WEEKLY LEARNING ACTIVITY SHEET
Basic Calculus, Grade 11, Quarter 3, Week 1
THE LIMIT OF A FUNCTION: Theorems and Examples
Most Essential Learning Competencies
1. illustrate the limit of a function using a table of values and the graph of the function
(STEM_BC11LCIIIa-1);
2. distinguish between lim 𝑓 (𝑥 ) and 𝑓 (𝑐 ) (STEM_BC11LCIIIa-2);
𝑥→𝑐
3. illustrate the limit laws (STEM_BC11LCIIIa-3); and
4. apply the limit laws in evaluating the limit of algebraic functions (polynomial, rational,
and radical) (STEM_BC11LCIIIa-4)
Specific Objectives
At the end of the lesson, the learner shall be able to:
1.
2.
3.
4.
define the limit of a function;
illustrate one-sided limits;
determine the limit of a function through tables of values and graphs;
compare lim 𝑓(𝑥 ) and 𝑓 (𝑐 );
𝑥→𝑐
5. demonstrate the limit theorems; and
6. evaluate the limits of polynomial, rational and radical functions.
Time Allotment: 4 hours
Key Concepts
THE LIMIT OF A FUNCTION
Definition
Consider a function 𝑓 of a single variable 𝑥 and a constant 𝑐 which the variable 𝑥 will
approach (𝑐 may or may not be in the domain of 𝑓). The limit of the function 𝑓 (𝑥 ) as 𝑥
approaches 𝑐 is equal to 𝐿, written as
lim 𝑓 (𝑥 ) = 𝐿
𝑥→𝑐
if for any 𝜖 > 0, however small, there exists a 𝛿 > 0 such that
|𝑓 (𝑥 ) − 𝐿| < 𝜖 whenever 0 < |𝑥 − 𝑐 | < 𝛿.
In other words, the definition states that the function values 𝑓 (𝑥 ) approach to a unique
real value 𝐿, i.e., the limit of the function 𝑓, as 𝑥 approaches to a number 𝑐 if the distance
between 𝑓 (𝑥 ) and 𝐿 can be made as small as we please by taking 𝑥 sufficiently near 𝑐, but
not equal to 𝑐.
Example 1: Investigate lim (1 + 2𝑥 ).
𝑥→1
Here, we have 𝑓 (𝑥 ) = 1 + 2𝑥 and the variable 𝑥 approaches to 𝑐 = 1. To evaluate the limit
of the given function, we will consider a table of values to look at the effect on 𝑓 (𝑥 ) when
𝑥 approaches to 1. Note that on the number line, 𝑥 may approach 𝑐 through values on its
left or its right. Hence, we can consider this table of values
Values close to 1 from the left
𝑥
0
0.5
0.9
0.99
𝑓(𝑥)
1
2
2.8
2.98
0.999
2.998
Values close to 1 from the right
𝑥
1.001
1.01
1.1
1.5
𝑓(𝑥) 3.0022 3.02
3.2
4
2
5
Observe that as the values of 𝑥 get closer and closer to 1,
regardless of the direction, the values of 𝑓(𝑥 ) gets closer
and closer to 3. Thus, the limit of 𝑓 (𝑥 ) = 1 + 2𝑥 as 𝑥
approaches to 1 is equal to 3, in symbols,
lim (1 + 2𝑥 ) = 3.
𝑥→1
Another way to evaluate the limit of the given function is
through its graph. Looking at the graph of 𝑓 (𝑥 ) = 1 + 2𝑥 in
the vicinity of 𝑥 = 1, it can be seen that the values of the
function approach the level where 𝑦 = 3, regardless of the
direction. Thus, the graph confirms that lim (1 + 2𝑥 ) = 3.
𝑥→1
Example 2: Investigate lim
𝑥 2 +𝑥−6
𝑥→2
𝑥−2
.
𝑥 2 +𝑥−6
Now, 𝑓 (𝑥 ) = 𝑥−2 and 𝑥 approaches to 𝑐 = 2. Note that when 𝑥 = 2, the function is not
defined, that is, 2 is not in the domain of 𝑓. But, this is not a problem since we are
evaluating the limit, which is the unique real value that 𝑓 (𝑥 ) is approaching as 𝑥
approaches to 2. Constructing the table of values from the left and the right of 2, we have
Values close to 2 from the left
𝑥
1
1.5
1.9
1.99
𝑓(𝑥)
4
4.5
4.9
4.99
1.999
4.999
Values close to 2 from the right
𝑥
2.001 2.01
2.1
2.5
𝑓(𝑥) 5.001 5.01
5.1
5.5
3
6
Observe that as 𝑥 approaches to 2, the values of 𝑓(𝑥 ) gets
closer and closer to 5, regardless of the direction. Hence,
we say that
𝑥2 + 𝑥 − 6
lim
= 5.
