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L5 infinite limits squeeze theorem
Jan 28, 2018
- 1. LIMITS OF FUNCTIONS
- 2. INFINITE LIMITS; VERTICAL AND HORIZONTAL ASYMPTOTES; SQUEEZE THEOREM OBJECTIVES: define infinite limits; illustrate the infinite limits ; and use the theorems to evaluate the limits of functions. determine vertical and horizontal asymptotes define squeeze theorem
- 3. DEFINITION: INFINITE LIMITS Sometimes one-sided or two-sided limits fail to exist because the value of the function increase or decrease without bound. For example, consider the behavior of for values of x near 0. It is evident from the table and graph in Fig 1.1.15 that as x values are taken closer and closer to 0 from the right, the values of are positive and increase without bound; and as x-values are taken closer and closer to 0 from the left, the values of are negative and decrease without bound. x 1 )x(f = x 1 )x(f = x 1 )x(f =
- 4. In symbols, we write =+= + x 1 limand x 1 lim 0x0x Note: The symbols here are not real numbers; they simply describe particular ways in which the limits fail to exist. Thus it is incorrect to write . + and ( ) ( ) 0=++
- 5. Figure 1.1.15 (p. 74)
- 6. 1.1.4 (p. 75) Infinite Limits (An Informal View)
- 7. Figure 1.2.2 (p. 84) Figure 1.1.2 illustrate graphically the limits for rational functions of the form . ( ) ( ) ( )22 ax 1 , ax 1 , ax 1
- 8. EXAMPLE: Evaluate the following limits: 4 0x x 1 lim.1 + 4 0x x 1 lim.2 5 0x x 1 lim.4 + == 0 1 x 1 lim 5 0x +== + + 0 1 x 1 lim 4 0x +== + 0 1 x 1 lim 4 0x +== + + 0 1 x 1 lim 5 0x 5 0x x 1 lim.5 += 40x x 1 lim.3 = 50x x 1 lim.6
- 9. 2x x3 lima..7 2x 2x x3 lim.b 2x + ( )( ) + = + == 0 6 0 23 2-x 3x lim- 2x 2.028.12x 8.1say,leftfrom2toclosexofvalue takewemeans2x == ( )( ) += + + = + ==+ 0 6 0 23 2-x 3x lim 2x 1.021.22x 1.2say,rightfrom2toclosexofvalue takewemeans2x +== + = 2x x3 lim.c 2x
- 10. )x(flim ax + )x(flim ax )x(flim ax + + + + + SUMMARY: )x(Q )x(R )x(fIf =
- 11. EXAMPLE ( ) += ++= + + = + += + ++ 3 1 3x 2 3x 2 lim,then 3 1 3x 2 limand 3x 2 lim.1 3x 3x3x ( ) ( ) =+ +=++ c c nSubtractio/Addition:Note
- 12. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) +==+ =++=++ cc cc :Note ( )( )( ) =+= + = + += + ++ 1 1x 3x 1x x2 lim,then 1 1x 3x limand 1x x2 lim.2 1x 1x1x ( ) += = + = + = 3 1 4x 6x2 2x x3 lim,then 3 1 4x 6x2 limand 2x x3 lim.3 2x 2x2x
- 13. VERTICAL AND HORIZONTAL ASYMPTOTES
- 14. DEFINITION: = += = += + + )x(flim.d )x(flim.c )x(flim.b )x(flim.a ax ax ax ax The line is a vertical asymptote of the graph of the function if at least one of the following statement is true: x a= ( )y f x=
- 15. x=a 0 +=+ )x(flim ax += )x(flim ax The following figures illustrate the vertical asymptote .x a= x=a 0
- 16. x=a 0 x=a = )x(flim ax =+ )x(flim ax The following figures illustrate the vertical asymptote .x a= 0
- 17. DEFINITION: b)x(flimorb)x(flim xx == + The line is a horizontal asymptote of the graph of the function if either by = ( )y f x=
- 18. y=b 0 y=b b)x(flim x = + The following figures illustrate the horizontal asymptote by = 0 b)x(flim x = +
- 19. y=b 0 y=b b)x(flimx = The following figures illustrate the horizontal asymptote by = 0 b)x(flimx =
- 20. Determine the horizontal and vertical asymptote of the function and sketch the graph.( ) 3 2 f x x = a. Vertical Asymptote: Equate the denominator to zero to solve for the vertical asymptote. 2x02x == Evaluate the limit as x approaches 2 2 3 3 3 lim 2 2 2 0x x = = = b. Horizontal Asymptote: Divide both the numerator and the denominator by the highest power of x to solve for the horizontal asymptote.
