SlideShare a Scribd company logo
Limit of an Exponential
Function
Let a and b be real numbers,
where b > 0 and b ≠ 1.
Then,
𝐥𝐢𝐦
𝒙→𝒂
𝒃 𝒙
= 𝒃 𝒂
.
Properties
There are four basic properties in limits,
which are used as formulas in evaluating the
limits of exponential functions.
1.Power Rule
2.Constant Base Power Rule
3.Constant Exponent Power Rule
4.Radical Power Rule
1. Power Rule
𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙 𝒈 𝒙
= 𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙
𝒍𝒊𝒎
𝒙→𝒂
𝒈 𝒙
It is a property of power rule, used to
find the limit of an exponential function
whose base and exponent are in a
function form.
1. Power Rule 𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙 𝒈 𝒙
= 𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙
𝒍𝒊𝒎
𝒙→𝒂
𝒈 𝒙
Example: Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟒, 𝒈 𝒙 = 𝟐𝒙 ; 𝐥𝐢𝐦
𝒙→𝟑
𝒇 𝒙 𝒈 𝒙
Solution: 𝐥𝐢𝐦
𝒙→3
𝒇 𝒙 𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→3
𝒇 𝒙
𝐥𝐢𝐦
𝒙→𝟑
𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→3
𝒇 𝒙2
− 4
𝐥𝐢𝐦
𝒙→3
𝒈 2𝒙
= 𝟑 𝟐
− 𝟒
𝟐 𝟑
= 𝟗 − 𝟒 𝟔
= 𝟓 𝟔 = 𝟏𝟓, 𝟔𝟐𝟓
2. Constant Base Power Rule
𝒍𝒊𝒎
𝒙→𝒂
𝒃 𝒇 𝒙
= 𝒍𝒊𝒎
𝒙→𝒂
𝒃
𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙
The limit of an exponential function is equal to
the limit of the exponent with same base. It is
called the limit rule of an exponential function.
2. Constant Base Power Rule
𝒍𝒊𝒎
𝒙→𝒂
𝒃 𝒇 𝒙
= 𝒍𝒊𝒎
𝒙→𝒂
𝒃
𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙
Example:
Solution:
Evaluate 𝑏 = 12, 𝑓 𝑥 = 𝑥 + 3; 𝐥𝐢𝐦
𝒙→𝟏
𝒃 𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟏
𝒃 )𝒇(𝒙
= 𝒍𝒊𝒎
𝒙→𝟏
𝟏𝟐 𝒙+𝟑
= 𝒍𝒊𝒎
𝒙→𝟏
𝟏𝟐
𝒍𝒊𝒎
𝒙→𝟏
𝒙+𝟑
= 𝟏𝟐 𝒙+𝟑
= 𝟏𝟐 𝟏+𝟑
= 𝟏𝟐 𝟒
= 𝟐𝟎, 𝟕𝟑𝟔
3. Constant Exponent Power Rule
𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→𝒂
𝒇 𝒙 𝒏
Example:
Solution:
Evaluate 𝑓 𝑥 = 16𝑥2
− 64, 𝑛 = 2; 𝐥𝐢𝐦
𝒙→𝟏
𝒇 𝒙
𝒏
𝒍𝒊𝒎
𝒙→𝟏
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→𝟏
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→𝟏
𝟏𝟔𝒙 𝟐
− 𝟔𝟒
𝟐
= 𝟏𝟔 𝟏 𝟐
− 𝟔𝟒
𝟐
= 𝟏𝟔 − 𝟔𝟒 𝟐
= −𝟒𝟖 𝟐 = 𝟐, 𝟑𝟎𝟒
4. Radical Power Rule
𝒍𝒊𝒎
𝒙→𝒂
𝒏
)𝒇(𝒙 = 𝒏
𝒍𝒊𝒎
𝒙→𝒂
)𝒇(𝒙
Example:
Solution:
Evaluate 𝒇 𝒙 = 𝒙 𝟐
− 𝟔𝒙 + 𝟗, 𝒏 = 𝟐; 𝒍𝒊𝒎
𝒙→𝟔
𝒏
𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟔
𝒏
)𝒇(𝒙 = 𝒏
𝒍𝒊𝒎
𝒙→𝟔
)𝒇(𝒙
= 𝐥𝐢𝐦
𝒙→6
𝒙2 − 6𝒙 + 9
= 𝒙 − 3 2 = 𝒙 − 3
= 6 − 3 = 𝟑
Let’s Practice!!!
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔, 𝒏 = 𝟐;
a. 𝒍𝒊𝒎
𝒙→−𝟑
𝒏
𝒇(𝒙) b. 𝒍𝒊𝒎
𝒙→𝟒
𝒇 𝒙 𝒏
2. Evaluate 𝒇 𝒙 = 𝒙 𝟑
− 𝟐𝟕, 𝒈 𝒙 = 𝟐𝒙 + 𝟏
a. 𝒍𝒊𝒎
𝒙→𝟐
𝒇 𝒙 𝒈 𝒙
3. Evaluate 𝑏 = 𝟑, 𝑓 𝑥 = 𝒙 − 𝟒
a. 𝐥𝐢𝐦
𝒙→𝟖
𝒃 𝒇(𝒙)
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔, 𝒏 = 𝟐;
a. 𝒍𝒊𝒎
𝒙→−𝟑
𝒏
𝒇(𝒙)
𝒍𝒊𝒎
𝒙→−𝟑
𝒏
)𝒇(𝒙 = 𝒏
𝒍𝒊𝒎
𝒙→−𝟑
)𝒇(𝒙
= 𝐥𝐢𝐦
𝒙→−𝟑
𝟒𝒙2 − 𝟐𝟒𝒙 + 𝟑𝟔
= 𝟐𝒙 − 𝟔 2 = 𝟐𝒙 − 𝟔
= 𝟐 −𝟑 − 𝟔
= −𝟏𝟐
= −𝟔 − 𝟔
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔, 𝒏 = 𝟐;
b. 𝒍𝒊𝒎
𝒙→𝟒
𝒇 𝒙 𝒏
𝒍𝒊𝒎
𝒙→𝟒
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→𝟒
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→𝟒
𝟒𝒙 𝟐
− 𝟐𝟒𝒙 + 𝟑𝟔
𝟐
= 𝟒 𝟒 𝟐
− 𝟐𝟒 𝟒 + 𝟑𝟔
𝟐
= 𝟒 𝟏𝟔 − 𝟗𝟔 + 𝟑𝟔 𝟐
= 𝟒 𝟐
= 𝟏𝟔
= 𝟔𝟒 − 𝟗𝟔 + 𝟑𝟔 𝟐
2. Evaluate 𝒇 𝒙 = 𝒙 𝟑 − 𝟐𝟕, 𝒈 𝒙 = 𝟐𝒙 + 𝟏
a. 𝒍𝒊𝒎
𝒙→𝟐
𝒇 𝒙 𝒈 𝒙
𝐥𝐢𝐦
𝒙→𝟐
𝒇 𝒙 𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→2
𝒇 𝒙
𝐥𝐢𝐦
𝒙→𝟐
𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→𝟐
𝒙 𝟑
− 𝟐𝟕
𝐥𝐢𝐦
𝒙→𝟐
𝟐𝐱+𝟏
= 𝟐 𝟑
− 𝟐𝟕
𝟐 𝟐 +𝟏
= 𝟖 − 𝟐𝟕 𝟓
= −𝟏𝟗 𝟓
= −𝟐, 𝟒𝟕𝟔, 𝟎𝟗𝟗
3. Evaluate 𝑏 = 𝟑, 𝑓 𝑥 = 𝒙 − 𝟒
a. 𝐥𝐢𝐦
𝒙→𝟖
𝒃 𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟖
𝒃 )𝒇(𝒙
= 𝒍𝒊𝒎
𝒙→𝟖
𝟑 𝒙−𝟒
= 𝒍𝒊𝒎
𝒙→𝟖
𝟑
𝒍𝒊𝒎
𝒙→𝟖
𝒙−𝟒
= 𝟑 𝒙−𝟒
= 𝟑 𝟖−𝟒
= 𝟑 𝟒
= 𝟖𝟏
Seatwork 3.1
1. Evaluate 𝒇 𝒙 = 𝒙 𝟐
+ 𝟏𝟔𝒙 + 𝟖, 𝒏 = 𝟑;
a. 𝒍𝒊𝒎
𝒙→𝟒
𝒏
𝒇(𝒙) b. 𝒍𝒊𝒎
𝒙→−𝟐
𝒇 𝒙 𝒏
2. Evaluate 𝒇 𝒙 = 𝒙 𝟐
− 𝟔𝟒, 𝒈 𝒙 = 𝒙 − 𝟓
a. 𝒍𝒊𝒎
𝒙→𝟗
𝒇 𝒙 𝒈 𝒙 b. 𝒍𝒊𝒎
𝒙→𝟗
𝒈 𝒙 𝒇 𝒙
3. Evaluate 𝑏 = 𝟏𝟓, 𝑓 𝑥 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏
a. 𝐥𝐢𝐦
𝒙→𝟐
𝒃 𝒇(𝒙)
b. 𝐥𝐢𝐦
𝒙→−𝟑
𝒃 𝒇(𝒙)
ANSWERS
1. Evaluate 𝒇 𝒙 = 𝒙 𝟐
+ 𝟏𝟔𝒙 + 𝟖, 𝒏 = 𝟑;
a. 𝒍𝒊𝒎
𝒙→𝟒
𝒏
𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟒
𝒏
)𝒇(𝒙 = 𝒏
𝒍𝒊𝒎
𝒙→𝟒
)𝒇(𝒙
=
𝟑
𝒍𝒊𝒎
𝒙→𝟒
𝒙 𝟐 + 𝟏𝟔𝒙 + 𝟖
=
3
𝟒 𝟐 + 𝟏𝟔 𝟒 + 𝟖
=
3
𝟏𝟔 + 𝟔𝟒 + 𝟖
= 𝟒. 𝟒𝟓
=
3
𝟖𝟖
1. Evaluate 𝒇 𝒙 = 𝒙 𝟐
+ 𝟏𝟔𝒙 + 𝟖, 𝒏 = 𝟑;
b. 𝒍𝒊𝒎
𝒙→−𝟐
𝒇 𝒙 𝒏
𝒍𝒊𝒎
𝒙→−𝟐
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→−𝟐
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→−𝟐
𝒙 𝟐
+ 𝟏𝟔𝒙 + 𝟖
𝟑
= −𝟐 𝟐
+ 𝟏𝟔 −𝟐 + 𝟖
𝟑
= 𝟏𝟔 − 𝟑𝟐 + 𝟖 𝟑
= −𝟖 𝟑
= −𝟓𝟏𝟐
2. Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟔𝟒, 𝒈 𝒙 = 𝒙 − 𝟓
a. 𝒍𝒊𝒎
𝒙→𝟗
𝒇 𝒙 𝒈 𝒙
𝐥𝐢𝐦
𝒙→𝟗
𝒇 𝒙 𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→9
𝒇 𝒙
𝐥𝐢𝐦
𝒙→𝟗
𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→𝟗
𝒙 𝟐
− 𝟔𝟒
𝐥𝐢𝐦
𝒙→𝟗
𝐱−𝟓
= 𝟗 𝟐
− 𝟔𝟒
𝟗−𝟓
= 𝟖𝟏 − 𝟔𝟒 𝟒
= 𝟏𝟕 𝟒
= 𝟖𝟑, 𝟓𝟐𝟏
2. Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟔𝟒, 𝒈 𝒙 = 𝒙 − 𝟓
b. 𝒍𝒊𝒎
𝒙→𝟗
𝒈 𝒙 𝒇 𝒙
𝐥𝐢𝐦
𝒙→𝟗
𝒈 𝒙 𝒇 𝒙
= 𝐥𝐢𝐦
𝒙→9
𝒈 𝒙
𝐥𝐢𝐦
𝒙→𝟗
𝒇 𝒙
= 𝐥𝐢𝐦
𝒙→𝟗
𝒙 − 𝟓
𝐥𝐢𝐦
𝒙→𝟗
𝐱 𝟐−𝟔𝟒
= (𝟗 − 𝟓) 𝟗 𝟐−𝟔𝟒
= (𝟒) 𝟖𝟏−𝟔𝟒
= 𝟒 𝟏𝟕
= 𝟏𝟕, 𝟏𝟕𝟗, 𝟖𝟔𝟗, 𝟏𝟖𝟒
3. Evaluate 𝑏 = 𝟏𝟓, 𝑓 𝑥 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏
a. 𝐥𝐢𝐦
𝒙→𝟐
𝒃 𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟐
𝒃 )𝒇(𝒙
= 𝒍𝒊𝒎
𝒙→𝟐
𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏
= 𝒍𝒊𝒎
𝒙→𝟐
𝟏𝟓
𝒍𝒊𝒎
𝒙→𝟐
𝒙 𝟐+𝟐𝒙+𝟏
= 𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏
= 𝟏𝟓 𝟐 𝟐+𝟐 𝟐 +𝟏
= 𝟏𝟓 𝟗
= 𝟑𝟖, 𝟒𝟒𝟑, 𝟑𝟓𝟗, 𝟑𝟕𝟓
3. Evaluate 𝑏 = 𝟏𝟓, 𝑓 𝑥 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏
b. 𝐥𝐢𝐦
𝒙→−𝟑
𝒃 𝒇(𝒙)
𝒍𝒊𝒎
𝒙→−𝟑
𝒃 )𝒇(𝒙
= 𝒍𝒊𝒎
𝒙→−𝟑
𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏
= 𝒍𝒊𝒎
𝒙→−𝟑
𝟏𝟓
𝒍𝒊𝒎
𝒙→−𝟑
𝒙 𝟐+𝟐𝒙+𝟏
= 𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏
= 𝟏𝟓 −𝟑 𝟐+𝟐 −𝟑 +𝟏
= 𝟏𝟓 𝟒
= 𝟓𝟎, 𝟔𝟐𝟓
Quiz 3.1
Evaluate 𝒇 𝒙 = 𝟏𝟔𝒙 𝟐 − 𝟖𝒙 + 𝟏,
𝒈 𝒙 = 𝟔𝟒𝒙 𝟐 + 𝟒𝟖𝒙 + 𝟗,
𝒏 = 𝟐;
a. 𝒍𝒊𝒎
𝒙→𝟓
𝒏
𝒇(𝒙) b. 𝒍𝒊𝒎
𝒙→−𝟓
𝒇 𝒙 𝒏
c. 𝒍𝒊𝒎
𝒙→𝟖
𝒏
𝒈(𝒙) d. 𝒍𝒊𝒎
𝒙→−𝟖
𝒈 𝒙 𝒏
Evaluate 𝒇 𝒙 = 𝟗𝒛 𝟐 − 𝟏𝟎𝟎,
𝒈 𝒙 = 𝒛 𝟐
− 𝟐𝟓
a. 𝒍𝒊𝒎
𝒛→−𝟑
𝒇 𝒙 𝒈 𝒙 b. 𝒍𝒊𝒎
𝒛→𝟑
𝒈 𝒙 𝒇 𝒙
Evaluate 𝒃 = 𝟑𝟔, 𝒇 𝒙 = 𝒙 𝟑 − 𝟐𝟕
a. 𝐥𝐢𝐦
𝒙→𝟑
𝒃 𝒇(𝒙) b. 𝐥𝐢𝐦
𝒙→𝟐
𝒃 𝒇(𝒙)
ANSWERS
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗,
𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑;
a. 𝒍𝒊𝒎
𝒙→𝟑
𝒏
𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟑
𝒏
)𝒇(𝒙 = 𝒏
𝒍𝒊𝒎
𝒙→𝟑
)𝒇(𝒙
=
𝟑
𝒍𝒊𝒎
