d i e
play

d i E Limits a l l u d Dr. Abdulla Eid b A College of - PowerPoint PPT Presentation

Jun 07, 2023 •276 likes •515 views

Section 10.2 d i E Limits a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Limits 1 / 22 Definition of a limit Example How does the function

  1. Section 10.2 d i E Limits a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Limits 1 / 22

  2. Definition of a limit Example How does the function f ( x ) = x 2 − 1 d x − 1 i E behave at x = 1? a l l u Solution: f ( 1 ) = 1 2 − 1 d 1 − 1 = 0 b undefined! A 0 Hence, we cannot substitute directly with x = 1, so instead we check . r D values that are very much close to x = 1 and we check the corresponding values of f ( x ) . x 0.9 0.99 0.9999 1 1.00001 1.001 1.01 f ( x ) - Dr. Abdulla Eid (University of Bahrain) Limits 2 / 22

  3. Continue... d i E So we have seen that as x approaches 1, f ( x ) approaches 2, we write a l l u x → 1 f ( x ) = 2 lim d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 3 / 22

  4. To find the limit lim x → a f ( x ) , we have d 1 Substitute directly by x = a in f ( x ) . If you get a real number, then i E that is the limit. a 2 If you get undefined values such as 0 l 0 , we use algebraic method to l u clear any problem. d b 3 If you get values such as non–zero number , then we stop and we say that A 0 the limit does not exist (DNE). . r D Dr. Abdulla Eid (University of Bahrain) Limits 4 / 22

  5. Zero denominator of a rational function A - Eliminating zero denominator by canceling common factor in the numerator and denominator ( 0 0 form). Example d Find i x 2 − 25 E lim a x − 5 x → 5 l l u Solution: Direct substitution gives d b 5 2 − 25 = 0 A undefined! 5 − 5 0 . r So we factor both the denominator and numerator to cancel the common D zero. x 2 − 5 ( x − 5 )( x + 5 ) lim x − 5 = lim ( x − 5 ) x → 5 x → 5 = lim x → 5 ( x + 5 ) = 5 + 5 = 10 Dr. Abdulla Eid (University of Bahrain) Limits 5 / 22

  6. Exercise Find x 2 − 2 x + 1 lim x − 1 x → 1 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 6 / 22

  7. Example Find 5 y 3 + 8 y 2 lim 3 y 4 − 16 y 2 y → 0 d Solution: Direct substitution gives i E 5 ( 0 ) 3 + 8 ( 0 ) 2 3 ( 0 ) 4 − 16 ( 0 ) 2 = 0 a undefined! l 0 l u d So we factor both the denominator and numerator to cancel the common b A zero. 5 y 3 + 8 y 2 . y 2 ( 5 y + 8 ) r D 3 y 4 − 16 y 2 = lim lim y 2 ( 3 y 2 − 16 ) y → 0 y → 0 5 y + 8 = lim 3 y 2 − 16 y → 0 5 ( 0 ) + 8 − 16 = − 1 8 = 3 ( 0 ) 2 − 16 = 2 Dr. Abdulla Eid (University of Bahrain) Limits 7 / 22

  8. Example Find x 2 − 3 x lim x 2 − 9 x → 3 d Solution: Direct substitution gives i E 0 2 − 3 ( 0 ) = 0 a undefined! l 0 2 − 9 l 0 u d So we factor both the denominator and numerator to cancel the common b A zero. . x 2 − 3 x r x ( x − 3 ) D lim x 2 − 9 = lim ( x − 3 )( x + 3 ) x → 3 x → 3 x = lim x + 3 x → 3 = 3 6 = 1 2 Dr. Abdulla Eid (University of Bahrain) Limits 8 / 22

  9. Exercise Find x 2 − 3 x + 4 lim x 2 − 6 x + 7 x → 1 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 9 / 22

  10. Example Find 2 x 2 + 3 x + 1 lim x 2 − 2 x − 3 x →− 1 Solution: Direct substitution gives d i E 2 ( 0 ) 2 + 3 ( 0 ) + 1 ( 0 ) 2 − 2 ( 0 ) − 3 = 0 a undefined! l 0 l u d So we factor both the denominator and numerator to cancel the common b A zero. 2 x 2 + 3 x + 1 . 2 ( x + 1 2 )( x + 1 ) r lim x 2 − 2 x − 3 = lim D ( x − 3 )( x + 1 ) x →− 1 x →− 1 2 ( x + 1 2 ) = lim x − 3 x →− 1 = 2 ( − 1 2 ) = 1 − 4 4 Dr. Abdulla Eid (University of Bahrain) Limits 10 / 22

