= 10
Thus, the limit of the function as x approaches 5 is 10.Sure, here
is a question and step-by-step solution on determining limits alge-
braically:
Question 18: Find the limit of the function as x approaches 5:
lim
x→5
x2−25
x−5
Solution: To find the limit, we can simplify the expression by
factoring the numerator:
lim
x→5
x2−25
x−5= lim
x→5
(x−5)(x+ 5)
x−5
Next, we can cancel out the common factor of (x - 5) from the
numerator and denominator:
lim
x→5(x+ 5)
Now we can substitute x = 5 into the expression:
5 + 5 = 10
Therefore,
lim
x→5
x2−25
x−5= 10
Thus, the limit of the function as x approaches 5 is 10.
17
Question 19
lim
x→1
x2−1
x−1
Step-by-step solution:
1. Substitute x= 1 into the expression:
12−1
1−1=0
0
2. Simplify the expression:
0
0
is an indeterminate form, so we need to simplify the expression fur-
ther.
3. Factor the numerator:
lim
x→1
(x−1)(x+ 1)
x−1
4. Cancel out the common factor of x−1:
lim
x→1(x+ 1)
5. Substitute x= 1 into the simplified expression:
1 + 1 = 2
Therefore,
lim
x→1
x2−1
x−1= 2
Question 19: Determine the limit algebraically.
lim
x→1
x2−1
x−1
Step-by-step solution:
1. Substitute x= 1 into the expression:
12−1
1−1=0
0
2. Simplify the expression:
0
0
is an indeterminate form, so we need to simplify the expression fur-
ther.
18
3. Factor the numerator:
lim
x→1
(x−1)(x+ 1)
x−1
4. Cancel out the common factor of x−1:
lim
x→1(x+ 1)
5. Substitute x= 1 into the simplified expression:
1 + 1 = 2
Therefore,
lim
x→1
x2−1
x−1= 2
Question 20
Question 20: Determine the limit algebraically:
lim
x→3
x2−9
x−3
Step-by-step solution:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3(Factor numerator)
= lim
x→3(x+ 3) (Cancel common factors)
= 3 + 3 (Substitute x= 3)
= 6
Therefore, the limit is 6.Certainly! Here is the question along with
the step-by-step solution in LaTeX code:
Question 20: Determine the limit algebraically:
lim
x→3
x2−9
x−3
Step-by-step solution:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3(Factor numerator)
= lim
x→3(x+ 3) (Cancel common factors)
= 3 + 3 (Substitute x= 3)
= 6
Therefore, the limit is 6.
19
Now, we can cancel out the common factor of (x−3) in the numer-
ator and denominator:
= lim
x→3(x+ 3)
Finally, we can substitute x= 3 into the simplified expression to
find the limit:
= 3 + 3 = 6
Therefore, the limit of x2−9
x−3as xapproaches 3 is 6.
Question 2
Solution:
We can simplify the given expression by factoring the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+3)(x−3)
x−3
Next, cancel out the common factor of (x−3):
= lim
x→3x+ 3
Now, we can directly substitute x= 3:
= 3 + 3 = 6
Therefore, the limit is 6.Question 2: Determine the limit alge-
braically: lim
x→3
x2−9
x−3.
Solution:
We can simplify the given expression by factoring the numerator:
lim
x→3
x2−9
x−3= lim
x→3
(x+3)(x−3)
x−3
Next, cancel out the common factor of (x−3):
= lim
x→3x+ 3
Now, we can directly substitute x= 3:
= 3 + 3 = 6
Therefore, the limit is 6.
Question 3
Step 1: Plug in x= 2 directly into the expression:
22−4
2−2=0
0
Step 2: Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
2
Step 3: Simplify the expression by canceling out the common fac-
tor of (x−2):
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
Step 4: Plug in x= 2 into the simplified expression:
lim
x→2(x+ 2) = 2 + 2 = 4
Answer: The limit of x2−4
x−2as xapproaches 2 is 4.Question 3: De-
termine the limit algebraically:
lim
x→2
x2−4
x−2
Step 1: Plug in x= 2 directly into the expression:
22−4
2−2=0
0
Step 2: Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
Step 3: Simplify the expression by canceling out the common fac-
tor of (x−2):
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
Step 4: Plug in x= 2 into the simplified expression:
lim
x→2(x+ 2) = 2 + 2 = 4
Answer: The limit of x2−4
x−2as xapproaches 2 is 4.