𝒙→𝟑
𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗
=
3
𝟒 𝟑 𝟐 − 𝟔 𝟑 + 𝟗
=
3
𝟑𝟔 − 𝟏𝟖 + 𝟗 = 𝟑=
3
𝟐𝟕
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗,
𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑;
b. 𝒍𝒊𝒎
𝒙→−𝟑
𝒇 𝒙 𝒏
𝒍𝒊𝒎
𝒙→−𝟑
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→−𝟑
𝒇 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→−𝟑
𝟒𝒙 𝟐
− 𝟔𝒙 + 𝟗
𝟑
= 𝟒 −𝟑 𝟐
− 𝟔 −𝟑 + 𝟗
𝟑
= 𝟑𝟔 + 𝟏𝟖 + 𝟗 𝟑
= 𝟔𝟑 𝟑
= 𝟐𝟓𝟎, 𝟎𝟒𝟕
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗,
𝒈 𝒙 = 𝒙 𝟑
+ 𝟖𝒙 𝟐
− 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑;
c. 𝒍𝒊𝒎
𝒙→𝟐
𝒏
𝒈(𝒙)
𝒍𝒊𝒎
𝒙→𝟐
𝒏
)𝒈(𝒙 = 𝒏
𝒍𝒊𝒎
𝒙→𝟐
)𝒈(𝒙
=
𝟑
𝒍𝒊𝒎
𝒙→𝟐
𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒
=
3
𝟐 𝟑 + 𝟖 𝟐 𝟐 − 𝟏𝟔 𝟐 + 𝟒
=
3
𝟖 + 𝟑𝟐 − 𝟑𝟐 + 𝟒 = 𝟐. 𝟐𝟗=
3
𝟏𝟐
1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗,
𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑;
d. 𝒍𝒊𝒎
𝒙→−𝟐
𝒈 𝒙 𝒏
𝒍𝒊𝒎
𝒙→−𝟐
𝒈 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→−𝟐
𝒈 𝒙 𝒏
= 𝒍𝒊𝒎
𝒙→−𝟐
𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒
𝟑
= −𝟐 𝟑
+ 𝟖 −𝟐 𝟐
− 𝟏𝟔 −𝟐 + 𝟒
𝟑
= −𝟖 + 𝟑𝟐 + 𝟑𝟐 + 𝟒 𝟑
= 𝟔𝟎 𝟑
= 𝟐𝟏𝟔, 𝟎𝟎𝟎
2. Evaluate 𝒇 𝒙 = 𝟏𝟔𝐱 𝟐 − 𝟐𝟓,
𝒈 𝒙 = 𝒙 𝟑
− 𝟏
a. 𝒍𝒊𝒎
𝒙→𝟑
𝒇 𝒙 𝒈 𝒙
𝐥𝐢𝐦
𝒙→𝟑
𝒇 𝒙 𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→3
𝒇 𝒙
𝐥𝐢𝐦
𝒙→𝟑
𝒈 𝒙
= 𝐥𝐢𝐦
𝒙→𝟑
𝟏𝟔𝒙 𝟐
− 𝟐𝟓
𝐥𝐢𝐦
𝒙→𝟑
𝐱 𝟑−𝟏
= 𝟏𝟔 𝟑 𝟐
− 𝟐𝟓
𝟑 𝟑−𝟏
= 𝟏𝟒𝟒 − 𝟐𝟓 𝟐𝟔
= 𝟏𝟏𝟗 𝟐𝟔
= 𝟗. 𝟐𝟏 × 𝟏𝟎 𝟓𝟑
2. Evaluate 𝒇 𝒙 = 𝟏𝟔𝐱 𝟐 − 𝟐𝟓,
𝒈 𝒙 = 𝒙 𝟑
− 𝟏
b. 𝒍𝒊𝒎
𝒙→𝟑
𝒈 𝒙 𝒇 𝒙
𝐥𝐢𝐦
𝒙→𝟑
𝒈 𝒙 𝒇 𝒙
= 𝐥𝐢𝐦
𝒙→3
𝒈 𝒙
𝐥𝐢𝐦
𝒙→𝟑
𝒇 𝒙
= 𝐥𝐢𝐦
𝒙→𝟑
𝒙 𝟑
− 𝟏
𝐥𝐢𝐦
𝒙→𝟑
𝟏𝟔𝐱 𝟐−𝟐𝟓
= ( 𝟑 𝟑
− 𝟏) 𝟏𝟔(𝟑) 𝟐−𝟐𝟓
= (𝟐𝟔) 𝟏𝟒𝟒−𝟐𝟓
= 𝟐𝟔 𝟏𝟏𝟗
= 𝟐. 𝟒𝟏 × 𝟏𝟎 𝟏𝟔𝟖
3. Evaluate 𝑏 = 𝟐𝟒, 𝑓 𝑥 = 𝒙 𝟑 − 𝟖
a. 𝐥𝐢𝐦
𝒙→𝟑
𝒃 𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟑
𝒃 )𝒇(𝒙
= 𝒍𝒊𝒎
𝒙→𝟑
𝟐𝟒 𝒙 𝟑−𝟖
= 𝒍𝒊𝒎
𝒙→𝟑
𝟐𝟒
𝒍𝒊𝒎
𝒙→𝟑
𝒙 𝟑−𝟖
= 𝟐𝟒 𝒙 𝟑−𝟖
= 𝟐𝟒 𝟑 𝟑−𝟖
= 𝟐𝟒 𝟏𝟗
= 𝟏. 𝟔𝟕 × 𝟏𝟎 𝟐𝟔
3. Evaluate 𝑏 = 𝟐𝟒, 𝑓 𝑥 = 𝒙 𝟑 − 𝟖
b. 𝐥𝐢𝐦
𝒙→𝟐
𝒃 𝒇(𝒙)
𝒍𝒊𝒎
𝒙→𝟐
𝒃 )𝒇(𝒙
= 𝒍𝒊𝒎
𝒙→𝟐
𝟐𝟒 𝒙 𝟑−𝟖
= 𝒍𝒊𝒎
𝒙→𝟐
𝟐𝟒
𝒍𝒊𝒎
𝒙→𝟐
𝒙 𝟑−𝟖
= 𝟐𝟒 𝒙 𝟑−𝟖
= 𝟐𝟒 𝟐 𝟑−𝟖
= 𝟐𝟒 𝟎
= 𝟏
Limit of Logarithmic
Function
Let a and b be real numbers,
where a > 0, b > 0 and b ≠ 1.
Then,
𝒍𝒊𝒎
𝒙→𝒂
( 𝒍𝒐𝒈 𝒃 𝒙) = 𝒍𝒐𝒈 𝒃 𝒂
Properties
There are three basic properties in limits,
which are used as formulas in evaluating the
limits of logarithmic functions.
1.Product Rule
2.Quotient Rule
3.Power Rule
1. Product Rule
Example:
Solution:
lim
𝑥→𝑎
(log 𝑏 𝐴𝐵) = lim
𝑥→𝑎
log 𝑏 𝐴 + log 𝑏 𝐵
Given 𝑨 = 𝒙 𝟐
+ 𝟐𝒙 + 𝟏, 𝑩 = 𝒙 − 𝟓, 𝒃 = 𝟓 𝒂𝒏𝒅 𝒙 = 𝟑,
find 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝒙→𝟑
𝐥𝐨𝐠 𝟓(𝒙 𝟐
+𝟐𝒙 + 𝟏) + 𝐥𝐨𝐠 𝟓 𝒙 − 𝟓)
= 𝐥𝐨𝐠 𝟓 𝟑 𝟐
+ 𝟐 𝟑 + 𝟏 + 𝟑 − 𝟓
= 𝐥𝐨𝐠 𝟓 𝟗 + 𝟔 + 𝟏 − 𝟐
= 𝐥𝐨𝐠 𝟓 𝟏𝟒 = 𝟏. 𝟔𝟒
2. Quotient Rule
Example:
Solution:
lim
𝑥→𝑎
log 𝑏(
𝐴
𝐵
) = lim
𝑥→𝑎
log 𝑏 𝐴 − log 𝑏 𝐵
Given 𝑨 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏, 𝑩 = 𝒙 − 𝟓, 𝒃 = 𝟓 𝒂𝒏𝒅 𝒙 = 𝟑,
find 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃
𝑨
𝑩
)
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝒙→𝟑
𝐥𝐨𝐠 𝟓(𝒙 𝟐
+𝟐𝒙 + 𝟏) − 𝐥𝐨𝐠 𝟓 𝒙 − 𝟓)
= 𝐥𝐨𝐠 𝟓 𝟑 𝟐
+ 𝟐 𝟑 + 𝟏 − 𝟑 − 𝟓
= 𝐥𝐨𝐠 𝟓 𝟗 + 𝟔 + 𝟏 + 𝟐
= 𝐥𝐨𝐠 𝟓 𝟏𝟖 = 𝟏. 𝟕𝟗𝟔
3. Power Rule
Example:
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
= 𝒍𝒊𝒎
𝒙→𝒂
𝒏 𝒍𝒐𝒈 𝒃 𝑨
Given 𝑨 = 𝒙 − 𝟓, 𝒏 = 𝟑, 𝒃 = 𝟐 𝒂𝒏𝒅 𝒙 = 𝟖,
find 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
𝒍𝒊𝒎
𝒙→𝟖
𝒍𝒐𝒈 𝟐 𝒙 − 𝟓 𝟑
= 𝒍𝒊𝒎
𝒙→𝟖
𝟑 𝒍𝒐𝒈 𝟐 𝒙 − 𝟓
= 𝟑 𝒍𝒐𝒈 𝟐 𝟖 − 𝟓
= 𝟑 𝒍𝒐𝒈 𝟐 𝟑
= 𝟑 𝟏. 𝟓𝟖𝟓 = 𝟒. 𝟕𝟓𝟓
Let’s Practice!!!
1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔,
𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ;
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝒙→𝟐
𝐥𝐨𝐠 𝟏𝟎(𝒙 𝟐
+ 𝟒𝒙 + 𝟒) + 𝐥𝐨𝐠 𝟏𝟎(𝒙 𝟐
− 𝟏𝟔)
= 𝐥𝐨𝐠 𝟏𝟎 𝟐 𝟐
+ 𝟒 𝟐 + 𝟒 + 𝟐 𝟐
− 𝟏𝟔
= 𝐥𝐨𝐠 𝟏𝟎 𝟒 + 𝟖 + 𝟒 + 𝟒 − 𝟏𝟔
= 𝐥𝐨𝐠 𝟏𝟎 𝟒 = 𝟎. 𝟔𝟎𝟐
1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔,
𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ;
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝒙→𝟐
𝐥𝐨𝐠 𝟏𝟎 𝒙 𝟐
+ 𝟒𝒙 + 𝟒 − 𝐥𝐨𝐠 𝟏𝟎(𝒙 𝟐
− 𝟏𝟔)
1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔,
𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ;
= 𝐥𝐨𝐠 𝟏𝟎 𝟐 𝟐
+ 𝟒 𝟐 + 𝟒 − 𝟐 𝟐
− 𝟏𝟔
= 𝐥𝐨𝐠 𝟏𝟎 𝟒 + 𝟖 + 𝟒 − 𝟒 + 𝟏𝟔
= 𝐥𝐨𝐠 𝟏𝟎 𝟐𝟖 = 𝟏. 𝟒𝟒𝟕
Solution:
𝒍𝒊𝒎
𝒙→𝟐
𝒍𝒐𝒈 𝟏𝟎 𝒙 𝟐
+ 𝟒𝒙 + 𝟒
𝟒
= 𝒍𝒊𝒎
𝒙→𝟐
𝟒 𝒍𝒐𝒈 𝟏𝟎 𝒙 𝟐
+ 𝟒𝒙 + 𝟒
= 𝟒 𝒍𝒐𝒈 𝟏𝟎 𝟐 𝟐
+ 𝟒 𝟐 + 𝟒
= 𝟒 𝒍𝒐𝒈 𝟏𝟎 𝟏𝟔
= 𝟒 𝟏. 𝟐𝟎𝟒 = 𝟒. 𝟖𝟏𝟔
1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔,
𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ;
Seatwork 3.2
1. Given 𝑨 = 𝟗𝒙 𝟐
+ 𝟑𝟎𝒙 + 𝟐𝟓, 𝑩 = 𝟒𝒙 𝟐
− 𝟏𝟔,
𝒙 = 𝟓, 𝒃 = 𝟑𝟓, 𝒏 = 𝟏𝟓 ;
2. Given 𝑨 = 𝒙 𝟐
− 𝟏𝟔𝒙 + 𝟔𝟒, 𝑩 = 𝒙 𝟐
− 𝟏,
𝒙 = 𝟑, 𝒃 = 𝟐𝟓, 𝒏 = 𝟐𝟓
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
2. Given 𝑨 = 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐 − 𝟑𝟔,
𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ;
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
ANSWERS
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝒙→𝟐
𝐥𝐨𝐠 𝟏𝟓(𝟐𝟓𝒙 𝟐
+ 𝟒𝟎𝒙 + 𝟏𝟔) + 𝐥𝐨𝐠 𝟏𝟓(𝒙 𝟐
− 𝟔𝟒)
= 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓 𝟐 𝟐
+ 𝟒𝟎 𝟐 + 𝟏𝟔 + 𝟐 𝟐
− 𝟔𝟒
= 𝐥𝐨𝐠 𝟏𝟓 𝟏𝟎𝟎 + 𝟖𝟎 + 𝟏𝟔 + 𝟒 − 𝟔𝟒
= 𝐥𝐨𝐠 𝟏𝟓 𝟏𝟑𝟔 = 𝟏. 𝟖𝟏𝟒
1. Given 𝑨 = 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔, 𝑩 = 𝒙 𝟐 − 𝟔𝟒,
𝒙 = 𝟐, 𝒃 = 𝟏𝟓, 𝒏 = 𝟓 ;
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝒙→𝟐
𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓𝒙 𝟐
+ 𝟒𝟎𝒙 + 𝟏𝟔 − 𝐥𝐨𝐠 𝟏𝟓(𝒙 𝟐
− 𝟔𝟒)
1. Given 𝑨 = 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔, 𝑩 = 𝒙 𝟐 − 𝟔𝟒,
𝒙 = 𝟐, 𝒃 = 𝟏𝟓, 𝒏 = 𝟓 ;
= 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓 𝟐 𝟐
+ 𝟒𝟎 𝟐 + 𝟏𝟔 − 𝟐 𝟐
− 𝟔𝟒
= 𝐥𝐨𝐠 𝟏𝟓 𝟏𝟎𝟎 + 𝟖𝟎 + 𝟏𝟔 − 𝟒 + 𝟔𝟒
= 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓𝟔 = 𝟐. 𝟎𝟒𝟖
Solution:
𝒍𝒊𝒎
𝒙→𝟐
𝒍𝒐𝒈 𝟏𝟓 𝟐𝟓𝒙 𝟐
+ 𝟒𝟎𝒙 + 𝟏𝟔
𝟓
= 𝒍𝒊𝒎
𝒙→𝟐
𝟓 𝒍𝒐𝒈 𝟏𝟓 𝟐𝟓𝒙 𝟐
+ 𝟒𝟎𝒙 + 𝟏𝟔
= 𝟓 𝒍𝒐𝒈 𝟏𝟓 𝟐𝟓 𝟐 𝟐
+ 𝟒𝟎 𝟐 + 𝟏𝟔
= 𝟓 𝒍𝒐𝒈 𝟏𝟓 𝟏𝟗𝟔
= 𝟓 𝟏. 𝟗𝟒𝟗 = 𝟗. 𝟕𝟒𝟓
1. Given 𝑨 = 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔, 𝑩 = 𝒙 𝟐 − 𝟔𝟒,
𝒙 = 𝟐, 𝒃 = 𝟏𝟓, 𝒏 = 𝟓 ;
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝒙→𝟏𝟐
𝐥𝐨𝐠 𝟑𝟔(𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎) + 𝐥𝐨𝐠 𝟑𝟔(𝒙 𝟐
− 𝟑𝟔)
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟐 𝟐
+ 𝟐𝟎 𝟏𝟐 − 𝟏𝟎𝟎 + 𝟏𝟐 𝟐
− 𝟑𝟔
