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Basic Calculus Lesson 2
- 27. 1. For 𝑓 𝑥 = 7−x 3− 𝑥+2 , evaluate: a. lim 𝑥→5 𝑓(𝑥) b. lim 𝑥→5+ 𝑓(𝑥) c. lim 𝑥→5− 𝑓(𝑥) lim 𝑥→5 𝑓 𝑥 = 7 − x 3 − 𝑥 + 2 = 𝟕 − 𝒙 𝟑 − 𝒙 + 𝟐 ∙ 𝟑 + 𝒙 + 𝟐 𝟑 + 𝒙 + 𝟐 = 𝟕 − 𝒙 𝟑 + 𝒙 + 𝟐 𝟗 − 𝒙 − 𝟐 = 𝟕 − 𝒙 𝟑 + 𝒙 + 𝟐 𝟕 − 𝒙 = 𝟑 + 𝒙 + 𝟐 a. lim 𝑥→5 𝑓(𝑥)= 𝟑 + 𝒙 + 𝟐 = lim 𝑥→5 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟓 + 𝟐 = 𝟑 + 𝟕 = 𝟑 + 𝟐. 𝟔𝟓 = 𝟓. 𝟔𝟓
- 28. x → 𝟓+ 6 7 8 f(x) =𝟑 + 𝒙 + 𝟐 𝟓. 𝟔𝟓 𝟓. 𝟖𝟑 𝟔 𝟔. 𝟏𝟔 𝒍𝒊𝒎 𝒙→𝟓+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟓+ 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟔 + 𝟐 b. lim 𝑥→5+ 𝑓(𝑥) = 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟖 = 𝟑 + 𝟐. 𝟖𝟑 = 𝟓. 𝟖𝟑 𝒍𝒊𝒎 𝒙→𝟓+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟓+ 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟕 + 𝟐 = 𝟑 + 𝟗 = 𝟑 + 𝟑 = 𝟔 𝒍𝒊𝒎 𝒙→𝟓+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟓+ 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟖 + 𝟐 = 𝟑 + 𝟏𝟎 = 𝟑 + 𝟑. 𝟏𝟔 = 𝟔. 𝟏𝟔
- 29. x 2 3 4 → 𝟓− f(x) =𝟑 + 𝒙 + 𝟐 𝟓. 𝟔𝟓𝟓 𝟓. 𝟐𝟒 𝟓. 𝟒𝟓 𝒍𝒊𝒎 𝒙→𝟓− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟓− 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟐 + 𝟐 c. lim 𝑥→5− 𝑓 𝑥 = 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟒 = 𝟑 + 𝟐 = 𝟓 𝒍𝒊𝒎 𝒙→𝟓− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟓− 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟑 + 𝟐 = 𝟑 + 𝟓 = 𝟑 + 𝟐. 𝟐𝟒 = 𝟓. 𝟐𝟒 𝒍𝒊𝒎 𝒙→𝟓− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟓− 𝟑 + 𝒙 + 𝟐 = 𝟑 + 𝟒 + 𝟐 = 𝟑 + 𝟔 = 𝟑 + 𝟐. 𝟒𝟓 = 𝟓. 𝟒𝟓