Question 4
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
lim
x→2
x2−4
x−2=4−4
0
3
2. Simplify the expression: Since the denominator is 0, we cannot
directly evaluate the limit. To simplify, factor the numerator as the
difference of squares:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
3. Evaluate the limit after cancellation:
lim
x→2
x2−4
x−2= lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the expression as xapproaches 2 is 4.Ques-
tion 4: Determine the limit algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution: 1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
lim
x→2
x2−4
x−2=4−4
0
2. Simplify the expression: Since the denominator is 0, we cannot
directly evaluate the limit. To simplify, factor the numerator as the
difference of squares:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2
lim
x→2
x2−4
x−2= lim
x→2(x+ 2)
3. Evaluate the limit after cancellation:
lim
x→2
x2−4
x−2= lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the expression as xapproaches 2 is 4.
Question 5
Find the limit algebraically:
lim
x→2
x2−4
x−2
4
Step-by-step solution:
1. Substitute x= 2 into the expression to see if it is indeterminate:
22−4
2−2=4−4
0=0
0
Since we obtained an indeterminate form, we proceed with alge-
braic simplification.
2. Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
3. Cancel out the common factor of (x−2):
lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the given expression as xapproaches 2 is
4.Question 5:
Find the limit algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution:
1. Substitute x= 2 into the expression to see if it is indeterminate:
22−4
2−2=4−4
0=0
0
Since we obtained an indeterminate form, we proceed with alge-
braic simplification.
2. Factor the numerator:
x2−4
x−2=(x+ 2)(x−2)
x−2
3. Cancel out the common factor of (x−2):
lim
x→2(x+ 2) = 2 + 2 = 4
Therefore, the limit of the given expression as xapproaches 2 is 4.
Question 6
Question 6: Find the limit of the function as x approaches 3:
lim
x→3
x2−9
x−3
Solution:
5
To find the limit of the function as x approaches 3, we can sim-
plify the expression by factoring the numerator and canceling out the
common factor in the denominator.
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
= lim
x→3(x+ 3)
Now, we can directly substitute x = 3 into the expression:
= 3 + 3 = 6
Therefore, the limit of the function as x approaches 3 is 6.
This is the solution to question 6.Certainly! Here is a question on
determining limits algebraically:
Question 6: Find the limit of the function as x approaches 3:
lim
x→3
x2−9
x−3
Solution:
To find the limit of the function as x approaches 3, we can sim-
plify the expression by factoring the numerator and canceling out the
common factor in the denominator.
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3
= lim
x→3(x+ 3)
Now, we can directly substitute x = 3 into the expression:
= 3 + 3 = 6
Therefore, the limit of the function as x approaches 3 is 6.
This is the solution to question 6.
Question 7
Solution: To find the limit, we can simplify the expression first by
factoring the numerator: lim
x→3
x2−9
x−3= lim
x→3
(x−3)(x+3)
x−3
Next, we can cancel out the common factor of (x−3) from the
numerator and the denominator: = lim
x→3(x+ 3)
Finally, evaluate the limit by substituting x= 3 into the simplified
expression: = 3 + 3 = 6
6
Therefore, lim
x→3
x2−9
x−3= 6. “‘“‘latex Question 7: Find the limit alge-
braically: lim
x→3
x2−9
x−3.
Solution: To find the limit, we can simplify the expression first by
factoring the numerator: lim
x→3
x2−9
x−3= lim
x→3
(x−3)(x+3)
x−3
Next, we can cancel out the common factor of (x−3) from the
numerator and the denominator: = lim
x→3(x+ 3)
Finally, evaluate the limit by substituting x= 3 into the simplified
expression: = 3 + 3 = 6
Therefore, lim
x→3
x2−9
x−3= 6. “‘
Question 8
Solution: To determine the limit algebraically, we first simplify
the expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
= lim
x→2(x+ 2)
Since this expression is now in a form that we can directly substi-
tute the value of x into, we can evaluate the limit:
= 2 + 2
= 4
Therefore,
lim
x→2
x2−4
x−2= 4
.Question 8: Determine the limit algebraically:
lim
x→2
x2−4
x−2
Solution: To determine the limit algebraically, we first simplify
the expression:
lim
x→2
x2−4
x−2= lim
x→2
(x−2)(x+ 2)
x−2
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
7
= lim
x→2(x+ 2)
Since this expression is now in a form that we can directly substi-
tute the value of x into, we can evaluate the limit:
= 2 + 2
= 4
Therefore,
lim
x→2
x2−4
x−2= 4
.