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟒𝟒 + 𝟐𝟒𝟎 − 𝟏𝟎𝟎 + 𝟏𝟒𝟒 − 𝟑𝟔
= 𝐥𝐨𝐠 𝟑𝟔 𝟑𝟗𝟐 = 𝟏. 𝟔𝟔𝟔
2. Given 𝑨 = 𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐
− 𝟑𝟔,
𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ;
Solution:
𝒍𝒊𝒎
𝒙→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝒙→𝟏𝟐
𝐥𝐨𝐠 𝟑𝟔 𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎 − 𝐥𝐨𝐠 𝟑𝟔(𝒙 𝟐
− 𝟑𝟔)
2. Given 𝑨 = 𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐
− 𝟑𝟔,
𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ;
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟐 𝟐
+ 𝟐𝟎 𝟏𝟐 − 𝟏𝟎𝟎 − 𝟏𝟐 𝟐
− 𝟑𝟔
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟒𝟒 + 𝟐𝟒𝟎 − 𝟏𝟎𝟎 − 𝟏𝟒𝟒 + 𝟑𝟔
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟕𝟔 = 𝟏. 𝟒𝟒𝟑
Solution:
𝒍𝒊𝒎
𝒙→𝟏𝟐
𝒍𝒐𝒈 𝟑𝟔 𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎
𝟐𝟓
= 𝒍𝒊𝒎
𝒙→𝟏𝟐
𝟐𝟓 𝒍𝒐𝒈 𝟑𝟔 𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎
= 𝟐𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟏𝟐 𝟐
+ 𝟐𝟎 𝟏𝟐 − 𝟏𝟎𝟎
= 𝟐𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟐𝟖𝟒
= 𝟐𝟓 𝟏. 𝟓𝟕𝟔 = 𝟑𝟗. 𝟒𝟎𝟗
2. Given 𝑨 = 𝒙 𝟐
+ 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐
− 𝟑𝟔,
𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ;
Quiz 3.2
Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒,
𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖 ;
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
Given 𝑨 = 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐
− 𝟐𝒎 + 𝟐,
𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ;
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏,
𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ;
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
ANSWERS
Solution:
𝒍𝒊𝒎
𝐲→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝐲→𝟓
𝐥𝐨𝐠 𝟐𝟓(𝟑𝟔𝒚 𝟐
− 𝟗𝟔𝒚 + 𝟔𝟒) + 𝐥𝐨𝐠 𝟐𝟓(𝒚 𝟐
− 𝟔𝟒)
= 𝐥𝐨𝐠 𝟐𝟓 𝟑𝟔 𝟓 𝟐
− 𝟗𝟔 𝟓 + 𝟔𝟒 + 𝟓 𝟐
− 𝟔𝟒
= 𝐥𝐨𝐠 𝟐𝟓 𝟗𝟎𝟎 − 𝟒𝟖𝟎 + 𝟔𝟒 + 𝟐𝟓 − 𝟔𝟒
= 𝐥𝐨𝐠 𝟐𝟓 𝟒𝟒𝟓 = 𝟏. 𝟖𝟗𝟒
1. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒,
𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖
Solution:
= 𝟏. 𝟗𝟒𝟓
1. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒,
𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖
𝒍𝒊𝒎
𝐲→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝐲→𝟓
𝐥𝐨𝐠 𝟐𝟓 𝟑𝟔𝒚 𝟐
− 𝟗𝟔𝒚 + 𝟔𝟒 − 𝐥𝐨𝐠 𝟐𝟓(𝒚 𝟐
− 𝟔𝟒)
= 𝐥𝐨𝐠 𝟐𝟓 𝟑𝟔 𝟓 𝟐
− 𝟗𝟔 𝟓 + 𝟔𝟒 − 𝟓 𝟐
− 𝟔𝟒
= 𝐥𝐨𝐠 𝟐𝟓 𝟗𝟎𝟎 − 𝟒𝟖𝟎 + 𝟔𝟒 − 𝟐𝟓 + 𝟔𝟒
= 𝐥𝐨𝐠 𝟐𝟓 𝟓𝟐𝟑
Solution:
𝒍𝒊𝒎
𝒚→𝟓
𝒍𝒐𝒈 𝟐𝟓 𝟑𝟔𝒚 𝟐
− 𝟗𝟔𝒚 + 𝟔𝟒
𝟖
= 𝒍𝒊𝒎
𝒚→𝟓
𝟖 𝒍𝒐𝒈 𝟐𝟓 𝟑𝟔𝒚 𝟐
− 𝟗𝟔𝒚 + 𝟔𝟒
= 𝟖 𝒍𝒐𝒈 𝟐𝟓 𝟑𝟔 𝟓 𝟐
− 𝟗𝟔 𝟓 + +𝟔𝟒
= 𝟖 𝒍𝒐𝒈 𝟐𝟓 𝟒𝟖𝟒
= 𝟖 𝟏. 𝟗𝟐𝟏 = 𝟏𝟓. 𝟑𝟔𝟓
1. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒,
𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖
Solution:
𝒍𝒊𝒎
𝐦→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝐦→𝟔
𝐥𝐨𝐠 𝟑𝟔(𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗) + 𝐥𝐨𝐠 𝟑𝟔(𝒎 𝟐
− 𝟐𝒎 + 𝟐)
= 𝐥𝐨𝐠 𝟑𝟔 𝟒𝟓 𝟔 𝟐
− 𝟑𝟔 𝟔 − 𝟗 + 𝟔 𝟐
− 𝟐 𝟔 + 𝟐
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟔𝟐𝟎 − 𝟐𝟏𝟔 − 𝟗 + 𝟑𝟔 − 𝟏𝟐 + 𝟐
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟒𝟐𝟏 = 𝟐. 𝟎𝟐𝟔
2. Given 𝑨 = 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐
− 𝟐𝒎 + 𝟐,
𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ;
Solution:
= 𝟐. 𝟎𝟏𝟓
2. Given 𝑨 = 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐
− 𝟐𝒎 + 𝟐,
𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ;
𝒍𝒊𝒎
𝐦→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝐦→𝟔
𝐥𝐨𝐠 𝟑𝟔 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗 − 𝐥𝐨𝐠 𝟑𝟔(𝒎 𝟐
− 𝟐𝒎 + 𝟐)
= 𝐥𝐨𝐠 𝟑𝟔 𝟒𝟓 𝟔 𝟐
− 𝟑𝟔 𝟔 − 𝟗 − 𝟔 𝟐
− 𝟐 𝟔 + 𝟐
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟔𝟐𝟎 − 𝟐𝟏𝟔 − 𝟗 − 𝟑𝟔 + 𝟏𝟐 − 𝟐
= 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟑𝟔𝟗
Solution:
𝒍𝒊𝒎
𝒎→𝟔
𝒍𝒐𝒈 𝟑𝟔 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗
𝟓
= 𝒍𝒊𝒎
𝒙→𝟐
𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗
= 𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟒𝟓 𝟔 𝟐
− 𝟑𝟔 𝟔 − 𝟗
= 𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟏𝟑𝟗𝟓
= 𝟓 𝟐. 𝟎𝟐𝟏 = 𝟏𝟎. 𝟏𝟎𝟑
2. Given 𝑨 = 𝟒𝟓𝒎 𝟐
− 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐
− 𝟐𝒎 + 𝟐,
𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ;
Solution:
𝒍𝒊𝒎
𝐲→𝒂
𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦
𝒚→𝟑
𝐥𝐨𝐠 𝟏𝟖(𝟏𝟔𝒚 𝟐
− 𝟒𝟔𝒚 + 𝟑𝟔) + 𝐥𝐨𝐠 𝟏𝟖(𝒚 𝟑
− 𝟏)
= 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟔 𝟑 𝟐
− 𝟒𝟖 𝟑 + 𝟑𝟔 + 𝟑 𝟑
− 𝟏
= 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟒𝟒 − 𝟏𝟒𝟒 + 𝟑𝟔 + 𝟐𝟕 − 𝟏
= 𝐥𝐨𝐠 𝟏𝟖 𝟔𝟐 = 𝟏. 𝟒𝟐𝟖
3. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏,
𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ;
Solution:
= 𝟎. 𝟕𝟗𝟕
3. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏,
𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ;
𝒍𝒊𝒎
𝐲→𝒂
𝐥𝐨𝐠 𝒃
𝑨
𝑩
= 𝐥𝐢𝐦
𝒚→𝟑
𝐥𝐨𝐠 𝟏𝟖 𝟏𝟔𝒚 𝟐
− 𝟒𝟔𝒚 + 𝟑𝟔 − 𝐥𝐨𝐠 𝟏𝟖(𝒚 𝟑
− 𝟏)
= 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟔 𝟑 𝟐
− 𝟒𝟖 𝟑 + 𝟑𝟔 − 𝟑 𝟐
− 𝟏
= 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟒𝟒 − 𝟏𝟒𝟒 + 𝟑𝟔 − 𝟐𝟕 + 𝟏
= 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟎
Solution:
𝒍𝒊𝒎
𝒚→𝟑
𝒍𝒐𝒈 𝟏𝟖 𝟏𝟔𝒚 𝟐
− 𝟒𝟖𝒚 + 𝟑𝟔
𝟔
= 𝒍𝒊𝒎
𝒚→𝟑
𝟔 𝒍𝒐𝒈 𝟏𝟖 𝟏𝟔𝒚 𝟐
− 𝟒𝟖𝒚 + 𝟑𝟔
= 𝟔 𝒍𝒐𝒈 𝟏𝟖 𝟏𝟔 𝟑 𝟐
− 𝟒𝟖 𝟑 + 𝟑𝟔
= 𝟔 𝒍𝒐𝒈 𝟏𝟖 𝟑𝟔
= 𝟔 𝟏. 𝟐𝟒𝟎 = 𝟕. 𝟒𝟑𝟗
3. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏,
𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ;
Pointers to Review
Limits of Non – Algebraic Functions
Limit of an Exponential Function
Limit of Logarithmic Function
Given 𝑨 = 𝒚 𝟐 + 𝟔𝒚 + 𝟗, 𝑩 = 𝟒𝒚 𝟐 − 𝟏,
𝒚 = 𝟖, 𝒃 = 𝟏𝟓, 𝒏 = 𝟔 ;
Find the following:
a. 𝒍𝒊𝒎
𝒚→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒚→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒚→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
Given 𝑨 = 𝟑𝒙 𝟐 + 𝟗𝒙 + 𝟗, 𝑩 = 𝒙 − 𝟓,
𝒙 = 𝟐, 𝒃 = 𝟏𝟑, 𝒏 = 𝟏𝟎 ;
Find the following:
a. 𝒍𝒊𝒎
𝒙→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒙→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏
Given 𝑨 = 𝟔𝒛 𝟐 − 𝟒𝒛 + 𝟓, 𝑩 = 𝟐𝒛 𝟐 + 𝟒𝒛 − 𝟑,
𝒛 = 𝟑, 𝒃 = 𝟖, 𝒏 = 𝟐;
Find the following:
a. 𝒍𝒊𝒎
𝒛→𝒂
(𝒍𝒐𝒈 𝒃 𝑨𝑩)
b. 𝒍𝒊𝒎
𝒛→𝒂
𝒍𝒐𝒈 𝒃(
𝑨
𝑩
)
c. 𝒍𝒊𝒎
𝒛→𝒂
𝒍𝒐𝒈 𝒃 𝑨 𝒏

