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# Introductory maths analysis chapter 10 official

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### Introductory maths analysis chapter 10 official

1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 10Chapter 10 Limits and ContinuityLimits and Continuity
2. 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
3. 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
4. 4. ©2007 Pearson Education Asia • To study limits and their basic properties. • To study one-sided limits, infinite limits, and limits at infinity. • To study continuity and to find points of discontinuity for a function. • To develop techniques for solving nonlinear inequalities. Chapter 10: Limits and Continuity Chapter ObjectivesChapter Objectives
5. 5. ©2007 Pearson Education Asia Limits Limits (Continued) Continuity Continuity Applied to Inequalities 10.1) 10.2) 10.3) Chapter 10: Limits and Continuity Chapter OutlineChapter Outline 10.4)
6. 6. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.1 Limits10.1 Limits Example 1 – Estimating a Limit from a Graph • The limit of f(x) as x approaches a is the number L, written as a. Estimate limx→1 f (x) from the graph. Solution: b. Estimate limx→1 f (x) from the graph. Solution: ( ) Lxf ax = → lim ( ) 2lim 1 = → xf x ( ) 2lim 1 = → xf x
7. 7. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.1 Limits Properties of Limits 1. 2. for any positive integer n 3. 4. 5. ( ) constantaiswherelimlim cccxf axax == →→ nn ax ax = → lim ( ) ( )[ ] ( ) ( )xgxfxgxf axaxax →→→ ±=± limlimlim ( ) ( )[ ] ( ) ( )xgxfxgxf axaxax →→→ ⋅=⋅ limlimlim ( )[ ] ( )xfcxcf axax →→ ⋅= limlim
8. 8. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.1 Limits Example 3 – Applying Limit Properties 1 and 2 Properties of Limits ( ) 162limc. 366limb. 77lim;77lima. 44 2 22 6 52 =−= == == → → −→→ t x t x xx ( ) ( ) ( ) ( ) ( ) 0limif lim lim lim6. ≠= → → → → xg xg xf xg xf ax ax ax ax ( ) ( )n ax n ax xfxf →→ = limlim7.
9. 9. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.1 Limits Example 5 – Limit of a Polynomial Function Find an expression for the polynomial function, Solution: where ( ) 01 1 1 ... cxcxcxcxf n n n n ++++= − − ( ) ( ) ( )af cacacac ccxcxc cxcxcxcxf n n n n axax n ax n n ax n n n n n axax = ++++= ++++= ++++= − − →→ − → − → − − →→ 01 1 1 01 1 1 01 1 1 ... limlim...limlim ...limlim ( ) ( )afxf ax = → lim
10. 10. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.1 Limits Example 7 – Finding a Limit Example 9 – Finding a Limit Find . Solution: If ,find . Solution: 1 1 lim 2 1 + − → x x x ( ) 2111lim 1 1 lim 1 2 1 −=−−=−= + − −→−→ x x x xx ( ) 12 += xxf ( ) ( ) h xfhxf h −+ →0 lim ( ) ( ) [ ] ( ) xhx h xhxhx h xfhxf h hh 22lim 112 limlim 0 222 00 =+= −−+++ = −+ → →→ Limits and Algebraic Manipulation • If f (x) = g(x) for all x ≠ a, then ( ) ( )xgxf axax →→ = limlim
11. 11. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.2 Limits (Continued)10.2 Limits (Continued) Example 1 – Infinite Limits Infinite Limits • Infinite limits are written as and . Find the limit (if it exists). Solution: a. The results are becoming arbitrarily large. The limit does not exist. b. The results are becoming arbitrarily large. The limit does not exist. ∞=+ −→ xx 1 lim 0 −∞=− −→ xx 1 lim 0 1 2 lima. 1 ++ −→ xx 4 2 limb. 22 − + → x x x
12. 12. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.2 Limits (Continued) Example 3 – Limits at Infinity Find the limit (if it exists). Solution: a. b. ( )3 5 4 lima. −∞→ xx ( ) 0 5 4 lim 3 = −∞→ xx ( )x x − ∞→ 4limb. ( ) ∞=− ∞→ x x 4lim Limits at Infinity for Rational Functions • If f (x) is a rational function, and( ) m m n n xx xb xa xf ∞→∞→ = limlim ( ) m m n n xx xb xa xf −∞→−∞→ = limlim
13. 13. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.2 Limits (Continued) Example 5 – Limits at Infinity for Polynomial Functions Find the limit (if it exists). Solution: Solution: ( ) ∞=−=+− −∞→−∞→ 33 2lim92lim xxx xx ( ) −∞==−+− −∞→−∞→ 323 lim2lim xxxx xx ( ) 33 2lim92limb. xxx xx −=+− −∞→−∞→ ( ) 323 lim2lima. xxxx xx −∞→−∞→ =−+−
14. 14. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.3 Continuity10.3 Continuity Example 1 – Applying the Definition of Continuity Definition • f(x) is continuous if three conditions are met: a. Show that f(x) = 5 is continuous at 7. Solution: Since , . b. Show that g(x) = x2 − 3 is continuous at −4. Solution: ( ) ( ) ( ) ( )afxf xf xf = → → ax ax lim3. existslim2. exists1. ( ) 55limlim 77 == →→ xx xf ( ) ( )75lim 7 fxf x == → ( ) ( ) ( )43limlim 2 44 −=−= −→−→ gxxg xx
15. 15. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.3 Continuity Example 3 – Discontinuities a. When does a function have infinite discontinuity? Solution: A function has infinite discontinuity at a when at least one of the one-sided limits is either ∞ or −∞ as x →a. b. Find discontinuity for Solution: f is defined at x = 0 but limx→0 f (x) does not exist. f is discontinuous at 0. ( )      <− = > = 0if1 0if0 0if1 x x x xf
16. 16. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.3 Continuity Example 5 – Locating Discontinuities in Case-Defined Functions For each of the following functions, find all points of discontinuity. ( )    < ≥+ = 3if 3if6 a. 2 xx xx xf ( )    < >+ = 2if 2if2 b. 2 xx xx xf
17. 17. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.3 Continuity Example 5 – Locating Discontinuities in Case-Defined Functions Solution: a. We know that f(3) = 3 + 6 = 9. Because and , the function has no points of discontinuity. ( ) ( ) 96limlim 33 =+= ++ →→ xxf xx ( ) 9limlim 2 33 == −→→ − xxf xx
18. 18. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.3 Continuity Example 5 – Locating Discontinuities in Case-Defined Functions Solution: b. It is discontinuous at 2, limx→2 f (x) exists. ( ) ( )xfxxxf xxxx +−−− →→→→ =+=== 22 2 22 lim2lim4limlim
19. 19. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.4 Continuity Applied to Inequalities10.4 Continuity Applied to Inequalities Example 1 – Solving a Quadratic Inequality Solve . Solution: Let . To find the real zeros of f, Therefore, x2 − 3x − 10 > 0 on (−∞,−2) ∪ (5,∞). 01032 >−− xx ( ) 1032 −−= xxxf ( )( ) 5,2 052 01032 −= =−+ =−− x xx xx
20. 20. ©2007 Pearson Education Asia Chapter 10: Limits and Continuity 10.4 Continuity Applied to Inequalities Example 3 – Solving a Rational Function Inequality Solve . Solution: Let . The zeros are 1 and 5. Consider the intervals: (−∞, 0) (0, 1) (1, 5) (5,∞) Thus, f(x) ≥ 0 on (0, 1] and [5,∞). 0 562 ≥ +− x xx ( ) ( )( ) x xx x xx xf 51562 −− = +− =