  11. Exercise (Old Final Exam Question) Find x 3 − 8 lim 3 x 2 − x − 10 x → 2 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 11 / 22

  12. Exercise Find x 2 − 3 x + 4 lim x 2 − 6 x + 7 x → 1 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 12 / 22

  13. Example Find 4 x 5 − 4 lim 5 x 2 − 5 x → 1 d Solution: Direct substitution gives i E 4 ( 0 ) 5 − 4 5 ( 0 ) 2 − 5 = 0 a undefined! l 0 l u d So we factor both the denominator and numerator to cancel the common b A zero. . 4 x 5 − 4 4 ( x − 1 )( x 4 + x 3 + x 2 + x + 1 ) r D 5 x 2 − 5 = lim lim 5 ( x − 1 )( x + 1 ) x → 1 x → 1 4 ( x 4 + x 3 + x 2 + x + 1 ) = lim 5 ( x + 1 ) x → 1 = 20 10 = 2 Dr. Abdulla Eid (University of Bahrain) Limits 13 / 22

  14. Exercise Find x 4 − 16 lim x 3 − 8 x → 2 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 14 / 22

  15. Example Find x − 2 lim x 2 − 5 x + 6 x → 2 Solution: Direct substitution gives d i E ( 2 ) 2 − 5 ( 2 ) + 6 = 0 2 − 2 a undefined! 0 l l u d So we factor both the denominator and numerator to cancel the common b A zero. . ( x − 2 ) x − 2 r D lim x 2 − 5 x + 6 = lim ( x − 2 )( x − 3 ) x → 2 x → 2 1 = lim ( x − 3 ) x → 2 = 1 0 Does Not Exist Dr. Abdulla Eid (University of Bahrain) Limits 15 / 22

  16. Exercise (Old Exam Question) Find 5 x − 25 lim x 2 − 25 x → 5 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 16 / 22

  17. Conjugate and multiply by 1 B - Eliminating zero denominator by multiplying with the conjugate. Example d i Find E √ x − 3 a lim l x − 9 x → 9 l u d Solution: Direct substitution gives b A √ 9 − 3 = 0 . undefined! r D 9 − 9 0 Since there is a square root and we cannot factor anything, we multiply both the numerator and denominator with the conjugate of the numerator (the one that contains the square root). Dr. Abdulla Eid (University of Bahrain) Limits 17 / 22

  18. Continue... √ x − 3 ( √ x − 3 ) ( √ x + 3 ) d i = lim ( √ x + 3 ) lim E x − 9 ( x − 9 ) x → 9 x → 9 a x − 9 l = lim ( x − 9 )( √ x + 3 ) l u x → 9 d b 1 A = lim ( √ x + 3 ) x → 9 . = 1 r D 6 Dr. Abdulla Eid (University of Bahrain) Limits 18 / 22

  19. Exercise (Old Exam Question) Find √ x − 5 − 2 lim x − 9 x → 9 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 19 / 22

  20. Exercise (Old Final Exam Question) Find x − 4 √ x − 2 lim x → 4 d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 20 / 22

  21. Properties of Limits Let lim x → a f ( x ) = L and lim x → a g ( x ) = K . d 1 lim x → a c = c . i E 2 lim x → a x n = a n . a 3 lim x → a [ f ( x ) + g ( x )] = lim x → a f ( x ) + lim x → a g ( x ) = L + K . l l u 4 lim x → a [ f ( x ) · g ( x )] = lim x → a f ( x ) · lim x → a g ( x ) = L · K . d b 5 lim x → a [ cf ( x )] = c lim x → a f ( x ) = cL . A � � f ( x ) = lim x → a f ( x ) 6 lim x → a lim x → a g ( x ) = L . K if K � = 0. r g ( x ) D √ 7 lim x → a � � n n f ( x ) = n lim x → a f ( x ) = L . If n is even, then L must be non–negative. Dr. Abdulla Eid (University of Bahrain) Limits 21 / 22

  22. Summary: These properties are telling us we can substitute directly with the value of a if there no problem. Exercise (Old Final Exam Question) Find d x + 1 i lim E x 2 + 9 x →− 3 a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Limits 22 / 22

Recommend

More recommend