More Related Content

What's hot

Representing Real-Life Situations Using Rational Functions.pptx
Representing Real-Life Situations Using Rational Functions.pptxRepresenting Real-Life Situations Using Rational Functions.pptx
Representing Real-Life Situations Using Rational Functions.pptx
EdelmarBenosa3
 
Chapter 2: Rational Function
Chapter 2: Rational FunctionChapter 2: Rational Function
Chapter 2: Rational Function
Jovic Rullepa
 
Statistics and probability lesson 1
Statistics and probability lesson 1Statistics and probability lesson 1
Statistics and probability lesson 1
MARIA CHRISTITA POLINAG
 
Lesson 4: Calcuating Limits (slides)
Lesson 4: Calcuating Limits (slides)Lesson 4: Calcuating Limits (slides)
Lesson 4: Calcuating Limits (slides)
Matthew Leingang
 
Annuities
AnnuitiesAnnuities
Statistics and probability lesson 4
Statistics and probability lesson 4Statistics and probability lesson 4
Statistics and probability lesson 4
MARIA CHRISTITA POLINAG
 
Gen. math g11 introduction to functions
Gen. math g11  introduction to functionsGen. math g11  introduction to functions
Gen. math g11 introduction to functions
liza magalso
 
General Mathematics - Rational Functions
General Mathematics - Rational FunctionsGeneral Mathematics - Rational Functions
General Mathematics - Rational Functions
Juan Miguel Palero
 
Math Powerpoint
Math PowerpointMath Powerpoint
Math Powerpoint
Minako De Leon
 
Math quiz bee
Math quiz beeMath quiz bee
Math quiz beeNo L's
 
Polynomial functions
Polynomial functionsPolynomial functions
Polynomial functionsdedearfandy
 
Rational function representation
Rational function representationRational function representation
Rational function representation
rey castro
 
Probability Distribution (Discrete Random Variable)
Probability Distribution (Discrete Random Variable)Probability Distribution (Discrete Random Variable)
Probability Distribution (Discrete Random Variable)
Cess011697
 
Pagbasa at Pagsulat
Pagbasa at PagsulatPagbasa at Pagsulat
Pagbasa at Pagsulat
menchu lacsamana
 
Basic concept of annuity
Basic concept of annuityBasic concept of annuity
Basic concept of annuity
rey castro
 
General Mathematics - Composition of Functions
General Mathematics - Composition of FunctionsGeneral Mathematics - Composition of Functions
General Mathematics - Composition of Functions
Juan Miguel Palero
 
Circular Functions
Circular FunctionsCircular Functions
Circular Functions
Jonalyn Asi
 
One to-one function (MATH 11)
One to-one function (MATH 11)One to-one function (MATH 11)
One to-one function (MATH 11)
majoydrew
 
Compound Interest
Compound InterestCompound Interest
Compound Interest
mxxaxx
 
Conic Section
Conic SectionConic Section
Conic Section
Roqui Gonzaga
 

What's hot (20)

Representing Real-Life Situations Using Rational Functions.pptx
Representing Real-Life Situations Using Rational Functions.pptxRepresenting Real-Life Situations Using Rational Functions.pptx
Representing Real-Life Situations Using Rational Functions.pptx
 
Chapter 2: Rational Function
Chapter 2: Rational FunctionChapter 2: Rational Function
Chapter 2: Rational Function
 
Statistics and probability lesson 1
Statistics and probability lesson 1Statistics and probability lesson 1
Statistics and probability lesson 1
 
Lesson 4: Calcuating Limits (slides)
Lesson 4: Calcuating Limits (slides)Lesson 4: Calcuating Limits (slides)
Lesson 4: Calcuating Limits (slides)
 
Annuities
AnnuitiesAnnuities
Annuities
 
Statistics and probability lesson 4
Statistics and probability lesson 4Statistics and probability lesson 4
Statistics and probability lesson 4
 
Gen. math g11 introduction to functions
Gen. math g11  introduction to functionsGen. math g11  introduction to functions
Gen. math g11 introduction to functions
 
General Mathematics - Rational Functions
General Mathematics - Rational FunctionsGeneral Mathematics - Rational Functions
General Mathematics - Rational Functions
 
Math Powerpoint
Math PowerpointMath Powerpoint
Math Powerpoint
 
Math quiz bee
Math quiz beeMath quiz bee
Math quiz bee
 
Polynomial functions
Polynomial functionsPolynomial functions
Polynomial functions
 
Rational function representation
Rational function representationRational function representation
Rational function representation
 
Probability Distribution (Discrete Random Variable)
Probability Distribution (Discrete Random Variable)Probability Distribution (Discrete Random Variable)
Probability Distribution (Discrete Random Variable)
 
Pagbasa at Pagsulat
Pagbasa at PagsulatPagbasa at Pagsulat
Pagbasa at Pagsulat
 
Basic concept of annuity
Basic concept of annuityBasic concept of annuity
Basic concept of annuity
 
General Mathematics - Composition of Functions
General Mathematics - Composition of FunctionsGeneral Mathematics - Composition of Functions
General Mathematics - Composition of Functions
 
Circular Functions
Circular FunctionsCircular Functions
Circular Functions
 
One to-one function (MATH 11)
One to-one function (MATH 11)One to-one function (MATH 11)
One to-one function (MATH 11)
 
Compound Interest
Compound InterestCompound Interest
Compound Interest
 
Conic Section
Conic SectionConic Section
Conic Section
 

Similar to Basic Calculus Lesson 3

Math 8 Lesson 3 - 2nd Quarter
Math 8 Lesson 3 - 2nd QuarterMath 8 Lesson 3 - 2nd Quarter
Math 8 Lesson 3 - 2nd Quarter
alicelagajino
 
6 DOF Rigid Body Equation of Motion 1
6 DOF Rigid Body Equation of Motion 16 DOF Rigid Body Equation of Motion 1
6 DOF Rigid Body Equation of Motion 1
kouhei1970
 
6 DOF Rigid Body Equation of Motion 3
6 DOF Rigid Body Equation of Motion 36 DOF Rigid Body Equation of Motion 3
6 DOF Rigid Body Equation of Motion 3
kouhei1970
 
Thai new testament-world_bible
Thai new testament-world_bibleThai new testament-world_bible
Thai new testament-world_bibleWorldBibles
 
ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2
ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2
ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2
san aye
 
Thai new testament-world_bible
Thai new testament-world_bibleThai new testament-world_bible
Thai new testament-world_bibleWorldBibles
 
Math20160415 epsilondelta
Math20160415 epsilondeltaMath20160415 epsilondelta
Math20160415 epsilondelta
Atsushi Kadotani
 
หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์
หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์
หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์หนุ่มน้อย ดาร์จีลิ่ง
 

Similar to Basic Calculus Lesson 3 (9)

Math 8 Lesson 3 - 2nd Quarter
Math 8 Lesson 3 - 2nd QuarterMath 8 Lesson 3 - 2nd Quarter
Math 8 Lesson 3 - 2nd Quarter
 
6 DOF Rigid Body Equation of Motion 1
6 DOF Rigid Body Equation of Motion 16 DOF Rigid Body Equation of Motion 1
6 DOF Rigid Body Equation of Motion 1
 
6 DOF Rigid Body Equation of Motion 3
6 DOF Rigid Body Equation of Motion 36 DOF Rigid Body Equation of Motion 3
6 DOF Rigid Body Equation of Motion 3
 
Thai new testament-world_bible
Thai new testament-world_bibleThai new testament-world_bible
Thai new testament-world_bible
 
ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2
ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2
ေဒါက္တာလွေဘ တရုတ္ႏိုင္ငံသြားေန႕စဥ္မွတ္စာတမ္း 2
 
Thai new testament-world_bible
Thai new testament-world_bibleThai new testament-world_bible
Thai new testament-world_bible
 
1 14
1 141 14
1 14
 
Math20160415 epsilondelta
Math20160415 epsilondeltaMath20160415 epsilondelta
Math20160415 epsilondelta
 
หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์
หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์
หนังสือพิมพ์งานพระราชทานเพลิงศพ สมเด็จพระพุฒโฆษาจารย์
 

More from alicelagajino

TLE 7 Lesson 5 - 2nd Quarter
TLE 7 Lesson 5 - 2nd QuarterTLE 7 Lesson 5 - 2nd Quarter
TLE 7 Lesson 5 - 2nd Quarter
alicelagajino
 
TLE 7 Lesson 4 - 2nd Quarter
TLE 7 Lesson 4 - 2nd QuarterTLE 7 Lesson 4 - 2nd Quarter
TLE 7 Lesson 4 - 2nd Quarter
alicelagajino
 
TLE 7 Lesson 3 - 2nd Quarter
TLE 7 Lesson 3 - 2nd QuarterTLE 7 Lesson 3 - 2nd Quarter
TLE 7 Lesson 3 - 2nd Quarter
alicelagajino
 
TLE 7 Lesson 2 - 2nd Quarter
TLE 7 Lesson 2 - 2nd QuarterTLE 7 Lesson 2 - 2nd Quarter
TLE 7 Lesson 2 - 2nd Quarter
alicelagajino
 
TLE 7 Lesson 1 - 2nd Quarter
TLE 7 Lesson 1 - 2nd QuarterTLE 7 Lesson 1 - 2nd Quarter
TLE 7 Lesson 1 - 2nd Quarter
alicelagajino
 
TLE 7 Lesson 6 - 1st Quarter
TLE 7 Lesson 6 - 1st QuarterTLE 7 Lesson 6 - 1st Quarter
TLE 7 Lesson 6 - 1st Quarter
alicelagajino
 
TLE 7 Lesson 5 - 1st Quarter
TLE 7 Lesson 5 - 1st QuarterTLE 7 Lesson 5 - 1st Quarter
TLE 7 Lesson 5 - 1st Quarter
alicelagajino
 
TLE 7 Lesson 4 - 1st Quarter
TLE 7 Lesson 4 - 1st QuarterTLE 7 Lesson 4 - 1st Quarter
TLE 7 Lesson 4 - 1st Quarter
alicelagajino
 
TLE 7 Lesson 3 - 1st Quarter
TLE 7 Lesson 3 - 1st QuarterTLE 7 Lesson 3 - 1st Quarter
TLE 7 Lesson 3 - 1st Quarter
alicelagajino
 
TLE 7 Lesson 2 - 1st Quarter
TLE 7 Lesson 2 - 1st QuarterTLE 7 Lesson 2 - 1st Quarter
TLE 7 Lesson 2 - 1st Quarter
alicelagajino
 
TLE 7 Lesson 1 - 1st Quarter
TLE 7 Lesson 1 - 1st QuarterTLE 7 Lesson 1 - 1st Quarter
TLE 7 Lesson 1 - 1st Quarter
alicelagajino
 
Math 8 Lesson 2 - 2nd Quarter
Math 8 Lesson 2 - 2nd QuarterMath 8 Lesson 2 - 2nd Quarter
Math 8 Lesson 2 - 2nd Quarter
alicelagajino
 
Math 8 Lesson 1 - 2nd Quarter
Math 8 Lesson 1 - 2nd QuarterMath 8 Lesson 1 - 2nd Quarter
Math 8 Lesson 1 - 2nd Quarter
alicelagajino
 
Math 8 Lesson 3 - 1st Quarter
Math 8 Lesson 3 - 1st QuarterMath 8 Lesson 3 - 1st Quarter
Math 8 Lesson 3 - 1st Quarter
alicelagajino
 
Math 8 Lesson 2 - 1st Quarter
Math 8 Lesson 2 - 1st QuarterMath 8 Lesson 2 - 1st Quarter
Math 8 Lesson 2 - 1st Quarter
alicelagajino
 
Math 8 Lesson 1 - 1st Quarter
Math 8 Lesson 1 - 1st QuarterMath 8 Lesson 1 - 1st Quarter
Math 8 Lesson 1 - 1st Quarter
alicelagajino
 
General Math Lesson 1
General Math Lesson 1General Math Lesson 1
General Math Lesson 1
alicelagajino
 
Empowerment Technology Lesson 5
Empowerment Technology Lesson 5Empowerment Technology Lesson 5
Empowerment Technology Lesson 5
alicelagajino
 
Empowerment Technology Lesson 4
Empowerment Technology Lesson 4Empowerment Technology Lesson 4
Empowerment Technology Lesson 4
alicelagajino
 
Empowerment Technology Lesson 3
Empowerment Technology Lesson 3Empowerment Technology Lesson 3
Empowerment Technology Lesson 3
alicelagajino
 

More from alicelagajino (20)

TLE 7 Lesson 5 - 2nd Quarter
TLE 7 Lesson 5 - 2nd QuarterTLE 7 Lesson 5 - 2nd Quarter
TLE 7 Lesson 5 - 2nd Quarter
 
TLE 7 Lesson 4 - 2nd Quarter
TLE 7 Lesson 4 - 2nd QuarterTLE 7 Lesson 4 - 2nd Quarter
TLE 7 Lesson 4 - 2nd Quarter
 
TLE 7 Lesson 3 - 2nd Quarter
TLE 7 Lesson 3 - 2nd QuarterTLE 7 Lesson 3 - 2nd Quarter
TLE 7 Lesson 3 - 2nd Quarter
 
TLE 7 Lesson 2 - 2nd Quarter
TLE 7 Lesson 2 - 2nd QuarterTLE 7 Lesson 2 - 2nd Quarter
TLE 7 Lesson 2 - 2nd Quarter
 
TLE 7 Lesson 1 - 2nd Quarter
TLE 7 Lesson 1 - 2nd QuarterTLE 7 Lesson 1 - 2nd Quarter
TLE 7 Lesson 1 - 2nd Quarter
 
TLE 7 Lesson 6 - 1st Quarter
TLE 7 Lesson 6 - 1st QuarterTLE 7 Lesson 6 - 1st Quarter
TLE 7 Lesson 6 - 1st Quarter
 
TLE 7 Lesson 5 - 1st Quarter
TLE 7 Lesson 5 - 1st QuarterTLE 7 Lesson 5 - 1st Quarter
TLE 7 Lesson 5 - 1st Quarter
 
TLE 7 Lesson 4 - 1st Quarter
TLE 7 Lesson 4 - 1st QuarterTLE 7 Lesson 4 - 1st Quarter
TLE 7 Lesson 4 - 1st Quarter
 
TLE 7 Lesson 3 - 1st Quarter
TLE 7 Lesson 3 - 1st QuarterTLE 7 Lesson 3 - 1st Quarter
TLE 7 Lesson 3 - 1st Quarter
 
TLE 7 Lesson 2 - 1st Quarter
TLE 7 Lesson 2 - 1st QuarterTLE 7 Lesson 2 - 1st Quarter
TLE 7 Lesson 2 - 1st Quarter
 
TLE 7 Lesson 1 - 1st Quarter
TLE 7 Lesson 1 - 1st QuarterTLE 7 Lesson 1 - 1st Quarter
TLE 7 Lesson 1 - 1st Quarter
 
Math 8 Lesson 2 - 2nd Quarter
Math 8 Lesson 2 - 2nd QuarterMath 8 Lesson 2 - 2nd Quarter
Math 8 Lesson 2 - 2nd Quarter
 
Math 8 Lesson 1 - 2nd Quarter
Math 8 Lesson 1 - 2nd QuarterMath 8 Lesson 1 - 2nd Quarter
Math 8 Lesson 1 - 2nd Quarter
 
Math 8 Lesson 3 - 1st Quarter
Math 8 Lesson 3 - 1st QuarterMath 8 Lesson 3 - 1st Quarter
Math 8 Lesson 3 - 1st Quarter
 
Math 8 Lesson 2 - 1st Quarter
Math 8 Lesson 2 - 1st QuarterMath 8 Lesson 2 - 1st Quarter
Math 8 Lesson 2 - 1st Quarter
 
Math 8 Lesson 1 - 1st Quarter
Math 8 Lesson 1 - 1st QuarterMath 8 Lesson 1 - 1st Quarter
Math 8 Lesson 1 - 1st Quarter
 
General Math Lesson 1
General Math Lesson 1General Math Lesson 1
General Math Lesson 1
 
Empowerment Technology Lesson 5
Empowerment Technology Lesson 5Empowerment Technology Lesson 5
Empowerment Technology Lesson 5
 
Empowerment Technology Lesson 4
Empowerment Technology Lesson 4Empowerment Technology Lesson 4
Empowerment Technology Lesson 4
 
Empowerment Technology Lesson 3
Empowerment Technology Lesson 3Empowerment Technology Lesson 3
Empowerment Technology Lesson 3
 