𝑥→2
𝑥−2
This is depicted in the graph of the function.
Example 3: Investigate lim √𝑥 − 3.
𝑥→3
We will evaluate the function 𝑓(𝑥 ) = √𝑥 − 3 at values close to 3 from the left and the right.
So, we have the following table of values
Values close to 3 from the left
𝑥
2
2.5
2.9
2.99
und
und
und
und
𝑓(𝑥)
2.999
und
Values close to 3 from the right
𝑥
3.001 3.01
3.1
3.5
𝑓(𝑥) 0.032
0.1
0.316 0.707
4
1
Notice that as 𝑥 approaches 3 from the right, the values of 𝑓 (𝑥 ) approach to 0. However,
as 𝑥 approaches 3 from the left, 𝑓 (𝑥 ) doesn’t converge to a real number. With this reason,
we say that lim √𝑥 − 3 does not exist (DNE). This behavior can also be seen in the graph of
𝑥→3
𝑓 (𝑥 ) = √𝑥 − 3. Observe that there is no graph on the left of 𝑥 = 3. That is, there is no real
value that 𝑓 (𝑥 ) will approach as 𝑥 approaches 3 from the left.
𝑥 + 1, 𝑥 < 4
Example 4: Investigate lim 𝑓 (𝑥 ) where 𝑓 (𝑥 ) = {
.
(𝑥 − 4)2 + 3, 𝑥 ≥ 4
𝑥→4
Since we have a piecewise function, we should evaluate the appropriate functional
expression of the values of 𝑥. That is, when 𝑥 approaches 4 from the left (or values less
than 4), we should consider the function 𝑓 (𝑥 ) = 𝑥 + 1. On the other hand, when 𝑥
approaches 4 from the right (or values greater than 4), we should use 𝑓(𝑥 ) = (𝑥 − 4)2 + 3.
Hence, we have
Values close to 4 from the left
𝑥
3
3.5
3.9
3.99
𝑓(𝑥)
4
4.5
4.9
4.99
3.999
4.999
Values close to 4 from the right
𝑥
4.001
4.01
4.1
4.5
𝑓(𝑥) 3.000001 3.0001 3.01 3.25
5
4
Observe that the values that 𝑓 (𝑥 ) approaches are not
equal as 𝑥 approaches 4 from the left and the right. The
values of 𝑓 (𝑥 ) approach to 5 as 𝑥 approaches 4 from the
left. Meanwhile, the values of 𝑓 (𝑥 ) approach to 3 as 𝑥
approaches 4 from the right. Since the limit of a function
is a unique real value, in this case, we say that the limit
of the given function does not exist (DNE). In symbols,
lim 𝑓 (𝑥 ) DNE.
𝑥→4
It can be seen in the graph of the function that the values
of 𝑓 (𝑥 ) approach to different values as 𝑥 approaches 4
from the left and the right.
Remarks
1. The left-hand limit is the value that 𝑓(𝑥 ) approaches as 𝑥 approaches to 𝑐 from the left,
or through values less than 𝑐. In symbols, lim− 𝑓 (𝑥 ).
𝑥→𝑐
2. The right-hand limit is the value that 𝑓(𝑥 ) approaches as 𝑥 approaches to 𝑐 from the
right or through values greater than 𝑐. In symbols, lim+ 𝑓 (𝑥 ).
𝑥→𝑐
3. lim− 𝑓(𝑥 ) and lim+ 𝑓(𝑥 ) are referred to as the one-sided limits.
𝑥→𝑐
𝑥→𝑐
lim 𝑓(𝑥) exists and is equal to 𝐿 if and only if lim− 𝑓(𝑥) = 𝐿 and lim+ 𝑓(𝑥) = 𝐿.
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
In other words, for a limit to exist, the limits from the left and the right must both exist
and are equal. Otherwise, the limit does not exist.
Thus, we say:
•
for Example 1, lim (1 + 2𝑥 ) = 3 since lim−(1 + 2𝑥 ) = 3 and lim+(1 + 2𝑥 ) = 3.
