Theorems on limits

Some theorems on limits, simplified

1. T H E O R E M S O N L I M I T S

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………………………………………………...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 14 Generalization……………………………………….….slide 15 Credits……………………………………………………….…slide 16

4. 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: Integral Calculus Differential Calculus Trivia: Calculus is a Latin word that means “small stones used for counting or pebbles”.

5. 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 19 th Century

6. T H E O R E M S O N L I M I T S

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

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

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

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

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

14. Limit #8 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 We used Limit #8. Then, we used Limit #3 to 2x only (just like on previous example). We substitute and used the indicated operation. Finally, we would get the square of it. The final answer is “the limit of 2x+3 squared as x approaches 3 is 81.

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

16. C R E D I T S JOHN ROME R. ARANAS Creator Sources and Resources Myself Math Notebook Ms. Charmaigne Marie Mahamis Wikepedia Google ☻☻☻☻☻☻☻☻☻