Basic Calculus Lesson 3

  • 1.
  • 2. Limit of an Exponential Function Let a and b be real numbers, where b > 0 and b ≠ 1. Then, 𝐥𝐢𝐦 𝒙→𝒂 𝒃 𝒙 = 𝒃 𝒂 .
  • 3. Properties There are four basic properties in limits, which are used as formulas in evaluating the limits of exponential functions. 1.Power Rule 2.Constant Base Power Rule 3.Constant Exponent Power Rule 4.Radical Power Rule
  • 4. 1. Power Rule 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝒈 𝒙 = 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝒍𝒊𝒎 𝒙→𝒂 𝒈 𝒙 It is a property of power rule, used to find the limit of an exponential function whose base and exponent are in a function form.
  • 5. 1. Power Rule 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝒈 𝒙 = 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝒍𝒊𝒎 𝒙→𝒂 𝒈 𝒙 Example: Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟒, 𝒈 𝒙 = 𝟐𝒙 ; 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 𝒈 𝒙 Solution: 𝐥𝐢𝐦 𝒙→3 𝒇 𝒙 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→3 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟑 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→3 𝒇 𝒙2 − 4 𝐥𝐢𝐦 𝒙→3 𝒈 2𝒙 = 𝟑 𝟐 − 𝟒 𝟐 𝟑 = 𝟗 − 𝟒 𝟔 = 𝟓 𝟔 = 𝟏𝟓, 𝟔𝟐𝟓
  • 6. 2. Constant Base Power Rule 𝒍𝒊𝒎 𝒙→𝒂 𝒃 𝒇 𝒙 = 𝒍𝒊𝒎 𝒙→𝒂 𝒃 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 The limit of an exponential function is equal to the limit of the exponent with same base. It is called the limit rule of an exponential function.
  • 7. 2. Constant Base Power Rule 𝒍𝒊𝒎 𝒙→𝒂 𝒃 𝒇 𝒙 = 𝒍𝒊𝒎 𝒙→𝒂 𝒃 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 Example: Solution: Evaluate 𝑏 = 12, 𝑓 𝑥 = 𝑥 + 3; 𝐥𝐢𝐦 𝒙→𝟏 𝒃 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟏 𝒃 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟏 𝟏𝟐 𝒙+𝟑 = 𝒍𝒊𝒎 𝒙→𝟏 𝟏𝟐 𝒍𝒊𝒎 𝒙→𝟏 𝒙+𝟑 = 𝟏𝟐 𝒙+𝟑 = 𝟏𝟐 𝟏+𝟑 = 𝟏𝟐 𝟒 = 𝟐𝟎, 𝟕𝟑𝟔
  • 8. 3. Constant Exponent Power Rule 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝒏 Example: Solution: Evaluate 𝑓 𝑥 = 16𝑥2 − 64, 𝑛 = 2; 𝐥𝐢𝐦 𝒙→𝟏 𝒇 𝒙 𝒏 𝒍𝒊𝒎 𝒙→𝟏 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→𝟏 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→𝟏 𝟏𝟔𝒙 𝟐 − 𝟔𝟒 𝟐 = 𝟏𝟔 𝟏 𝟐 − 𝟔𝟒 𝟐 = 𝟏𝟔 − 𝟔𝟒 𝟐 = −𝟒𝟖 𝟐 = 𝟐, 𝟑𝟎𝟒
  • 9. 4. Radical Power Rule 𝒍𝒊𝒎 𝒙→𝒂 𝒏 )𝒇(𝒙 = 𝒏 𝒍𝒊𝒎 𝒙→𝒂 )𝒇(𝒙 Example: Solution: Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟔𝒙 + 𝟗, 𝒏 = 𝟐; 𝒍𝒊𝒎 𝒙→𝟔 𝒏 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟔 𝒏 )𝒇(𝒙 = 𝒏 𝒍𝒊𝒎 𝒙→𝟔 )𝒇(𝒙 = 𝐥𝐢𝐦 𝒙→6 𝒙2 − 6𝒙 + 9 = 𝒙 − 3 2 = 𝒙 − 3 = 6 − 3 = 𝟑
  • 10. Let’s Practice!!! 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔, 𝒏 = 𝟐; a. 𝒍𝒊𝒎 𝒙→−𝟑 𝒏 𝒇(𝒙) b. 𝒍𝒊𝒎 𝒙→𝟒 𝒇 𝒙 𝒏 2. Evaluate 𝒇 𝒙 = 𝒙 𝟑 − 𝟐𝟕, 𝒈 𝒙 = 𝟐𝒙 + 𝟏 a. 𝒍𝒊𝒎 𝒙→𝟐 𝒇 𝒙 𝒈 𝒙 3. Evaluate 𝑏 = 𝟑, 𝑓 𝑥 = 𝒙 − 𝟒 a. 𝐥𝐢𝐦 𝒙→𝟖 𝒃 𝒇(𝒙)
  • 11. 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔, 𝒏 = 𝟐; a. 𝒍𝒊𝒎 𝒙→−𝟑 𝒏 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→−𝟑 𝒏 )𝒇(𝒙 = 𝒏 𝒍𝒊𝒎 𝒙→−𝟑 )𝒇(𝒙 = 𝐥𝐢𝐦 𝒙→−𝟑 𝟒𝒙2 − 𝟐𝟒𝒙 + 𝟑𝟔 = 𝟐𝒙 − 𝟔 2 = 𝟐𝒙 − 𝟔 = 𝟐 −𝟑 − 𝟔 = −𝟏𝟐 = −𝟔 − 𝟔
  • 12. 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔, 𝒏 = 𝟐; b. 𝒍𝒊𝒎 𝒙→𝟒 𝒇 𝒙 𝒏 𝒍𝒊𝒎 𝒙→𝟒 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→𝟒 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→𝟒 𝟒𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔 𝟐 = 𝟒 𝟒 𝟐 − 𝟐𝟒 𝟒 + 𝟑𝟔 𝟐 = 𝟒 𝟏𝟔 − 𝟗𝟔 + 𝟑𝟔 𝟐 = 𝟒 𝟐 = 𝟏𝟔 = 𝟔𝟒 − 𝟗𝟔 + 𝟑𝟔 𝟐
  • 13. 2. Evaluate 𝒇 𝒙 = 𝒙 𝟑 − 𝟐𝟕, 𝒈 𝒙 = 𝟐𝒙 + 𝟏 a. 𝒍𝒊𝒎 𝒙→𝟐 𝒇 𝒙 𝒈 𝒙 𝐥𝐢𝐦 𝒙→𝟐 𝒇 𝒙 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→2 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟐 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→𝟐 𝒙 𝟑 − 𝟐𝟕 𝐥𝐢𝐦 𝒙→𝟐 𝟐𝐱+𝟏 = 𝟐 𝟑 − 𝟐𝟕 𝟐 𝟐 +𝟏 = 𝟖 − 𝟐𝟕 𝟓 = −𝟏𝟗 𝟓 = −𝟐, 𝟒𝟕𝟔, 𝟎𝟗𝟗
  • 14. 3. Evaluate 𝑏 = 𝟑, 𝑓 𝑥 = 𝒙 − 𝟒 a. 𝐥𝐢𝐦 𝒙→𝟖 𝒃 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟖 𝒃 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟖 𝟑 𝒙−𝟒 = 𝒍𝒊𝒎 𝒙→𝟖 𝟑 𝒍𝒊𝒎 𝒙→𝟖 𝒙−𝟒 = 𝟑 𝒙−𝟒 = 𝟑 𝟖−𝟒 = 𝟑 𝟒 = 𝟖𝟏
  • 16. 1. Evaluate 𝒇 𝒙 = 𝒙 𝟐 + 𝟏𝟔𝒙 + 𝟖, 𝒏 = 𝟑; a. 𝒍𝒊𝒎 𝒙→𝟒 𝒏 𝒇(𝒙) b. 𝒍𝒊𝒎 𝒙→−𝟐 𝒇 𝒙 𝒏 2. Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟔𝟒, 𝒈 𝒙 = 𝒙 − 𝟓 a. 𝒍𝒊𝒎 𝒙→𝟗 𝒇 𝒙 𝒈 𝒙 b. 𝒍𝒊𝒎 𝒙→𝟗 𝒈 𝒙 𝒇 𝒙 3. Evaluate 𝑏 = 𝟏𝟓, 𝑓 𝑥 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏 a. 𝐥𝐢𝐦 𝒙→𝟐 𝒃 𝒇(𝒙) b. 𝐥𝐢𝐦 𝒙→−𝟑 𝒃 𝒇(𝒙)
  • 18. 1. Evaluate 𝒇 𝒙 = 𝒙 𝟐 + 𝟏𝟔𝒙 + 𝟖, 𝒏 = 𝟑; a. 𝒍𝒊𝒎 𝒙→𝟒 𝒏 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟒 𝒏 )𝒇(𝒙 = 𝒏 𝒍𝒊𝒎 𝒙→𝟒 )𝒇(𝒙 = 𝟑 𝒍𝒊𝒎 𝒙→𝟒 𝒙 𝟐 + 𝟏𝟔𝒙 + 𝟖 = 3 𝟒 𝟐 + 𝟏𝟔 𝟒 + 𝟖 = 3 𝟏𝟔 + 𝟔𝟒 + 𝟖 = 𝟒. 𝟒𝟓 = 3 𝟖𝟖
  • 19. 1. Evaluate 𝒇 𝒙 = 𝒙 𝟐 + 𝟏𝟔𝒙 + 𝟖, 𝒏 = 𝟑; b. 𝒍𝒊𝒎 𝒙→−𝟐 𝒇 𝒙 𝒏 𝒍𝒊𝒎 𝒙→−𝟐 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→−𝟐 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→−𝟐 𝒙 𝟐 + 𝟏𝟔𝒙 + 𝟖 𝟑 = −𝟐 𝟐 + 𝟏𝟔 −𝟐 + 𝟖 𝟑 = 𝟏𝟔 − 𝟑𝟐 + 𝟖 𝟑 = −𝟖 𝟑 = −𝟓𝟏𝟐
  • 20. 2. Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟔𝟒, 𝒈 𝒙 = 𝒙 − 𝟓 a. 𝒍𝒊𝒎 𝒙→𝟗 𝒇 𝒙 𝒈 𝒙 𝐥𝐢𝐦 𝒙→𝟗 𝒇 𝒙 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→9 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟗 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→𝟗 𝒙 𝟐 − 𝟔𝟒 𝐥𝐢𝐦 𝒙→𝟗 𝐱−𝟓 = 𝟗 𝟐 − 𝟔𝟒 𝟗−𝟓 = 𝟖𝟏 − 𝟔𝟒 𝟒 = 𝟏𝟕 𝟒 = 𝟖𝟑, 𝟓𝟐𝟏
  • 21. 2. Evaluate 𝒇 𝒙 = 𝒙 𝟐 − 𝟔𝟒, 𝒈 𝒙 = 𝒙 − 𝟓 b. 𝒍𝒊𝒎 𝒙→𝟗 𝒈 𝒙 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟗 𝒈 𝒙 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→9 𝒈 𝒙 𝐥𝐢𝐦 𝒙→𝟗 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟗 𝒙 − 𝟓 𝐥𝐢𝐦 𝒙→𝟗 𝐱 𝟐−𝟔𝟒 = (𝟗 − 𝟓) 𝟗 𝟐−𝟔𝟒 = (𝟒) 𝟖𝟏−𝟔𝟒 = 𝟒 𝟏𝟕 = 𝟏𝟕, 𝟏𝟕𝟗, 𝟖𝟔𝟗, 𝟏𝟖𝟒
  • 22. 3. Evaluate 𝑏 = 𝟏𝟓, 𝑓 𝑥 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏 a. 𝐥𝐢𝐦 𝒙→𝟐 𝒃 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟐 𝒃 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟐 𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏 = 𝒍𝒊𝒎 𝒙→𝟐 𝟏𝟓 𝒍𝒊𝒎 𝒙→𝟐 𝒙 𝟐+𝟐𝒙+𝟏 = 𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏 = 𝟏𝟓 𝟐 𝟐+𝟐 𝟐 +𝟏 = 𝟏𝟓 𝟗 = 𝟑𝟖, 𝟒𝟒𝟑, 𝟑𝟓𝟗, 𝟑𝟕𝟓
  • 23. 3. Evaluate 𝑏 = 𝟏𝟓, 𝑓 𝑥 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏 b. 𝐥𝐢𝐦 𝒙→−𝟑 𝒃 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→−𝟑 𝒃 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→−𝟑 𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏 = 𝒍𝒊𝒎 𝒙→−𝟑 𝟏𝟓 𝒍𝒊𝒎 𝒙→−𝟑 𝒙 𝟐+𝟐𝒙+𝟏 = 𝟏𝟓 𝒙 𝟐+𝟐𝒙+𝟏 = 𝟏𝟓 −𝟑 𝟐+𝟐 −𝟑 +𝟏 = 𝟏𝟓 𝟒 = 𝟓𝟎, 𝟔𝟐𝟓
  • 25. Evaluate 𝒇 𝒙 = 𝟏𝟔𝒙 𝟐 − 𝟖𝒙 + 𝟏, 𝒈 𝒙 = 𝟔𝟒𝒙 𝟐 + 𝟒𝟖𝒙 + 𝟗, 𝒏 = 𝟐; a. 𝒍𝒊𝒎 𝒙→𝟓 𝒏 𝒇(𝒙) b. 𝒍𝒊𝒎 𝒙→−𝟓 𝒇 𝒙 𝒏 c. 𝒍𝒊𝒎 𝒙→𝟖 𝒏 𝒈(𝒙) d. 𝒍𝒊𝒎 𝒙→−𝟖 𝒈 𝒙 𝒏