Question 9
Question 9: Determine the limit algebraically:
lim
x→0
3x2−2x
4x
Step-by-step solution: We can simplify the expression by factoring
out an xfrom the numerator:
lim
x→0
3x2−2x
4x= lim
x→0
x(3x−2)
4x
Now we can cancel out the xterms:
lim
x→0
3x−2
4
Next, we substitute x= 0 into the expression:
3(0) −2
4=−1
2
Therefore, the limit is −1
2.Sure, here is the question and step-by-
step solution formatted in LateX code:
Question 9: Determine the limit algebraically:
lim
x→0
3x2−2x
4x
Step-by-step solution: We can simplify the expression by factoring
out an xfrom the numerator:
lim
x→0
3x2−2x
4x= lim
x→0
x(3x−2)
4x
8
Now we can cancel out the xterms:
lim
x→0
3x−2
4
Next, we substitute x= 0 into the expression:
3(0) −2
4=−1
2
Therefore, the limit is −1
2.
Question 10
Find the limit of the function algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution:
1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
2. Simplify the expression:
=4−4
0=0
0
3. Factor out the numerator:
=(x−2)(x+ 2)
x−2
4. Cancel out the common factor of x−2:
=x+ 2
5. Substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, the limit of the function as xapproaches 2 is 4.Question
10:
Find the limit of the function algebraically:
lim
x→2
x2−4
x−2
Step-by-step solution:
9
1. Substitute x= 2 into the expression:
lim
x→2
x2−4
x−2=22−4
2−2
2. Simplify the expression:
=4−4
0=0
0
3. Factor out the numerator:
=(x−2)(x+ 2)
x−2
4. Cancel out the common factor of x−2:
=x+ 2
5. Substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, the limit of the function as xapproaches 2 is 4.
Question 11
Question 11: Find the limit of the function f(x) = 3x2−4x+1
2x2−x−3as x
approaches 1.
Solution: To find the limit as xapproaches 1, we can directly
substitute x= 1 into the function:
f(1) = 3(1)2−4(1) + 1
2(1)2−1−3
f(1) = 3−4+1
2−1−3
f(1) = 0
−2
f(1) = 0
Therefore, the limit of the function as xapproaches 1 is 0.Sure!
Here is a question on determining limits algebraically:
Question 11: Find the limit of the function f(x) = 3x2−4x+1
2x2−x−3as x
approaches 1.
Solution: To find the limit as xapproaches 1, we can directly
substitute x= 1 into the function:
10
f(1) = 3(1)2−4(1) + 1
2(1)2−1−3
f(1) = 3−4+1
2−1−3
f(1) = 0
−2
f(1) = 0
Therefore, the limit of the function as xapproaches 1 is 0.
Question 12
Question 12: Find the limit of the function
f(x) = x2−4
x−2
as xapproaches 2.
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can try to simplify the expression by factoring the
numerator:
f(x) = x2−4
x−2=(x+ 2)(x−2)
x−2
Now, we see that x−2appears in both the numerator and denom-
inator, so we can cancel it to simplify the expression:
f(x) = x+ 2
Now we can find the limit as xapproaches 2 by directly substitut-
ing the value 2 into the simplified expression:
f(2) = 2 + 2 = 4
Therefore, the limit of the function f(x)as xapproaches 2 is 4.
You can use this solution to practice calculating limits algebraically.Certainly!
Here is a question on determining limits algebraically:
Question 12: Find the limit of the function
f(x) = x2−4
x−2
as xapproaches 2.
11
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can try to simplify the expression by factoring the
numerator:
f(x) = x2−4
x−2=(x+ 2)(x−2)
x−2
Now, we see that x−2appears in both the numerator and denom-
inator, so we can cancel it to simplify the expression:
f(x) = x+ 2
Now we can find the limit as xapproaches 2 by directly substitut-
ing the value 2 into the simplified expression:
f(2) = 2 + 2 = 4
Therefore, the limit of the function f(x)as xapproaches 2 is 4.
You can use this solution to practice calculating limits algebraically.
Question 13
Question 13: Find the limit:
lim
x→2
x2−4
x−2.
Solution: To find the limit, we can simplify the expression using
factorization:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2.
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
lim
x→2(x+ 2) = 2 + 2 = 4.