- 30. 𝐥𝐢𝐦 𝒙→𝟓+ 𝟑 + 𝒙 + 𝟐 𝐥𝐢𝐦 𝒙→𝟓− 𝟑 + 𝒙 + 𝟐= x → 𝟓+ 6 7 8 f(x) =𝟑 + 𝒙 + 𝟐 𝟓. 𝟔𝟓 𝟓. 𝟖𝟑 𝟔 𝟔. 𝟏𝟔 x 2 3 4 → 𝟓− f(x) =𝟑 + 𝒙 + 𝟐 𝟓. 𝟔𝟓𝟓 𝟓. 𝟐𝟒 𝟓. 𝟒𝟓
- 31. 2. For 𝑓 𝑥 = 1 x − 1 11 x−11 , evaluate: a. lim 𝑥→11 𝑓(𝑥) b. lim 𝑥→11+ 𝑓(𝑥) c. lim 𝑥→11− 𝑓(𝑥) a. lim 𝑥→11 𝑓(𝑥) = 1 x − 1 11 x − 11 = 11 − 𝑥 11𝑥 x − 11 = −(𝑥 − 11) 11𝑥 x − 11 ∙ 1 𝑥 − 11 1 𝑥 − 11 = −𝟏 𝟏𝟏𝒙 a. 𝒍𝒊𝒎 𝐱→𝟏𝟏 𝐟(𝐱) = − 𝟏 𝟏𝟏𝐱 = − 𝟏 𝟏𝟏(𝟏𝟏) = − 𝟏 𝟏𝟐𝟏 𝒐𝒓 − 𝟎. 𝟎𝟎𝟖
- 32. x → 𝟏𝟏+ 12 13 14 f(x) =− 𝟏 𝟏𝟏𝒙 − 𝟏 𝟏𝟐𝟏 − 𝟏 𝟏𝟑𝟐 − 𝟏 𝟏𝟒𝟑 − 𝟏 𝟏𝟓𝟒 𝒍𝒊𝒎 𝒙→𝟏𝟏+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏𝟏+ − 𝟏 𝟏𝟏𝒙 b. lim 𝑥→11+ 𝑓 𝑥 = − 𝟏 𝟏𝟏𝒙 = − 𝟏 𝟏𝟏(𝟏𝟐) = − 𝟏 𝟏𝟑𝟐 𝒍𝒊𝒎 𝒙→𝟏𝟏+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏𝟏+ − 𝟏 𝟏𝟏𝒙 = − 𝟏 𝟏𝟏(𝟏𝟑) = − 𝟏 𝟏𝟒𝟑 𝒍𝒊𝒎 𝒙→𝟏𝟏+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏𝟏+ − 𝟏 𝟏𝟏𝒙 = − 𝟏 𝟏𝟏(𝟏𝟒) = − 𝟏 𝟏𝟓𝟒
- 33. x 8 9 10 → 𝟏𝟏− f(x) =− 𝟏 𝟏𝟏𝒙 − 𝟏 𝟏𝟐𝟏 − 𝟏 𝟖𝟖 − 𝟏 𝟗𝟗 − 𝟏 𝟏𝟏𝟎 𝒍𝒊𝒎 𝒙→𝟏𝟏− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏𝟏− − 𝟏 𝟏𝟏𝒙 c. lim 𝑥→11− 𝑓 𝑥 = − 𝟏 𝟏𝟏𝒙 = − 𝟏 𝟏𝟏(𝟖) = − 𝟏 𝟖𝟖 𝒍𝒊𝒎 𝒙→𝟏𝟏− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏𝟏− − 𝟏 𝟏𝟏𝒙 = − 𝟏 𝟏𝟏(𝟗) = − 𝟏 𝟗𝟗 𝒍𝒊𝒎 𝒙→𝟏𝟏− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏𝟏− − 𝟏 𝟏𝟏𝒙 = − 𝟏 𝟏𝟏(𝟏𝟎) = − 𝟏 𝟏𝟏𝟎