  • 26. Evaluate 𝒇 𝒙 = 𝟗𝒛 𝟐 − 𝟏𝟎𝟎, 𝒈 𝒙 = 𝒛 𝟐 − 𝟐𝟓 a. 𝒍𝒊𝒎 𝒛→−𝟑 𝒇 𝒙 𝒈 𝒙 b. 𝒍𝒊𝒎 𝒛→𝟑 𝒈 𝒙 𝒇 𝒙 Evaluate 𝒃 = 𝟑𝟔, 𝒇 𝒙 = 𝒙 𝟑 − 𝟐𝟕 a. 𝐥𝐢𝐦 𝒙→𝟑 𝒃 𝒇(𝒙) b. 𝐥𝐢𝐦 𝒙→𝟐 𝒃 𝒇(𝒙)
  • 28. 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗, 𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑; a. 𝒍𝒊𝒎 𝒙→𝟑 𝒏 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟑 𝒏 )𝒇(𝒙 = 𝒏 𝒍𝒊𝒎 𝒙→𝟑 )𝒇(𝒙 = 𝟑 𝒍𝒊𝒎 𝒙→𝟑 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗 = 3 𝟒 𝟑 𝟐 − 𝟔 𝟑 + 𝟗 = 3 𝟑𝟔 − 𝟏𝟖 + 𝟗 = 𝟑= 3 𝟐𝟕
  • 29. 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗, 𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑; b. 𝒍𝒊𝒎 𝒙→−𝟑 𝒇 𝒙 𝒏 𝒍𝒊𝒎 𝒙→−𝟑 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→−𝟑 𝒇 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→−𝟑 𝟒𝒙 𝟐 − 𝟔𝒙 + 𝟗 𝟑 = 𝟒 −𝟑 𝟐 − 𝟔 −𝟑 + 𝟗 𝟑 = 𝟑𝟔 + 𝟏𝟖 + 𝟗 𝟑 = 𝟔𝟑 𝟑 = 𝟐𝟓𝟎, 𝟎𝟒𝟕
  • 30. 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗, 𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑; c. 𝒍𝒊𝒎 𝒙→𝟐 𝒏 𝒈(𝒙) 𝒍𝒊𝒎 𝒙→𝟐 𝒏 )𝒈(𝒙 = 𝒏 𝒍𝒊𝒎 𝒙→𝟐 )𝒈(𝒙 = 𝟑 𝒍𝒊𝒎 𝒙→𝟐 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒 = 3 𝟐 𝟑 + 𝟖 𝟐 𝟐 − 𝟏𝟔 𝟐 + 𝟒 = 3 𝟖 + 𝟑𝟐 − 𝟑𝟐 + 𝟒 = 𝟐. 𝟐𝟗= 3 𝟏𝟐
  • 31. 1. Evaluate 𝒇 𝒙 = 𝟒 𝒙 𝟐 − 𝟔𝒙 + 𝟗, 𝒈 𝒙 = 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒, 𝒏 = 𝟑; d. 𝒍𝒊𝒎 𝒙→−𝟐 𝒈 𝒙 𝒏 𝒍𝒊𝒎 𝒙→−𝟐 𝒈 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→−𝟐 𝒈 𝒙 𝒏 = 𝒍𝒊𝒎 𝒙→−𝟐 𝒙 𝟑 + 𝟖𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟒 𝟑 = −𝟐 𝟑 + 𝟖 −𝟐 𝟐 − 𝟏𝟔 −𝟐 + 𝟒 𝟑 = −𝟖 + 𝟑𝟐 + 𝟑𝟐 + 𝟒 𝟑 = 𝟔𝟎 𝟑 = 𝟐𝟏𝟔, 𝟎𝟎𝟎
  • 32. 2. Evaluate 𝒇 𝒙 = 𝟏𝟔𝐱 𝟐 − 𝟐𝟓, 𝒈 𝒙 = 𝒙 𝟑 − 𝟏 a. 𝒍𝒊𝒎 𝒙→𝟑 𝒇 𝒙 𝒈 𝒙 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→3 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟑 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑 𝟏𝟔𝒙 𝟐 − 𝟐𝟓 𝐥𝐢𝐦 𝒙→𝟑 𝐱 𝟑−𝟏 = 𝟏𝟔 𝟑 𝟐 − 𝟐𝟓 𝟑 𝟑−𝟏 = 𝟏𝟒𝟒 − 𝟐𝟓 𝟐𝟔 = 𝟏𝟏𝟗 𝟐𝟔 = 𝟗. 𝟐𝟏 × 𝟏𝟎 𝟓𝟑
  • 33. 2. Evaluate 𝒇 𝒙 = 𝟏𝟔𝐱 𝟐 − 𝟐𝟓, 𝒈 𝒙 = 𝒙 𝟑 − 𝟏 b. 𝒍𝒊𝒎 𝒙→𝟑 𝒈 𝒙 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟑 𝒈 𝒙 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→3 𝒈 𝒙 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟑 − 𝟏 𝐥𝐢𝐦 𝒙→𝟑 𝟏𝟔𝐱 𝟐−𝟐𝟓 = ( 𝟑 𝟑 − 𝟏) 𝟏𝟔(𝟑) 𝟐−𝟐𝟓 = (𝟐𝟔) 𝟏𝟒𝟒−𝟐𝟓 = 𝟐𝟔 𝟏𝟏𝟗 = 𝟐. 𝟒𝟏 × 𝟏𝟎 𝟏𝟔𝟖
  • 34. 3. Evaluate 𝑏 = 𝟐𝟒, 𝑓 𝑥 = 𝒙 𝟑 − 𝟖 a. 𝐥𝐢𝐦 𝒙→𝟑 𝒃 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟑 𝒃 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟑 𝟐𝟒 𝒙 𝟑−𝟖 = 𝒍𝒊𝒎 𝒙→𝟑 𝟐𝟒 𝒍𝒊𝒎 𝒙→𝟑 𝒙 𝟑−𝟖 = 𝟐𝟒 𝒙 𝟑−𝟖 = 𝟐𝟒 𝟑 𝟑−𝟖 = 𝟐𝟒 𝟏𝟗 = 𝟏. 𝟔𝟕 × 𝟏𝟎 𝟐𝟔
  • 35. 3. Evaluate 𝑏 = 𝟐𝟒, 𝑓 𝑥 = 𝒙 𝟑 − 𝟖 b. 𝐥𝐢𝐦 𝒙→𝟐 𝒃 𝒇(𝒙) 𝒍𝒊𝒎 𝒙→𝟐 𝒃 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟐 𝟐𝟒 𝒙 𝟑−𝟖 = 𝒍𝒊𝒎 𝒙→𝟐 𝟐𝟒 𝒍𝒊𝒎 𝒙→𝟐 𝒙 𝟑−𝟖 = 𝟐𝟒 𝒙 𝟑−𝟖 = 𝟐𝟒 𝟐 𝟑−𝟖 = 𝟐𝟒 𝟎 = 𝟏
  • 36. Limit of Logarithmic Function Let a and b be real numbers, where a > 0, b > 0 and b ≠ 1. Then, 𝒍𝒊𝒎 𝒙→𝒂 ( 𝒍𝒐𝒈 𝒃 𝒙) = 𝒍𝒐𝒈 𝒃 𝒂
  • 37. Properties There are three basic properties in limits, which are used as formulas in evaluating the limits of logarithmic functions. 1.Product Rule 2.Quotient Rule 3.Power Rule
  • 38. 1. Product Rule Example: Solution: lim 𝑥→𝑎 (log 𝑏 𝐴𝐵) = lim 𝑥→𝑎 log 𝑏 𝐴 + log 𝑏 𝐵 Given 𝑨 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏, 𝑩 = 𝒙 − 𝟓, 𝒃 = 𝟓 𝒂𝒏𝒅 𝒙 = 𝟑, find 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝒙→𝟑 𝐥𝐨𝐠 𝟓(𝒙 𝟐 +𝟐𝒙 + 𝟏) + 𝐥𝐨𝐠 𝟓 𝒙 − 𝟓) = 𝐥𝐨𝐠 𝟓 𝟑 𝟐 + 𝟐 𝟑 + 𝟏 + 𝟑 − 𝟓 = 𝐥𝐨𝐠 𝟓 𝟗 + 𝟔 + 𝟏 − 𝟐 = 𝐥𝐨𝐠 𝟓 𝟏𝟒 = 𝟏. 𝟔𝟒
  • 39. 2. Quotient Rule Example: Solution: lim 𝑥→𝑎 log 𝑏( 𝐴 𝐵 ) = lim 𝑥→𝑎 log 𝑏 𝐴 − log 𝑏 𝐵 Given 𝑨 = 𝒙 𝟐 + 𝟐𝒙 + 𝟏, 𝑩 = 𝒙 − 𝟓, 𝒃 = 𝟓 𝒂𝒏𝒅 𝒙 = 𝟑, find 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨 𝑩 ) 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝒙→𝟑 𝐥𝐨𝐠 𝟓(𝒙 𝟐 +𝟐𝒙 + 𝟏) − 𝐥𝐨𝐠 𝟓 𝒙 − 𝟓) = 𝐥𝐨𝐠 𝟓 𝟑 𝟐 + 𝟐 𝟑 + 𝟏 − 𝟑 − 𝟓 = 𝐥𝐨𝐠 𝟓 𝟗 + 𝟔 + 𝟏 + 𝟐 = 𝐥𝐨𝐠 𝟓 𝟏𝟖 = 𝟏. 𝟕𝟗𝟔
  • 40. 3. Power Rule Example: Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏 = 𝒍𝒊𝒎 𝒙→𝒂 𝒏 𝒍𝒐𝒈 𝒃 𝑨 Given 𝑨 = 𝒙 − 𝟓, 𝒏 = 𝟑, 𝒃 = 𝟐 𝒂𝒏𝒅 𝒙 = 𝟖, find 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏 𝒍𝒊𝒎 𝒙→𝟖 𝒍𝒐𝒈 𝟐 𝒙 − 𝟓 𝟑 = 𝒍𝒊𝒎 𝒙→𝟖 𝟑 𝒍𝒐𝒈 𝟐 𝒙 − 𝟓 = 𝟑 𝒍𝒐𝒈 𝟐 𝟖 − 𝟓 = 𝟑 𝒍𝒐𝒈 𝟐 𝟑 = 𝟑 𝟏. 𝟓𝟖𝟓 = 𝟒. 𝟕𝟓𝟓
  • 41. Let’s Practice!!! 1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔, 𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ; Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 42. Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝒙→𝟐 𝐥𝐨𝐠 𝟏𝟎(𝒙 𝟐 + 𝟒𝒙 + 𝟒) + 𝐥𝐨𝐠 𝟏𝟎(𝒙 𝟐 − 𝟏𝟔) = 𝐥𝐨𝐠 𝟏𝟎 𝟐 𝟐 + 𝟒 𝟐 + 𝟒 + 𝟐 𝟐 − 𝟏𝟔 = 𝐥𝐨𝐠 𝟏𝟎 𝟒 + 𝟖 + 𝟒 + 𝟒 − 𝟏𝟔 = 𝐥𝐨𝐠 𝟏𝟎 𝟒 = 𝟎. 𝟔𝟎𝟐 1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔, 𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ;
  • 43. Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝒙→𝟐 𝐥𝐨𝐠 𝟏𝟎 𝒙 𝟐 + 𝟒𝒙 + 𝟒 − 𝐥𝐨𝐠 𝟏𝟎(𝒙 𝟐 − 𝟏𝟔) 1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔, 𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ; = 𝐥𝐨𝐠 𝟏𝟎 𝟐 𝟐 + 𝟒 𝟐 + 𝟒 − 𝟐 𝟐 − 𝟏𝟔 = 𝐥𝐨𝐠 𝟏𝟎 𝟒 + 𝟖 + 𝟒 − 𝟒 + 𝟏𝟔 = 𝐥𝐨𝐠 𝟏𝟎 𝟐𝟖 = 𝟏. 𝟒𝟒𝟕
  • 44. Solution: 𝒍𝒊𝒎 𝒙→𝟐 𝒍𝒐𝒈 𝟏𝟎 𝒙 𝟐 + 𝟒𝒙 + 𝟒 𝟒 = 𝒍𝒊𝒎 𝒙→𝟐 𝟒 𝒍𝒐𝒈 𝟏𝟎 𝒙 𝟐 + 𝟒𝒙 + 𝟒 = 𝟒 𝒍𝒐𝒈 𝟏𝟎 𝟐 𝟐 + 𝟒 𝟐 + 𝟒 = 𝟒 𝒍𝒐𝒈 𝟏𝟎 𝟏𝟔 = 𝟒 𝟏. 𝟐𝟎𝟒 = 𝟒. 𝟖𝟏𝟔 1. Given 𝑨 = 𝒙 𝟐 + 𝟒𝒙 + 𝟒, 𝑩 = 𝒙 𝟐 − 𝟏𝟔, 𝒙 = 𝟐, 𝒃 = 𝟏𝟎, 𝒏 = 𝟒 ;
  • 46. 1. Given 𝑨 = 𝟗𝒙 𝟐 + 𝟑𝟎𝒙 + 𝟐𝟓, 𝑩 = 𝟒𝒙 𝟐 − 𝟏𝟔, 𝒙 = 𝟓, 𝒃 = 𝟑𝟓, 𝒏 = 𝟏𝟓 ; 2. Given 𝑨 = 𝒙 𝟐 − 𝟏𝟔𝒙 + 𝟔𝟒, 𝑩 = 𝒙 𝟐 − 𝟏, 𝒙 = 𝟑, 𝒃 = 𝟐𝟓, 𝒏 = 𝟐𝟓 Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 47. 2. Given 𝑨 = 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐 − 𝟑𝟔, 𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ; Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 49. Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝒙→𝟐 𝐥𝐨𝐠 𝟏𝟓(𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔) + 𝐥𝐨𝐠 𝟏𝟓(𝒙 𝟐 − 𝟔𝟒) = 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓 𝟐 𝟐 + 𝟒𝟎 𝟐 + 𝟏𝟔 + 𝟐 𝟐 − 𝟔𝟒 = 𝐥𝐨𝐠 𝟏𝟓 𝟏𝟎𝟎 + 𝟖𝟎 + 𝟏𝟔 + 𝟒 − 𝟔𝟒 = 𝐥𝐨𝐠 𝟏𝟓 𝟏𝟑𝟔 = 𝟏. 𝟖𝟏𝟒 1. Given 𝑨 = 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔, 𝑩 = 𝒙 𝟐 − 𝟔𝟒, 𝒙 = 𝟐, 𝒃 = 𝟏𝟓, 𝒏 = 𝟓 ;
  • 50. Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝒙→𝟐 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔 − 𝐥𝐨𝐠 𝟏𝟓(𝒙 𝟐 − 𝟔𝟒) 1. Given 𝑨 = 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔, 𝑩 = 𝒙 𝟐 − 𝟔𝟒, 𝒙 = 𝟐, 𝒃 = 𝟏𝟓, 𝒏 = 𝟓 ; = 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓 𝟐 𝟐 + 𝟒𝟎 𝟐 + 𝟏𝟔 − 𝟐 𝟐 − 𝟔𝟒 = 𝐥𝐨𝐠 𝟏𝟓 𝟏𝟎𝟎 + 𝟖𝟎 + 𝟏𝟔 − 𝟒 + 𝟔𝟒 = 𝐥𝐨𝐠 𝟏𝟓 𝟐𝟓𝟔 = 𝟐. 𝟎𝟒𝟖
  • 51. Solution: 𝒍𝒊𝒎 𝒙→𝟐 𝒍𝒐𝒈 𝟏𝟓 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔 𝟓 = 𝒍𝒊𝒎 𝒙→𝟐 𝟓 𝒍𝒐𝒈 𝟏𝟓 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔 = 𝟓 𝒍𝒐𝒈 𝟏𝟓 𝟐𝟓 𝟐 𝟐 + 𝟒𝟎 𝟐 + 𝟏𝟔 = 𝟓 𝒍𝒐𝒈 𝟏𝟓 𝟏𝟗𝟔 = 𝟓 𝟏. 𝟗𝟒𝟗 = 𝟗. 𝟕𝟒𝟓 1. Given 𝑨 = 𝟐𝟓𝒙 𝟐 + 𝟒𝟎𝒙 + 𝟏𝟔, 𝑩 = 𝒙 𝟐 − 𝟔𝟒, 𝒙 = 𝟐, 𝒃 = 𝟏𝟓, 𝒏 = 𝟓 ;
