This presentation introduces limits and provides examples of applying 8 theorems of limits. It begins with an introduction explaining that calculus and limits are challenging topics for college students. It then covers the definitions of limits and 8 theorems for evaluating different types of limits, including limits of constants, variables, products, quotients, and composite functions. Example problems demonstrate how to use each theorem to find the limit as the variable approaches a value. The presentation aims to make solving limits easier by explaining the theorems.

2. INTRODUCTION
Calculus is one of the hardest part of Mathematics.
Almost all college students says that it is a real pain.
As fourth year high school students, we should have
preparations about Calculus in order to be familiarized to it
when we get to college.
This presentation would teach you a basic topic about
Calculus, the Limits and its theorems. This also intends to
make solving limits from a hard stuff to an easy one.

3. TABLE OF CONTENTS
Title…………………………………………………
……………..slide 1
Introduction………………………………………
……....slide 2
Table of
Contents………………………………………slide
3
Introduction to
Calculus………………………...slide 4
What are Limits?.......................................slide 5
Theorems on
Limits…………………………..……..slide 6-14
Limit
#1………………………………………………...sli
de 7
Limit
#2…………………………………………….…slid

5. LIMITS are used to describe
the value of the function at a
certain input in terms of its
values at nearby input. Limits
also replaced Infinitesimals in
19th Century

7. Lim c =
c
xa
Example: Find the lim 4
x2
* Since 4 is a constant(c), the limit of 4 as x
approaches 2 is 4.

8. Lim x = a
xa
Example: Find lim x
x7
* By observing the given and the limit, we
could obviously say that “the limit of x as x
approaches 7 is 7. But the real explanation is
that we substitute “lim x” by the value of a
which is 7.

9. lim cx = c lim
x
xa xa
Example: Find lim 5x
x6
Solution:
= 5 lim x
x6
= 5(6)
= 30
1) We used the limit format to convert it.
2) Now that we have “lim x”, we can substitute it by the value of a which
is 6.
3) Then perform the indicated operation.
4) So we the answer is “the limit of 5x as x approaches 6 is 30”.

10. Lim f(x)+g(x) = lim f(x)+lim
g(x)
xa xa xaExample: Find lim 3x2+4x
x2
Solution:
= lim 3x2 + lim 4x
x2 x2
= 3 lim x2 + 4 lim x  limit #3 & #3
x2 x2
= 3(2)2 + 4(2)
= 12+8
= 20
1) We use limit #4 to have a new operation.
2) Now we have 2 equations or functions which we need to simplify. Based on the
functions, we could use limit #3 on both of them to solve it.
3) After we used Limit #3 on both functions, we get “lim x2” and “lim x” which we
could substitute by the value of a.
4) Perform the indicated operations.
5) Then, the final answer is “the limit of 3x2+4x as x approaches 2 is 20.

11. Lim f(x)•g(x) = lim f(x)•lim
g(x)
xa xa xaExample: Find lim 3x2•4x
x2
Solution:
= lim 3x2•lim 4x
x2 x2
= 3 lim x2•4 lim x Limit #3 & #3
x2 x2
= 3(2)2•4(2)
= 12•8
= 96
1) We first use Limit #5 and then Limit #3 to the 2 new functions.
2) We use substitute the value of a(2) to “lim x2” and “lim x”.
3) Perform the indicated operations.
4) Then, the final answer is “the limit of 3x2•4x as x approaches 2 is 96.

12. Lim f(x) = lim f(x)
g(x) xa
xa lim g(x)
xa
Example: Find lim 2x
4
x8
Solution:
= lim 2x = 2 lim x Limit #3 = 2(8)/4
x8 x8 = 16/4
lim 4 4 Limit #1 = 4
x8
1) We use Limit #6 to convert the given.
2) We derived to 2 different limits. We used Limit #3 on f(x) and Limit #1 on g(x).
3) It is now simplified, so we can substitute and perform the indicated operations.
4) Then, the final answer is “the limit of 2x over 4 as x approaches 8 is 4.

13. Lim nf(x) = nlim
f(x)
xa xaExample: Find lim 2x+5
x2
Solution:
= lim 2x+5
x2
= 2 lim x +5
x2
= 2(2) + 5
= 4+5
= 9
= 3
1) We first used Limit #7.
2) Then, we used Limit #3 in the term “2x” only and we just copy 5.
3) Then, we substitute. And add 5.
4) We get its square root.
5) So, the final answer would be “the limit of square root of 2x+5 as x approaches 2 is 3.

14. Lim f(x)n = lim
f(x)n
xa xa
Example: Find lim2x+32
x3
Solution:
= lim 2x+32
x3
= 2 lim x + (3)2
x3
= 2(3) + 32
= (6+3)2
= (9)2
= 81
1) We used Limit #8.
2) Then, we used Limit #3 to 2x
only (just like on previous
example).
3) We substitute and used the
indicated operation.
4) Finally, we would get the square
of it.
5) The final answer is “the limit of
2x+3 squared as x approaches 3
is 81.

15. Limit is one of the
basic topics in
Calculus.
Limits have 8
theorems.

16. JOHN ROME R. ARANAS
Creator
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