- 34. lim 𝑥→11− − 𝟏 𝟏𝟏𝒙 = x → 𝟏𝟏+ 12 13 14 f(x) =− 𝟏 𝟏𝟏𝒙 − 𝟏 𝟏𝟐𝟏 − 𝟏 𝟏𝟑𝟐 − 𝟏 𝟏𝟒𝟑 − 𝟏 𝟏𝟓𝟒 x 8 9 10 → 𝟏𝟏− f(x) =− 𝟏 𝟏𝟏𝒙 − 𝟏 𝟏𝟐𝟏 − 𝟏 𝟖𝟖 − 𝟏 𝟗𝟗 − 𝟏 𝟏𝟏𝟎 lim 𝑥→11+ − 𝟏 𝟏𝟏𝒙
- 35. 3. For 𝑓 𝑥 = x−7 2−49 𝑥 , evaluate: a. lim 𝑥→0 𝑓(𝑥) b. lim 𝑥→0+ 𝑓(𝑥) c. lim 𝑥→0− 𝑓(𝑥) a. 𝒍𝒊𝒎 𝐱→𝟎 𝐟(𝐱) = 𝐱 − 𝟕 𝟐 − 𝟒𝟗 𝒙 = 𝒙 𝟐 − 𝟏𝟒𝒙 + 𝟒𝟗 − 𝟒𝟗 𝒙 = 𝒙 𝟐 − 𝟏𝟒𝒙 𝒙 = 𝒙(𝒙 − 𝟏𝟒) 𝒙 = 𝒙 − 𝟏𝟒 a. 𝒍𝒊𝒎 𝐱→𝟎 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = 𝟎 − 𝟏𝟒 = −𝟏𝟒
- 36. x → 𝟎+ 1 2 3 f(x) =− 𝟏 𝟏𝟏𝒙 −𝟏𝟒 −𝟏𝟑 −𝟏𝟐 −𝟏𝟏 b. 𝒍𝒊𝒎 𝐱→𝟎+ 𝐟(𝐱) = 𝒙 − 𝟏𝟒 b. 𝒍𝒊𝒎 𝐱→𝟎+ 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = 𝟏 − 𝟏𝟒 = −𝟏𝟑 b. 𝒍𝒊𝒎 𝐱→𝟎+ 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = 𝟐 − 𝟏𝟒 = −𝟏𝟐 b. 𝒍𝒊𝒎 𝐱→𝟎+ 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = 𝟑 − 𝟏𝟒 = −𝟏𝟏
- 37. x -3 -2 -1 → 𝟎− f(x) =− 𝟏 𝟏𝟏𝒙 −𝟏𝟒−𝟏𝟕 −𝟏𝟔 −𝟏𝟓 c. 𝒍𝒊𝒎 𝐱→𝟎− 𝐟(𝐱) = 𝒙 − 𝟏𝟒 b. 𝒍𝒊𝒎 𝐱→𝟎− 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = −𝟑 − 𝟏𝟒 = −𝟏𝟕 b. 𝒍𝒊𝒎 𝐱→𝟎− 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = −𝟐 − 𝟏𝟒 = −𝟏𝟔 b. 𝒍𝒊𝒎 𝐱→𝟎− 𝐟(𝐱) = 𝒙 − 𝟏𝟒 = −𝟏 − 𝟏𝟒 = −𝟏𝟓
- 38. lim 𝑥→0− 𝒙 − 𝟏𝟒=lim 𝑥→0+ 𝒙 − 𝟏𝟒 x -3 -2 -1 → 𝟎− f(x) =− 𝟏 𝟏𝟏𝒙 −𝟏𝟒−𝟏𝟕 −𝟏𝟔 −𝟏𝟓 x → 𝟎+ 1 2 3 f(x) =− 𝟏 𝟏𝟏𝒙 −𝟏𝟒 −𝟏𝟑 −𝟏𝟐 −𝟏𝟏
- 40. 𝐥𝐢𝐦 𝒙→𝟎 𝟒 + 𝒙 𝟐 − 𝟏𝟔 𝒙 𝐥𝐢𝐦 𝒙→𝟒 𝟐𝒙 𝟐 − 𝟏𝟐𝒙 − 𝟏𝟔 𝒙 𝟐 − 𝒙 − 𝟏𝟐 𝐥𝐢𝐦 𝒙→𝟑𝟔 𝟔 − 𝒙 𝟑𝟔 − 𝒙 𝐥𝐢𝐦 𝒙→𝟑𝟔 𝟔 − 𝒙 𝟑𝟔 − 𝒙
- 45. Get the value of 𝐥𝐢𝐦 𝒙→𝟑− 𝒇(𝒙 + 𝟑) using a table of function values. x 2.9 2.99 2.999 → 𝟑− f(x) = x+3 𝒍𝒊𝒎 𝒙→𝟑− )𝒇(𝒙 + 𝟑 = lim 𝑥→3− )𝑓(2.9 + 3 = 2.9 + 3 = 5.9 𝒍𝒊𝒎 𝒙→𝟑− )𝒇(𝒙 + 𝟑 𝟓. 𝟗 = lim 𝑥→3− )𝑓(2.9𝟗 + 3 = 2. 𝟗9 + 3 = 5.9𝟗 𝟓. 𝟗𝟗 𝒍𝒊𝒎 𝒙→𝟑− )𝒇(𝒙 + 𝟑 = lim 𝑥→3− )𝑓(2.9𝟗𝟗 + 3 = 2. 𝟗9𝟗 + 3 = 5.9𝟗𝟗 𝟓. 𝟗𝟗9 𝒍𝒊𝒎 𝒙→𝟑− )𝒇(𝒙 + 𝟑 = lim 𝑥→3− )𝑓(𝟑 + 3 = 𝟑 + 3 = 𝟔 𝟔