𝑥→1
𝑥 2 +𝑥−6
𝑥→1
𝑥 2 +𝑥−6
𝑥→1
𝑥 2 +𝑥−6
•
for Example 2, lim
•
for Example 3, lim √𝑥 − 3 DNE since lim− √𝑥 − 3 DNE.
•
for Example 4, lim 𝑓 (𝑥 ) DNE since lim 𝑓 (𝑥 ) = 5 and lim 𝑓 (𝑥 ) = 3.
𝑥→2
𝑥−2
= 5 since lim−
𝑥→2
𝑥→3
𝑥−2
= 5 and lim+
𝑥→2
𝑥−2
= 5.
𝑥→3
𝑥→4
𝑥→4
𝑥→4
THE LIMIT OF A FUNCTION AND THE VALUE OF THE FUNCTION
Now, consider the previous three examples and compare the limit of the function 𝑓 (𝑥 ) as
𝑥 approaches to 𝑐, i.e., lim 𝑓 (𝑥 ) and the value of the function at 𝑥 = 𝑐, i.e., 𝑓 (𝑐 ).
𝑥→𝑐
In Example 1,
lim 𝑓(𝑥 )
𝑓 (𝑐 )
lim (1 + 2𝑥 ) = 3
𝑓 (1) = 1 + 2(1) = 3
𝑥→𝑐
𝑥→1
Here, notice that lim 𝑓 (𝑥 ) and 𝑓(1) are equal.
𝑥→1
In Example 2,
lim 𝑓 (𝑥 )
𝑓 (𝑐 )
𝑥→𝑐
𝑥2 + 𝑥 − 6
=5
𝑥→2
𝑥−2
lim
𝑓(2) =
(2)2 + 2 − 6
(und)
2−2
Hence, lim 𝑓 (𝑥 ) and 𝑓 (2) are not the same.
𝑥→2
In Example 3,
lim 𝑓 (𝑥 )
𝑓 (𝑐 )
lim √𝑥 − 3 DNE
𝑓 (3) = √3 − 3 = 0
𝑥→𝑐
𝑥→3
So, lim 𝑓 (𝑥 ) and 𝑓 (3) are not the same.
𝑥→3
In Example 4,
lim 𝑓 (𝑥 )
𝑓 (𝑐 )
lim 𝑓 (𝑥 ) DNE
𝑓 (4) = (4 − 4)2 + 3 = 3
𝑥→𝑐
𝑥→4
Once again, in this example, lim 𝑓 (𝑥 ) and 𝑓 (4) are not the same.
𝑥→4
Example 5
Consider the graph of the function below.
From the given graph, we can conclude that
•
lim 𝑓 (𝑥 ) = 1 since lim − 𝑓 (𝑥 ) = 1 and lim + 𝑓 (𝑥 ) = 1, and 𝑓(−2) = 1.
𝑥→−2
•
𝑥→−2
𝑥→−2
𝑥→0
𝑥→0
𝑥→0
𝑥→3
𝑥→3
lim 𝑓 (𝑥 ) = 3 since
lim− 𝑓 (𝑥 ) = 3 and lim+ 𝑓 (𝑥 ) = 3, and 𝑓 (0) does not exist (or
undefined).
lim 𝑓 (𝑥 ) DNE since lim− 𝑓 (𝑥 ) = 0 and lim+ 𝑓(𝑥 ) = 4, and 𝑓 (3) = 2.
•
Remark:
𝑥→3
lim 𝑓(𝑥 ) is NOT NECESSARILY the same as 𝑓 (𝑐 ).
𝑥→𝑐
THE LIMIT THEOREMS
Limit Theorem 1
(The limit of a constant is itself)
If 𝑘 is any constant, then lim 𝑘 = 𝑘.
𝑥→𝑐
Example 6:
a. lim 2 = 2
c. lim 𝜋 = 𝜋
b. lim −5.12 = −5.12
d. lim 1000 = 1000
𝑥→0
𝑥→1
𝑥→100
𝑥→𝑐
Limit Theorem 2
(The limit of 𝑥 as 𝑥 approaches 𝑐 is equal to 𝑐)
lim 𝑥 = 𝑐.
𝑥→𝑐
Example 7:
a. lim 𝑥 = 5
c.
b.
d. lim 𝑥 = 𝑒
𝑥→5
lim 𝑥 = 0.25
𝑥→0.25
lim 𝑥 = −100
𝑥→−100
𝑥→𝑒
For the remaining theorems, let lim 𝑓(𝑥 ) = 𝐿 and lim 𝑔(𝑥 ) = 𝑀.