  • 52. Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝒙→𝟏𝟐 𝐥𝐨𝐠 𝟑𝟔(𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎) + 𝐥𝐨𝐠 𝟑𝟔(𝒙 𝟐 − 𝟑𝟔) = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟐 𝟐 + 𝟐𝟎 𝟏𝟐 − 𝟏𝟎𝟎 + 𝟏𝟐 𝟐 − 𝟑𝟔 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟒𝟒 + 𝟐𝟒𝟎 − 𝟏𝟎𝟎 + 𝟏𝟒𝟒 − 𝟑𝟔 = 𝐥𝐨𝐠 𝟑𝟔 𝟑𝟗𝟐 = 𝟏. 𝟔𝟔𝟔 2. Given 𝑨 = 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐 − 𝟑𝟔, 𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ;
  • 53. Solution: 𝒍𝒊𝒎 𝒙→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝒙→𝟏𝟐 𝐥𝐨𝐠 𝟑𝟔 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎 − 𝐥𝐨𝐠 𝟑𝟔(𝒙 𝟐 − 𝟑𝟔) 2. Given 𝑨 = 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐 − 𝟑𝟔, 𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ; = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟐 𝟐 + 𝟐𝟎 𝟏𝟐 − 𝟏𝟎𝟎 − 𝟏𝟐 𝟐 − 𝟑𝟔 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟒𝟒 + 𝟐𝟒𝟎 − 𝟏𝟎𝟎 − 𝟏𝟒𝟒 + 𝟑𝟔 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟕𝟔 = 𝟏. 𝟒𝟒𝟑
  • 54. Solution: 𝒍𝒊𝒎 𝒙→𝟏𝟐 𝒍𝒐𝒈 𝟑𝟔 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎 𝟐𝟓 = 𝒍𝒊𝒎 𝒙→𝟏𝟐 𝟐𝟓 𝒍𝒐𝒈 𝟑𝟔 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎 = 𝟐𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟏𝟐 𝟐 + 𝟐𝟎 𝟏𝟐 − 𝟏𝟎𝟎 = 𝟐𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟐𝟖𝟒 = 𝟐𝟓 𝟏. 𝟓𝟕𝟔 = 𝟑𝟗. 𝟒𝟎𝟗 2. Given 𝑨 = 𝒙 𝟐 + 𝟐𝟎𝒙 − 𝟏𝟎𝟎, 𝑩 = 𝒙 𝟐 − 𝟑𝟔, 𝒙 = 𝟏𝟐, 𝒃 = 𝟑𝟔, 𝒏 = 𝟐𝟓 ;
  • 56. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒, 𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖 ; Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 57. Given 𝑨 = 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐 − 𝟐𝒎 + 𝟐, 𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ; Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 58. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏, 𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ; Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 60. Solution: 𝒍𝒊𝒎 𝐲→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝐲→𝟓 𝐥𝐨𝐠 𝟐𝟓(𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒) + 𝐥𝐨𝐠 𝟐𝟓(𝒚 𝟐 − 𝟔𝟒) = 𝐥𝐨𝐠 𝟐𝟓 𝟑𝟔 𝟓 𝟐 − 𝟗𝟔 𝟓 + 𝟔𝟒 + 𝟓 𝟐 − 𝟔𝟒 = 𝐥𝐨𝐠 𝟐𝟓 𝟗𝟎𝟎 − 𝟒𝟖𝟎 + 𝟔𝟒 + 𝟐𝟓 − 𝟔𝟒 = 𝐥𝐨𝐠 𝟐𝟓 𝟒𝟒𝟓 = 𝟏. 𝟖𝟗𝟒 1. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒, 𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖
  • 61. Solution: = 𝟏. 𝟗𝟒𝟓 1. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒, 𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖 𝒍𝒊𝒎 𝐲→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝐲→𝟓 𝐥𝐨𝐠 𝟐𝟓 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒 − 𝐥𝐨𝐠 𝟐𝟓(𝒚 𝟐 − 𝟔𝟒) = 𝐥𝐨𝐠 𝟐𝟓 𝟑𝟔 𝟓 𝟐 − 𝟗𝟔 𝟓 + 𝟔𝟒 − 𝟓 𝟐 − 𝟔𝟒 = 𝐥𝐨𝐠 𝟐𝟓 𝟗𝟎𝟎 − 𝟒𝟖𝟎 + 𝟔𝟒 − 𝟐𝟓 + 𝟔𝟒 = 𝐥𝐨𝐠 𝟐𝟓 𝟓𝟐𝟑
  • 62. Solution: 𝒍𝒊𝒎 𝒚→𝟓 𝒍𝒐𝒈 𝟐𝟓 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒 𝟖 = 𝒍𝒊𝒎 𝒚→𝟓 𝟖 𝒍𝒐𝒈 𝟐𝟓 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒 = 𝟖 𝒍𝒐𝒈 𝟐𝟓 𝟑𝟔 𝟓 𝟐 − 𝟗𝟔 𝟓 + +𝟔𝟒 = 𝟖 𝒍𝒐𝒈 𝟐𝟓 𝟒𝟖𝟒 = 𝟖 𝟏. 𝟗𝟐𝟏 = 𝟏𝟓. 𝟑𝟔𝟓 1. Given 𝑨 = 𝟑𝟔𝒚 𝟐 − 𝟗𝟔𝒚 + 𝟔𝟒, 𝑩 = 𝒚 𝟐 − 𝟔𝟒, 𝒚 = 𝟓, 𝒃 = 𝟐𝟓, 𝒏 = 𝟖
  • 63. Solution: 𝒍𝒊𝒎 𝐦→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝐦→𝟔 𝐥𝐨𝐠 𝟑𝟔(𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗) + 𝐥𝐨𝐠 𝟑𝟔(𝒎 𝟐 − 𝟐𝒎 + 𝟐) = 𝐥𝐨𝐠 𝟑𝟔 𝟒𝟓 𝟔 𝟐 − 𝟑𝟔 𝟔 − 𝟗 + 𝟔 𝟐 − 𝟐 𝟔 + 𝟐 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟔𝟐𝟎 − 𝟐𝟏𝟔 − 𝟗 + 𝟑𝟔 − 𝟏𝟐 + 𝟐 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟒𝟐𝟏 = 𝟐. 𝟎𝟐𝟔 2. Given 𝑨 = 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐 − 𝟐𝒎 + 𝟐, 𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ;
  • 64. Solution: = 𝟐. 𝟎𝟏𝟓 2. Given 𝑨 = 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐 − 𝟐𝒎 + 𝟐, 𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ; 𝒍𝒊𝒎 𝐦→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝐦→𝟔 𝐥𝐨𝐠 𝟑𝟔 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗 − 𝐥𝐨𝐠 𝟑𝟔(𝒎 𝟐 − 𝟐𝒎 + 𝟐) = 𝐥𝐨𝐠 𝟑𝟔 𝟒𝟓 𝟔 𝟐 − 𝟑𝟔 𝟔 − 𝟗 − 𝟔 𝟐 − 𝟐 𝟔 + 𝟐 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟔𝟐𝟎 − 𝟐𝟏𝟔 − 𝟗 − 𝟑𝟔 + 𝟏𝟐 − 𝟐 = 𝐥𝐨𝐠 𝟑𝟔 𝟏𝟑𝟔𝟗
  • 65. Solution: 𝒍𝒊𝒎 𝒎→𝟔 𝒍𝒐𝒈 𝟑𝟔 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗 𝟓 = 𝒍𝒊𝒎 𝒙→𝟐 𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗 = 𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟒𝟓 𝟔 𝟐 − 𝟑𝟔 𝟔 − 𝟗 = 𝟓 𝒍𝒐𝒈 𝟑𝟔 𝟏𝟑𝟗𝟓 = 𝟓 𝟐. 𝟎𝟐𝟏 = 𝟏𝟎. 𝟏𝟎𝟑 2. Given 𝑨 = 𝟒𝟓𝒎 𝟐 − 𝟑𝟔𝒎 − 𝟗, 𝑩 = 𝒎 𝟐 − 𝟐𝒎 + 𝟐, 𝒎 = 𝟔, 𝒃 = 𝟑𝟔, 𝒏 = 𝟓 ;
  • 66. Solution: 𝒍𝒊𝒎 𝐲→𝒂 𝐥𝐨𝐠 𝒃 𝑨𝑩 = 𝐥𝐢𝐦 𝒚→𝟑 𝐥𝐨𝐠 𝟏𝟖(𝟏𝟔𝒚 𝟐 − 𝟒𝟔𝒚 + 𝟑𝟔) + 𝐥𝐨𝐠 𝟏𝟖(𝒚 𝟑 − 𝟏) = 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟔 𝟑 𝟐 − 𝟒𝟖 𝟑 + 𝟑𝟔 + 𝟑 𝟑 − 𝟏 = 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟒𝟒 − 𝟏𝟒𝟒 + 𝟑𝟔 + 𝟐𝟕 − 𝟏 = 𝐥𝐨𝐠 𝟏𝟖 𝟔𝟐 = 𝟏. 𝟒𝟐𝟖 3. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏, 𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ;
  • 67. Solution: = 𝟎. 𝟕𝟗𝟕 3. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏, 𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ; 𝒍𝒊𝒎 𝐲→𝒂 𝐥𝐨𝐠 𝒃 𝑨 𝑩 = 𝐥𝐢𝐦 𝒚→𝟑 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟔𝒚 𝟐 − 𝟒𝟔𝒚 + 𝟑𝟔 − 𝐥𝐨𝐠 𝟏𝟖(𝒚 𝟑 − 𝟏) = 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟔 𝟑 𝟐 − 𝟒𝟖 𝟑 + 𝟑𝟔 − 𝟑 𝟐 − 𝟏 = 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟒𝟒 − 𝟏𝟒𝟒 + 𝟑𝟔 − 𝟐𝟕 + 𝟏 = 𝐥𝐨𝐠 𝟏𝟖 𝟏𝟎
  • 68. Solution: 𝒍𝒊𝒎 𝒚→𝟑 𝒍𝒐𝒈 𝟏𝟖 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔 𝟔 = 𝒍𝒊𝒎 𝒚→𝟑 𝟔 𝒍𝒐𝒈 𝟏𝟖 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔 = 𝟔 𝒍𝒐𝒈 𝟏𝟖 𝟏𝟔 𝟑 𝟐 − 𝟒𝟖 𝟑 + 𝟑𝟔 = 𝟔 𝒍𝒐𝒈 𝟏𝟖 𝟑𝟔 = 𝟔 𝟏. 𝟐𝟒𝟎 = 𝟕. 𝟒𝟑𝟗 3. Given 𝑨 = 𝟏𝟔𝒚 𝟐 − 𝟒𝟖𝒚 + 𝟑𝟔, 𝑩 = 𝒚 𝟑 − 𝟏, 𝒚 = 𝟑, 𝒃 = 𝟏𝟖, 𝒏 = 𝟔 ;
  • 69. Pointers to Review Limits of Non – Algebraic Functions Limit of an Exponential Function Limit of Logarithmic Function
  • 70. Given 𝑨 = 𝒚 𝟐 + 𝟔𝒚 + 𝟗, 𝑩 = 𝟒𝒚 𝟐 − 𝟏, 𝒚 = 𝟖, 𝒃 = 𝟏𝟓, 𝒏 = 𝟔 ; Find the following: a. 𝒍𝒊𝒎 𝒚→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒚→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒚→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 71. Given 𝑨 = 𝟑𝒙 𝟐 + 𝟗𝒙 + 𝟗, 𝑩 = 𝒙 − 𝟓, 𝒙 = 𝟐, 𝒃 = 𝟏𝟑, 𝒏 = 𝟏𝟎 ; Find the following: a. 𝒍𝒊𝒎 𝒙→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒙→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏
  • 72. Given 𝑨 = 𝟔𝒛 𝟐 − 𝟒𝒛 + 𝟓, 𝑩 = 𝟐𝒛 𝟐 + 𝟒𝒛 − 𝟑, 𝒛 = 𝟑, 𝒃 = 𝟖, 𝒏 = 𝟐; Find the following: a. 𝒍𝒊𝒎 𝒛→𝒂 (𝒍𝒐𝒈 𝒃 𝑨𝑩) b. 𝒍𝒊𝒎 𝒛→𝒂 𝒍𝒐𝒈 𝒃( 𝑨 𝑩 ) c. 𝒍𝒊𝒎 𝒛→𝒂 𝒍𝒐𝒈 𝒃 𝑨 𝒏