THEOREMSON

LIMITS

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.

TABLE OF CONTENTS

Title………………………………………………………………..slide 1Introduction……………………………………………....slide 2Table of Contents………………………………………slide 3Introduction to Calculus………………………...slide 4What are Limits?.......................................slide 5Theorems on Limits…………………………..……..slide 6-14 Limit #1………………………………………………...slide 7 Limit #2…………………………………………….…slide 8 Limit #3……………………………………………....slide 9 Limit #4……………………………………………….slide 10 Limit #5…………………………………………….…slide 11 Limit #6………………………………….………..…slide 12 Limit #7……………………………………………….slide 13 Limit #8…………………………………………….…slide 14Generalization……………………………………….….slide 15Credits……………………………………………………….…slide 16

Introduction to Calculus

CALCULUS is a branch of Mathematics focused on limits, functions, derivatives, integrals and infinite series. This subject constitutes a major part of modern Mathematics education.It has 2 major branches:1)Integral Calculus2)Differential CalculusTrivia: Calculus is a Latin word that means “small stones used for counting or pebbles”.

What are Limits?

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

THEOREMSON

LIMITS

Limit #1Lim c = c xa

Example: Find the lim 4 x2

* Since 4 is a constant(c), the limit of 4 as x approaches 2 is 4.

Limit #2Lim x = axa

Example: Find lim x x7

* 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.

Limit #3lim cx = c lim x xa xaExample: Find lim 5x x6Solution: = 5 lim x x6 = 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”.

Limit #4Lim f(x)+g(x) = lim f(x)+lim g(x) xa xa xa

Example: Find lim 3x2+4x x2Solution: = lim 3x2 + lim 4x x2 x2 = 3 lim x2 + 4 lim x limit #3 & #3 x2 x2 = 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.

Limit #5Lim f(x)•g(x) = lim f(x)•lim g(x) xa xa xaExample: Find lim 3x2•4x

x2 Solution: = lim 3x2•lim 4x x2 x2 = 3 lim x2•4 lim x Limit #3 & #3 x2 x2 = 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.

Limit #6Lim f(x) = lim f(x) g(x) xa xa lim g(x) xa

Example: Find lim 2x 4 x8Solution: = lim 2x = 2 lim x Limit #3 = 2(8)/4 x8 x8 = 16/4 lim 4 4 Limit #1 = 4 x8

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.

Limit #7Lim nf(x) = nlim f(x) xa xa Example: Find lim

2x+5 x2Solution: = lim 2x+5 x2 = 2 lim x +5 x2 = 2(2) + 5 = 4+5 = 9 = 31) 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.

Limit #8Lim f(x)n = lim f(x)n

xa xaExample: Find lim2x+32

x3Solution: = lim 2x+32

x3 = 2 lim x + (3)2

x3 = 2(3) + 32

= (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.

GENERALIZATION

Limit is one of the basic topics

in Calculus.

Limits have 8 theorems.

CREDITSJOHN ROME R. ARANAS

Creator

Sources and ResourcesMyself

Math Notebook

Ms. Charmaigne Marie Mahamis

Wikepedia

Google

☻☻☻☻☻☻☻☻☻