- 21. 3 3 0 lim 0 2 2 1 01 x x x x x + += = = + 3 3 0 lim 0 2 2 1 01 x x x x x = = = ( ) erceptintxnoistheretherefore 30; 2x 3 0,0)xf(If 2 3 20 3 xf,0xIf :Intercepts == = == .asymptotehorizontalais0,Thus
- 22. 2 3 ,0 VA: x=2 HA:y=0 0 ( ) 3 2 f x x =
- 23. Determine the horizontal and vertical asymptote of the function and sketch the graph.( ) 3x 1x2 xf + = a. Vertical Asymptote: b. Horizontal Asymptote: 3x03x == == + 0 7 3x 1x2 lim 3x 2 1 2 x 3 x x x 1 x x2 lim x == + asymptotehorizontalais2y =asymptoteverticalais3x = ( ) 2 1 x; 3x 1x2 0,0)xf(If 3 1 30 10 xf,0xIf :Intercepts = + == = + ==
- 24. HA:y=2 VA:x=3 o ( ) 3x 1x2 xf + =
- 25. SQUEEZE THEOREM
- 26. LIMITS OF FUNCTIONS USING THE SQUEEZE PRINCIPLE The Squeeze Principle is used on limit problems where the usual algebraic methods (factoring, conjugation, algebraic manipulation, etc.) are not effective. However, it requires that you be able to ``squeeze'' your problem in between two other ``simpler'' functions whose limits are easily computable and equal. The use of the Squeeze Principle requires accurate analysis, algebra skills, and careful use of inequalities. The method of squeezing is used to prove that f(x)L as xc by trapping or squeezing f between two functions, g and h, whose limits as xc are known with certainty to be L.
- 27. SQUEEZE PRINCIPLE : Lf(x)limthen h(x)limLg(x)limand h(x)f(x)g(x)satisfyhand,g,ffunctionsthatAssume ax axax = ==
- 28. Theorem 1.6.5 (p. 123) Figure 1.6.3 (p. 123)
- 29. EXAMPLE: x cos4x-cos3x-2 lim4. x5sin x3sin lim.3 x x2sin lim.2 x xtan lim.1 limits.followingtheEvaluate 0x0x 0x0x ( )( ) 111 xcos 1 lim x xsin lim xcos 1 x xsin lim x xtan lim.1 0x0x 0x0x == = = ( )( ) 212 x2 x2sin lim2 2 2 x x2sin lim x x2sin lim.2 0x 0x0x == = = SOLUTION:
- 30. 5 3 15 13 x5 x5sin 5 x3 x3sin 3 lim x x5sin x x3sin lim x5sin x3sin lim.3 0x 0x0x = = = = ( ) ( ) ( ) 00403 4x cos4x-1 lim4 3x cos3x-1 lim3 x cos4x-1 lim x cos3x-1 lim x x4cosx3cos11 lim x cos4x-cos3x-2 lim4. 0x0x 0x0x 0x 0x =+= + = + = + =
- 31. 3x2x 2xx lim.4 4x x16 lim.3 4t 2t lim.2 x9 x4 lim.1 2 2 3x 2 4x 2 2t 2 2 3x ++ + + EXERCISES: Evaluate the following limits:
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