𝑥→𝑐
𝑥→𝑐
Limit Theorem 3
(The Constant Multiple Theorem)
If 𝑘 is any constant, then lim 𝑘 ⋅ 𝑓(𝑥) = 𝑘 ⋅ lim 𝑓(𝑥) = 𝑘 ⋅ 𝐿.
𝑥→𝑐
𝑥→𝑐
Example 8: Let lim 𝑓 (𝑥 ) = 5. Then
𝑥→𝑐
a. lim 3 ⋅ 𝑓 (𝑥 ) = 3 ⋅ lim 𝑓 (𝑥 ) = 3 ⋅ 5 = 15
𝑥→𝑐
𝑥→𝑐
b. lim −2 ⋅ 𝑓(𝑥 ) = −2 ⋅ lim 𝑓 (𝑥 ) = −2 ⋅ 5 =
𝑥→𝑐
c. lim 0.5 ⋅ 𝑓 (𝑥 ) = 0.5 ⋅ lim 𝑓 (𝑥 ) = 0.5 ⋅ 5 =
𝑥→𝑐
𝑥→𝑐
2.5
4
4
4
d. lim 5 ⋅ 𝑓 (𝑥 ) = 5 ⋅ lim 𝑓 (𝑥 ) = 5 ⋅ 5 = 4
𝑥→𝑐
−10
𝑥→𝑐
𝑥→𝑐
Limit Theorem 4
(The Addition/Subtraction Theorem)
lim (𝑓(𝑥) + 𝑔(𝑥)) = lim 𝑓(𝑥) + lim 𝑔(𝑥) = 𝐿 + 𝑀
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
lim (𝑓(𝑥) − 𝑔(𝑥)) = lim 𝑓(𝑥) − lim 𝑔(𝑥) = 𝐿 − 𝑀
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
Example 9: Let lim 𝑓 (𝑥 ) = 2 and lim 𝑔(𝑥 ) = −3. Then
𝑥→𝑐
𝑥→𝑐
a. lim(𝑓 (𝑥 ) + 𝑔(𝑥 )) = lim 𝑓(𝑥 ) + lim 𝑔(𝑥 ) = 2 + (−3) = −1
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
b. lim(𝑓(𝑥 ) − 𝑔(𝑥 )) = lim 𝑓 (𝑥 ) − lim 𝑔(𝑥 ) = 2 − (−3) = 5
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
Limit Theorem 5
(The Multiplication Theorem)
lim (𝑓(𝑥) ⋅ 𝑔(𝑥)) = lim 𝑓(𝑥) ⋅ lim 𝑔(𝑥) = 𝐿 ⋅ 𝑀
𝑥→𝑐
Example 10:
𝑥→𝑐
𝑥→𝑐
Let lim 𝑓 (𝑥 ) = 5 and lim 𝑔(𝑥 ) = −2. Then
𝑥→𝑐
𝑥→𝑐
lim(𝑓(𝑥 ) ⋅ 𝑔(𝑥 )) = lim 𝑓 (𝑥 ) ⋅ lim 𝑔(𝑥 ) = 5 ⋅ (−2) = −10
𝑥→𝑐
Remark:
𝑥→𝑐
𝑥→𝑐
Limit Theorems 4 and 5 still hold to sums, differences, and products of more
than two functions
Limit Theorem 6
(The Division Theorem)
lim 𝑓(𝑥) 𝐿
𝑓(𝑥) 𝑥→𝑐
=
= ,
𝑥→𝑐 𝑔(𝑥)
lim 𝑔(𝑥) 𝑀
lim
𝑥→𝑐
Example 11:
a. If lim 𝑓 (𝑥 ) = 3 and lim 𝑔(𝑥 ) = −6, then
𝑥→𝑐
𝑥→𝑐
𝑀≠0
lim 𝑓 (𝑥 )
𝑓 (𝑥 ) 𝑥→𝑐
3
1
=
=
=− .
𝑥→𝑐 𝑔(𝑥 )
lim 𝑔(𝑥 ) −6
2
lim
𝑥→𝑐
b. If lim 𝑓 (𝑥 ) = 0 and lim 𝑔(𝑥 ) = 4, then
𝑥→𝑐
𝑥→𝑐
lim 𝑓 (𝑥 ) 0
𝑓 (𝑥 ) 𝑥→𝑐
=
= = 0.