Therefore,
lim
x→2
x2−4
x−2= 4.
The limit of the expression as x approaches 2 is 4.Sure, here is a
question along with the step-by-step solution on determining limits
algebraically:
Question 13: Find the limit:
lim
x→2
x2−4
x−2.
12
Solution: To find the limit, we can simplify the expression using
factorization:
lim
x→2
x2−4
x−2= lim
x→2
(x+ 2)(x−2)
x−2.
Now, we can cancel out the common factor of (x - 2) in the nu-
merator and denominator:
lim
x→2(x+ 2) = 2 + 2 = 4.
Therefore,
lim
x→2
x2−4
x−2= 4.
The limit of the expression as x approaches 2 is 4.
Question 14
Question 14: Find the limit algebraically: limx→2
x2−4
x−2
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can simplify the expression by factoring the numer-
ator:
limx→2
x2−4
x−2= limx→2(x+2)(x−2)
x−2
Now, we can cancel out the common factor of (x−2):
= limx→2(x+ 2)
Now, substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, limx→2
x2−4
x−2= 4.Sure, here is a question with step-by-
step solutions on Determining Limits Algebraically:
Question 14: Find the limit algebraically: limx→2
x2−4
x−2
Step-by-step solution: To find the limit of the given function as x
approaches 2, we can simplify the expression by factoring the numer-
ator:
limx→2
x2−4
x−2= limx→2(x+2)(x−2)
x−2
Now, we can cancel out the common factor of (x−2):
= limx→2(x+ 2)
Now, substitute x= 2 into the simplified expression:
= 2 + 2 = 4
Therefore, limx→2
x2−4
x−2= 4.
Question 15
lim
h→0
3h2+ 2h
h
13
Step-by-step solution: To find the limit limh→03h2+2h
h, we can sim-
plify the expression by factoring out an hfrom the numerator:
= lim
h→0
h(3h+ 2)
h
Next, cancel out the common factor of hin the numerator and
denominator:
= lim
h→0(3h+ 2)
Now, substitute h= 0 into the expression:
= 3(0) + 2
= 2
Therefore, the limit is equal to 2.Question 15: Determine the limit
algebraically:
lim
h→0
3h2+ 2h
h
Step-by-step solution: To find the limit limh→03h2+2h
h, we can sim-
plify the expression by factoring out an hfrom the numerator:
= lim
h→0
h(3h+ 2)
h
Next, cancel out the common factor of hin the numerator and
denominator:
= lim
h→0(3h+ 2)
Now, substitute h= 0 into the expression:
= 3(0) + 2
= 2
Therefore, the limit is equal to 2.
14
Question 16
Question 16: Find the limit algebraically. You may need to ratio-
nalize the denominator if there is a radical expression.
lim
x→4
√x−2
x−4
Step-by-step solution: 1. Substitute x= 4 into the expression:
√4−2
4−4=2−2
0=0
0
2. We have an indeterminate form of 0
0, so we need to simplify the
expression further. 3. Multiply the numerator and denominator by
the conjugate of the numerator to rationalize the denominator:
lim
x→4
√x−2
x−4= lim
x→4
√x−2
x−4×√x+ 2
√x+ 2 = lim
x→4
x−4
(x−4)(√x+ 2)
4. Simplify the expression:
= lim
x→4
1
√x+ 2
5. Substitute x= 4 into the simplified expression:
=1
√4+2 =1
2+2 =1
4
Therefore, the limit of √x−2
x−4as xapproaches 4 is 1
4.Sure, here is a ques-
tion and step-by-step solution on determining limits algebraically:
Question 16: Find the limit algebraically. You may need to ratio-
nalize the denominator if there is a radical expression.
lim
x→4
√x−2
x−4
Step-by-step solution: 1. Substitute x= 4 into the expression:
√4−2
4−4=2−2
0=0
0
2. We have an indeterminate form of 0
0, so we need to simplify the
expression further. 3. Multiply the numerator and denominator by
the conjugate of the numerator to rationalize the denominator:
lim
x→4
√x−2
x−4= lim
x→4
√x−2
x−4×√x+ 2
√x+ 2 = lim
x→4
x−4
(x−4)(√x+ 2)
15
4. Simplify the expression:
= lim
x→4
1
√x+ 2
5. Substitute x= 4 into the simplified expression:
=1
√4+2 =1
2+2 =1
4
Therefore, the limit of √x−2
x−4as xapproaches 4 is 1
4.
