# Lesson 5: Continuity

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<ul><li> 1. Section 2.4 Continuity Math 1a October 3, 2007 </li></ul>
<p> 2. Questions True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature. 3. Questions True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature. True or FalseAt one point in your life your height in inches equaled your weightin pounds. 4. Questions True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature. True or FalseAt one point in your life your height in inches equaled your weightin pounds. True or FalseAt one point in your life you were exactly three feet tall. 5. Direct Substitution PropertyTheorem (The Direct Substitution Property)If f is a polynomial or a rational function and a is in the domain off , then lim f (x) = f (a)xa 6. Denition of ContinuityDenitionLet f be a function dened near a. We say that f is continuous ata if lim f (x) = f (a).xa 7. Free Theorems Theorem(a) Any polynomial is continuous everywhere; that is, it iscontinuous on R = (, ). (b) Any rational function is continuous wherever it is dened; that is, it is continuous on its domain. 8. Showing a function is continuousExampleLet f (x) = 4x + 1. Show that f is continuous at 2. 9. Showing a function is continuousExampleLet f (x) = 4x + 1. Show that f is continuous at 2. SolutionWe have lim f (x) = lim4x + 1xax2= lim (4x + 1) x2 = 9 = 3. Each step comes from the limit laws. 10. Showing a function is continuousExampleLet f (x) = 4x + 1. Show that f is continuous at 2. SolutionWe have lim f (x) = lim4x + 1xax2= lim (4x + 1) x2 = 9 = 3. Each step comes from the limit laws.In fact, f is continuous on its whole domain, which is 1 , . 4 11. The Limit Laws give Continuity Laws TheoremIf f and g are continuous at a and c is a constant, then thefollowing functions are also continuous at a: 1. f + g 2. f g 3. cf 4. fgf 5.(if g (a) = 0)g 12. Transcendental functions are continuous, tooTheoremThe following functions are continuous wherever they are dened: 1. sin, cos, tan, cot sec, csc 2. x ax , loga , ln 3. sin1 , tan1 , sec1 13. What could go wrong? In what ways could a function f fail to be continuous at a point a? Look again at the denition: lim f (x) = f (a)xa 14. Pitfall #1ExampleLet x2if 0 x 1 f (x) = if 1 < x 2 2xAt which points is f continuous? 15. Pitfall #1: The limit does not existExampleLet x2if 0 x 1 f (x) = if 1 < x 2 2xAt which points is f continuous? SolutionAt any point a in [0, 2] besides 1, lim f (x) = f (a) because f isxarepresented by a polynomial near a, and polynomials have thedirect substitution property. However, lim f (x) = lim x 2 = 12 = 1 x1x1 lim f (x) = lim+ 2x = 2(1) = 2x1+ x1 So f has no limit at 1. Therefore f is not continuous at 1. 16. Pitfall #2 ExampleLet x 2 + 2x + 1 f (x) = x +1At which points is f continuous? 17. Pitfall #2: The function has no value ExampleLet x 2 + 2x + 1 f (x) = x +1At which points is f continuous? SolutionBecause f is rational, it is continuous on its whole domain. Notethat 1 is not in the domain of f , so f is not continuous there. 18. Pitfall #3 ExampleLet 46 if x = 1 f (x) = if x = 1At which points is f continuous? 19. Pitfall #3: function value = limit ExampleLet46 if x = 1f (x) =if x = 1At which points is f continuous? Solutionf is not continuous at 1 because f (1) = but lim f (x) = 46.x1 20. Special types of discontinuitesremovable discontinuity The limit lim f (x) exists, but f is notxa dened at a or its value at a is not equal to the limit at a.jump discontinuity The limits lim f (x) and lim+ f (x) exist, but xaxa are dierent. f (a) is one of these limits. 21. Special types of discontinuitesremovable discontinuity The limit lim f (x) exists, but f is notxa dened at a or its value at a is not equal to the limit at a.jump discontinuity The limits lim f (x) and lim+ f (x) exist, butxaxaare dierent. f (a) is one of these limits.The greatest integer function f (x) = [[x]] has jump discontinuities. 22. A Big Time Theorem Theorem (The Intermediate Value Theorem) Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N. 23. Illustrating the IVT f (x)x 24. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] f (x) x 25. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]f (x) f (b) f (a) xa b 26. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) = f (b). f (x) f (b)Nf (a) xab 27. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) = f (b). Thenthere exists a number c in (a, b) such that f (c) = N. f (x) f (b)Nf (a) xac b 28. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) = f (b). Thenthere exists a number c in (a, b) such that f (c) = N. f (x) f (b)Nf (a) xab 29. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) = f (b). Thenthere exists a number c in (a, b) such that f (c) = N. f (x) f (b)Nf (a) xa c1c2 c3 b 30. Using the IVTExampleProve that the square root of two exists. 31. Using the IVTExampleProve that the square root of two exists. Proof.Let f (x) = x 2 , a continuous function on [1, 2]. 32. Using the IVTExampleProve that the square root of two exists. Proof.Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such thatf (c) = c 2 = 2. 33. Using the IVTExampleProve that the square root of two exists. Proof.Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such thatf (c) = c 2 = 2.In fact, we can narrow in on the square root of 2 by the methodof bisections. </p>