𝑥→𝑐 𝑔(𝑥 )
lim 𝑔(𝑥 ) 4
lim
𝑥→𝑐
c. If lim 𝑓 (𝑥 ) = 2 and lim 𝑔(𝑥 ) = 0, then
𝑥→𝑐
𝑥→𝑐
𝑓(𝑥 )
lim 𝑔(𝑥) DNE since lim 𝑔(𝑥 ) = 0.
𝑥→𝑐
𝑥→𝑐
Limit Theorem 7
(The Power Theorem)
𝑛
lim [𝑓(𝑥)]𝑛 = [lim 𝑓(𝑥)] = 𝐿𝑛 , for any integer 𝑛.
𝑥→𝑐
Example 12:
𝑥→𝑐
Let lim 𝑓 (𝑥 ) = 2. Then
𝑥→𝑐
3
a. lim[𝑓(𝑥 )]3 = [lim 𝑓 (𝑥 )] = (2)3 = 8
𝑥→𝑐
𝑥→𝑐
b. lim[𝑓(𝑥 )]−4 = [lim 𝑓(𝑥 )]
𝑥→𝑐
𝑥→𝑐
−4
1
1
= (2)−4 = 24 = 16
Limit Theorem 8
(The Radical Theorem)
𝑛
𝑛
lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) = √𝐿, for any positive integer 𝑛
𝑥→𝑐
𝑥→𝑐
and provided that 𝐿 > 0 when 𝑛 is even.
Example 13:
a. If lim 𝑓 (𝑥 ) = 16, then
𝑥→𝑐
lim √𝑓 (𝑥 ) = √lim 𝑓 (𝑥 ) = √16 = 4
𝑥→𝑐
𝑥→𝑐
b. If lim 𝑓 (𝑥 ) = −16, then
𝑥→𝑐
lim √𝑓(𝑥 ) = √lim 𝑓 (𝑥 ) = √−16.
𝑥→𝑐
𝑥→𝑐
Since √−16 is not a real number, lim √𝑓 (𝑥 ) DNE.
𝑥→𝑐
c. If lim 𝑓 (𝑥 ) = 8, then
𝑥→𝑐
3
3
lim √𝑓(𝑥 ) = 3√lim 𝑓(𝑥 ) = √8 = 2.
𝑥→𝑐
𝑥→𝑐
d. If lim 𝑓 (𝑥 ) = −8, then
𝑥→𝑐
3
3
lim √𝑓(𝑥 ) = 3√lim 𝑓(𝑥 ) = √−8 = −2.
𝑥→𝑐
𝑥→𝑐
Example 14:
Compute the following limits if
2
lim 𝑓(𝑥 ) = , lim 𝑔(𝑥 ) = 8, and lim ℎ(𝑥 ) = −4
3
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
a. lim(5 ⋅ 𝑓 (𝑥 ) − 2 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ))
𝑥→𝑐
Solution:
lim(5 ⋅ 𝑓 (𝑥 ) − 2 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 )) = lim 5 ⋅ 𝑓 (𝑥 ) − lim 2 ⋅ 𝑔(𝑥 ) +
𝑥→𝑐
𝑥→𝑐
lim ℎ(𝑥 )
(Limit Theorem 4)
lim(5 ⋅ 𝑓 (𝑥 ) − 2 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 )) = 5 ⋅ lim 𝑓 (𝑥 ) − 2 ⋅ lim 𝑔(𝑥 ) +
𝑥→𝑐
𝑥→𝑐
lim ℎ(𝑥 )
(Limit Theorem 3)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
2
lim(5 ⋅ 𝑓 (𝑥 ) − 2 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 )) = 5 (3) − 2(8) + (−4)
𝑥→𝑐
lim(5 ⋅ 𝑓 (𝑥 ) − 2 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 )) = −
𝑥→𝑐
b. lim
𝑔(𝑥)
2
𝑥→𝑐 (ℎ(𝑥))
(Given)
50
3
⋅ 𝑓 (𝑥 )
Solution:
lim
𝑔(𝑥)
2
𝑥→𝑐 (ℎ(𝑥))
lim
𝑔(𝑥)
2
𝑥→𝑐 (ℎ(𝑥))
lim
lim 𝑔(𝑥)
𝑥→𝑐
2
lim(ℎ(𝑥))
lim 𝑔(𝑥)
⋅ 𝑓 (𝑥 ) =
𝑥→𝑐
2
(lim ℎ(𝑥))
8
⋅ 𝑓 (𝑥 ) = (−4)2 ⋅ 3
2
⋅ 𝑓 (𝑥 ) = 3
𝑥→𝑐 (ℎ(𝑥))
(Limit Theorem 5 & 6)
⋅ lim 𝑓 (𝑥 )
(Limit Theorem 7)
𝑥→𝑐
𝑥→𝑐
2
2
𝑥→𝑐 (ℎ(𝑥))
𝑔(𝑥)
⋅ lim 𝑓 (𝑥 )
𝑥→𝑐
𝑥→𝑐
𝑔(𝑥)
lim
⋅ 𝑓 (𝑥 ) =
(Given)
1
c. lim √3 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ) + 5
𝑥→𝑐
Solution:
lim √3 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ) + 5 = √lim(3 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ) + 5)
(Limit Theorem 8)
lim √3 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ) + 5 = √3 lim 𝑔(𝑥 ) + lim ℎ(𝑥 ) + 5
(Limit Theorem 1, 3 & 4)
lim √3 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ) + 5 = √3(8) + (−4) + 5
(Given)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
lim √3 ⋅ 𝑔(𝑥 ) + ℎ(𝑥 ) + 5 = 5
𝑥→𝑐
LIMITS OF ALGEBRAIC FUNCTIONS
Example 15:
a. lim (3𝑥 + 1)
𝑥→1
Evaluate the limit of the following functions.