Question 17
Question 17: Find the limit of the following function algebraically:
lim
x→4
x2−16
x−4
Step-by-step solution: To find the limit of the function, we can try
simplifying the expression by factoring out the common factor in the
numerator:
lim
x→4
x2−16
x−4= lim
x→4
(x+ 4)(x−4)
x−4
Next, we can cancel out the common factor of x−4in the numer-
ator and denominator:
= lim
x→4x+ 4
Now, substitute x= 4 into the simplified expression:
= 4 + 4
Therefore, the limit of the given function as xapproaches 4 is
8.Sure, here is a question on Determining Limits Algebraically:
Question 17: Find the limit of the following function algebraically:
lim
x→4
x2−16
x−4
Step-by-step solution: To find the limit of the function, we can try
simplifying the expression by factoring out the common factor in the
numerator:
lim
x→4
x2−16
x−4= lim
x→4
(x+ 4)(x−4)
x−4
Next, we can cancel out the common factor of x−4in the numer-
ator and denominator:
= lim
x→4x+ 4
Now, substitute x= 4 into the simplified expression:
= 4 + 4
Therefore, the limit of the given function as xapproaches 4 is 8.
16
Question 18
Question 18: Find the limit of the function as x approaches 5:
lim
x→5
x2−25
x−5
Solution: To find the limit, we can simplify the expression by
factoring the numerator:
lim
x→5
x2−25
x−5= lim
x→5
(x−5)(x+ 5)
x−5
Next, we can cancel out the common factor of (x - 5) from the
numerator and denominator:
lim
x→5(x+ 5)
Now we can substitute x = 5 into the expression:
5 + 5 = 10
Therefore,
lim
x→5
x2−25
x−5= 10
Thus, the limit of the function as x approaches 5 is 10.Sure, here
is a question and step-by-step solution on determining limits alge-
braically:
Question 18: Find the limit of the function as x approaches 5:
lim
x→5
x2−25
x−5
Solution: To find the limit, we can simplify the expression by
factoring the numerator:
lim
x→5
x2−25
x−5= lim
x→5
(x−5)(x+ 5)
x−5
Next, we can cancel out the common factor of (x - 5) from the
numerator and denominator:
lim
x→5(x+ 5)
Now we can substitute x = 5 into the expression:
5 + 5 = 10
Therefore,
lim
x→5
x2−25
x−5= 10
Thus, the limit of the function as x approaches 5 is 10.
17
Question 19
lim
x→1
x2−1
x−1
Step-by-step solution:
1. Substitute x= 1 into the expression:
12−1
1−1=0
0
2. Simplify the expression:
0
0
is an indeterminate form, so we need to simplify the expression fur-
ther.
3. Factor the numerator:
lim
x→1
(x−1)(x+ 1)
x−1
4. Cancel out the common factor of x−1:
lim
x→1(x+ 1)
5. Substitute x= 1 into the simplified expression:
1 + 1 = 2
Therefore,
lim
x→1
x2−1
x−1= 2
Question 19: Determine the limit algebraically.
lim
x→1
x2−1
x−1
Step-by-step solution:
1. Substitute x= 1 into the expression:
12−1
1−1=0
0
2. Simplify the expression:
0
0
is an indeterminate form, so we need to simplify the expression fur-
ther.
18
3. Factor the numerator:
lim
x→1
(x−1)(x+ 1)
x−1
4. Cancel out the common factor of x−1:
lim
x→1(x+ 1)
5. Substitute x= 1 into the simplified expression:
1 + 1 = 2
Therefore,
lim
x→1
x2−1
x−1= 2
Question 20
Question 20: Determine the limit algebraically:
lim
x→3
x2−9
x−3
Step-by-step solution:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3(Factor numerator)
= lim
x→3(x+ 3) (Cancel common factors)
= 3 + 3 (Substitute x= 3)
= 6
Therefore, the limit is 6.Certainly! Here is the question along with
the step-by-step solution in LaTeX code:
Question 20: Determine the limit algebraically:
lim
x→3
x2−9
x−3
Step-by-step solution:
lim
x→3
x2−9
x−3= lim
x→3
(x+ 3)(x−3)
x−3(Factor numerator)
= lim
x→3(x+ 3) (Cancel common factors)
= 3 + 3 (Substitute x= 3)
= 6
Therefore, the limit is 6.
19