Solution:
lim (3𝑥 + 1) = lim 3𝑥 + lim 1
𝑥→1
𝑥→1
lim (3𝑥 + 1) = 3 lim 𝑥 + lim 1
(Limit Theorem 4)
lim (3𝑥 + 1) = 3(1) + 1
lim (3𝑥 + 1) = 4
(Limit Theorem 1 & 2)
𝑥→1
𝑥→1
𝑥→1
(Limit Theorem 3)
𝑥→1
𝑥→1
𝑥→1
b. lim
2𝑥+3
𝑥→3 4−𝑥
Solution:
lim
2𝑥+3
𝑥→3 4−𝑥
lim
2𝑥+3
𝑥→3 4−𝑥
2𝑥+3
lim (2𝑥+3)
𝑥→3
lim 2𝑥+lim 3
𝑥→3
2(3)+3
=
lim
=9
𝑥→3 4−𝑥
(Limit Theorem 4)
𝑥→3
= 𝑥→3
lim 4−lim 𝑥
lim
𝑥→3 4−𝑥
2𝑥+3
(Limit Theorem 6)
= 𝑥→3
lim (4−𝑥)
𝑥→3
(Limit Theorem 1, 2 & 3)
4−3
Observe that from Example 14, if we evaluate the given function in (a) at 𝑥 = 1, 𝑓(1) =
2(3)+3
3(1) + 1 = 4 and in (b) at 𝑥 = 3, 𝑓 (3) =
= 9. Hence, the function's values are equal to
4−3
their limits as 𝑥 approaches 1 and 3, respectively. With this, we have the following
theorems.
Theorem 9
Let 𝑓 be a polynomial function of the form
𝑓(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + 𝑎𝑛−2 𝑥 𝑛−2 + ⋯ + 𝑎1 𝑥 + 𝑎0 .
If 𝑐 is a real number, then
lim 𝑓(𝑥) = 𝑓(𝑐).
𝑥→𝑐
Example 16:
a. lim (3𝑥 4 + 4𝑥 3 − 5𝑥 2 + 𝑥 − 3)
𝑥→−1
Solution:
Note that the function 𝑓 (𝑥 ) = 3𝑥 4 + 4𝑥 3 − 5𝑥 2 + 𝑥 − 3 is a polynomial, and when 𝑥 = −1, we
have
𝑓 (−1) = 3(−1)4 + 4(−1)3 − 5(−1)2 + (−1) − 3 = −10.
Thus, by Theorem 9,
lim (3𝑥 4 + 4𝑥 3 − 5𝑥 2 + 𝑥 − 3) = 𝑓(−1) = −10.
𝑥→−1
b. lim (2𝑥 2 + 1)(3 − 2𝑥 )
𝑥→2
Solution:
By Limit Theorem 5 and Theorem 9, we have
lim (2𝑥 2 + 1)(3 − 2𝑥 ) = (2(2)2 + 1)(3 − 2(2)) = −9.
𝑥→2
Theorem 10
Let ℎ be a rational function of the form ℎ(𝑥) =
If 𝑐 is a real number and 𝑔(𝑐) ≠ 0, then
𝑓(𝑥)
𝑔(𝑥)
where 𝑓 and 𝑔 are polynomial functions.
lim 𝑓(𝑥) 𝑓(𝑐)
𝑓(𝑥) 𝑥→𝑐
=
=
.
𝑥→𝑐 𝑔(𝑥)
lim 𝑔(𝑥) 𝑔(𝑐)
lim ℎ(𝑥) = lim
𝑥→𝑐
𝑥→𝑐
Example 17:
1+3𝑥
a. lim 3𝑥2 −2𝑥+4
𝑥→1
Solution:
Here, 𝑓 (𝑥 ) = 1 + 3𝑥 and 𝑔(𝑥 ) = 3𝑥 2 − 2𝑥 + 4. Evaluating the two functions at 𝑥 = 1, we have
𝑓 (1) = 1 + 3(1) = 4 and 𝑔(𝑥 ) = 3(1)2 − 2(1) + 4 = 5.
Hence, by Theorem 10,
1 + 3𝑥
4
lim 2
= .
𝑥→1 3𝑥 − 2𝑥 + 4
5
b. lim
(𝑥−1)(𝑥 2 +1)
1+3𝑥 2
𝑥→2
Solution:
Using Theorem 10, we have
(𝑥 − 1)(𝑥 2 + 1) (2 − 1)((2)2 + 1)
5
=
=
.
2
2
𝑥→2
1 + 3𝑥
1 + 3(2)
13
lim
For radical functions, apply Limit Theorem 8.
Example 18:
a. lim √3𝑥 + 4
𝑥→4
Solution:
lim √3𝑥 + 4 = √lim (3𝑥 + 4)
(Limit Theorem 8)
lim √3𝑥 + 4 = √3(4) + 4
(Theorem 9)
𝑥→4
𝑥→4
𝑥→4
lim √3𝑥 + 4 = 4
𝑥→4
3
b. lim √𝑥 2 + 3𝑥 − 6
𝑥→−2
Solution:
Applying the same technique as Example 16(a), we have
3
3
3
lim √𝑥 2 + 3𝑥 − 6 = √(−2)2 + 3(−2) − 6 = √−8 = −2.
𝑥→−2
√2𝑥+5
𝑥→2 1−3𝑥
c. lim
Solution:
3
√2𝑥 + 5 √2(2) + 5 √9
=
=
=− .
𝑥→2 1 − 3𝑥
1 − 3(2)
−5
5
lim
Activity No. 1 Where Am I Going?
What you need: Paper and Ballpen
What to do: Fill in the tables below using the specified values of 𝑥 and illustrate the limit of
the function by answering the given questions.
A. Consider the function defined by 𝑓 (𝑥 ) = √𝑥 2 − 3𝑥 + 2 as 𝑥 approaches to 2.
Values close to 2 from the left
1
𝑥
1.5
1.9
1.99
𝑓(𝑥)
1.999
Values close to 2 from the right
𝑥
2.001 2.01
2.1
2.5
𝑓(𝑥)
3
1. What happened to the values of 𝑓 (𝑥 ) as 𝑥 gets closer and closer to 2 from the left?
a. The values of 𝑓(𝑥 ) gets closer and closer to 0.
b. The values of 𝑓(𝑥 ) gets closer and closer to 1.
c. The values of 𝑓(𝑥 ) gets closer and closer to 2.
d. The values of 𝑓(𝑥 ) is not approaching to a real number.
2. What is lim− 𝑓(𝑥 )?
𝑥→2
a. 0
b. 1
c. 2
d. DNE
3. What happened to the values of 𝑓 (𝑥 ) as 𝑥 gets closer and closer to 2 from the right?
a. The values of 𝑓(𝑥 ) gets closer and closer to 0.
b. The values of 𝑓(𝑥 ) gets closer and closer to 1.
c. The values of 𝑓(𝑥 ) gets closer and closer to 2.
d. The values of 𝑓(𝑥 ) is not approaching to a real number.
4. What is lim+ 𝑓(𝑥 )?
𝑥→2
a. 0
b. 1
c. 2
d. DNE
c. 2
d. DNE
5. What is lim 𝑓 (𝑥 )? Why?
𝑥→2
a. 0
b. 1
3𝑥 − 2, 𝑥 < 1
B. Consider the function defined by 𝑓 (𝑥 ) = {
as 𝑥 approaches to 1.
(𝑥 − 4)2 − 8, 𝑥 ≥ 1
Values close to 1 from the left
𝑥
0
0.5
0.9
0.99
𝑓(𝑥)
0.999
Values close to 1 from the right
𝑥
1.001 1.01
1.1
1.5
𝑓(𝑥)
2
1. What happened to the values of 𝑓 (𝑥 ) as 𝑥 gets closer and closer to 1 from the left?
a. The values of 𝑓(𝑥 ) gets closer and closer to 1.
b. The values of 𝑓(𝑥 ) gets closer and closer to 2.
c. The values of 𝑓(𝑥 ) gets closer and closer to 3.
d. The values of 𝑓(𝑥 ) is not approaching to a real number.
2. What is lim− 𝑓 (𝑥 )?
𝑥→1
a. 1
b. 2
c. 3
d. DNE
3. What happened to the values of 𝑓 (𝑥 ) as 𝑥 gets closer and closer to 1 from the right?
a. The values of 𝑓(𝑥 ) gets closer and closer to 1.
b. The values of 𝑓(𝑥 ) gets closer and closer to 2.
c. The values of 𝑓(𝑥 ) gets closer and closer to 3.
d. The values of 𝑓(𝑥 ) is not approaching to a real number.
4. What is lim+ 𝑓 (𝑥 )?
𝑥→1
a. 1
b. 2
c. 3
d. DNE
c. 3
d. DNE
5. What is lim 𝑓 (𝑥 )? Why?
𝑥→1
a. 1
b. 2
Activity No. 2 Investigate Me!
What you need: Paper and Ballpen
What to do: Consider the graph of the function below and investigate the limit of the function
as 𝑥 approaches 𝑐 and the value of the function at 𝑥 = 𝑐 by completing the given table.
𝑐
−2
−0.5
0
1
3
4
lim 𝑓(𝑥 )
𝑥→𝑐
𝑓 (𝑐 )
Equal or
Unequal?
Activity No. 3 The Limit Maze
What you need: Coloring material, Paper, and Ballpen
What to do: Find your way through the end of the maze by evaluating the given limits.
Shade the path that corresponds to your answer. Show your complete solution
Reflection
On a separate sheet of paper, write a short reflective essay (one to two paragraphs) detailing
your experiences in completing the activities. You may summarize the things you have
learned, their applications in your daily lives, and the things you enjoy or dislike.
RUBRICS
10 – 9 points
The reflection explains
the student's thinking
and learning
experiences and
implications for future
learning.
8 – 6 points
The reflection explains
the student's thinking
about his/her own
learning experiences.
5 – 3 points
The reflection attempts
to demonstrate thinking
about learning but is
vague and/or unclear
about the personal
learning experiences.
2 – 0 points
The reflection does not
address the student's
thinking and/or
learning.
References
Balmecada, J. P. et al. (2016). Basic Calculus Teacher's Guide (1st ed.). Philippines:
Department of Education.
Leithold, L. (1976). The Calculus with Analytic Geometry (3rd ed.). New York: Harper &
Row.
Writer:
Reviewers:
JEREMIAH A. ATENTA
ELMER R. ANDEBOR
AMALIA B. RINGOR, DevEdD
Special Science Teacher I
Agusan National High School
STEM Group Head
Agusan National High School
Academic Track Head
Agusan National High School
RUTH A. CASTROMAYOR
ISRAEL B. REVECHE, PhD
Principal IV
SHS Assistant Principal
Agusan National High School
Education Program Supervisor
Butuan City Division SHS Coordinator
KEY TO CORRECTION
Activity No. 1 (Where am I going?)
A.
Values close to 2 from the left
und
0
𝑓(𝑥)
B.
1.
2.
3.
4.
5.
B.
Values
1.5
1
𝑥
1.9
und
1.99
und
Values close to 2 from the right
𝑓(𝑥)
und
𝑥
1.999
2.001
0.0316
2.01
0.1004
2.1
0.3317
1.414
0.866
3
2.5
D
D
A
A
D
close to 1 from the left
−2
𝑓(𝑥)
C.
0
𝑥
1.
2.
3.
4.
5.
0.5
−0.5
0.9
0.7
Values close to 2 from the right
0.99
0.97
𝑓(𝑥)
0.997
𝑥
0.999
1.001
0.994
1.01
0.94
1.1
0.41
−4
−1.75
2
1.5
A
A
A
A
A
Activity No. 2 (Investigate me!)
lim 𝑓(𝑥)
𝑐
𝑥→𝑐
Equal or
Unequal?
𝑓(𝑐)
5
5
4
4
DNE
3
0
DNE
1
3
DNE
0
1
3.5
−0.5
2
2
−2
Equal
Unequal
Unequal
Unequal
Unequal
Equal
Activity No. 3 The Limit Maze