- 46. Get the value of 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇(𝒙 + 𝟑) using a table of function values. x → 𝟑+ 3.001 3.01 3.1 f(x) = x+3 𝒍𝒊𝒎 𝒙→𝟑+ )𝒇(𝒙 + 𝟑 = lim 𝑥→3+ )𝑓(𝟑 + 3 = 𝟑 + 3 = 𝟔 𝒍𝒊𝒎 𝒙→𝟑+ )𝒇(𝒙 + 𝟑 𝟔 = lim 𝑥→3+ )𝑓(𝟑. 𝟎𝟎𝟏 + 3 = 𝟑. 𝟎𝟎𝟏 + 3 = 𝟔. 𝟎𝟎𝟏 𝟔. 𝟎𝟎𝟏 𝒍𝒊𝒎 𝒙→𝟑+ )𝒇(𝒙 + 𝟑 = lim 𝑥→3+ )𝑓(𝟑. 𝟎𝟏 + 3 = 𝟑. 𝟎𝟏 + 3 = 𝟔. 𝟎𝟏 𝟔. 𝟎𝟏 𝒍𝒊𝒎 𝒙→𝟑+ )𝒇(𝒙 + 𝟑 = lim 𝑥→3+ )𝑓(𝟑. 𝟏 + 3 = 𝟑. 𝟏 + 3 = 𝟔. 𝟏 𝟔.1
- 47. x → 𝟑+ 3.001 3.01 3.1 f(x) = x+3 𝒍𝒊𝒎 𝒙→𝟑+ )𝒇(𝒙 + 𝟑 x 2.9 2.99 2.999 → 𝟑− f(x) = x+3 𝒍𝒊𝒎 𝒙→𝟑− )𝒇(𝒙 + 𝟑 𝟓. 𝟗 𝟓. 𝟗𝟗 𝟓. 𝟗𝟗9 𝟔 𝟔 𝟔. 𝟎𝟎𝟏 𝟔. 𝟎𝟏 𝟔.1 𝒍𝒊𝒎 𝒙→𝟑− )𝒇(𝒙 + 𝟑 𝒍𝒊𝒎 𝒙→𝟑+ )𝒇(𝒙 + 𝟑=
- 48. For 𝑓 𝑥 = 9 𝑥−3 5, evaluate: a. lim 𝑥→3 𝑓(𝑥) b. lim 𝑥→3+ 𝑓(𝑥) c. lim 𝑥→3− 𝑓(𝑥) a. lim 𝑥→3 𝑓 𝑥 = lim 𝑥→3 𝑓 9 𝑥 − 3 5 = 𝟗 𝟑−𝟑 𝟓 = 9 0 5 = 9 0
- 49. Get the value of 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇 9 𝑥−3 5 using a table of function values. x → 𝟑+ 4 5 6 f(x) = 9 𝑥−3 5 𝐃𝐍𝐄 𝟗 𝟎. 𝟐𝟖 𝟎. 𝟎𝟑𝟕 𝒍𝒊𝒎 𝒙→𝟑+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇 𝟗 𝒙 − 𝟑 𝟓 = 9 4 − 3 5 = 9 1 5 = 9 1 = 9 𝒍𝒊𝒎 𝒙→𝟑+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇 𝟗 𝒙 − 𝟑 𝟓 = 9 𝟓 − 3 5 = 9 𝟐 5 = 9 𝟑𝟐 = 𝟎. 𝟐𝟖 𝒍𝒊𝒎 𝒙→𝟑+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇 𝟗 𝒙 − 𝟑 𝟓 = 9 𝟔 − 3 5 = 9 𝟑 5 = 9 𝟐𝟒𝟑 = 𝟏 𝟐𝟕 𝐨𝐫 𝟎. 𝟎𝟑𝟕
- 50. Get the value of 𝐥𝐢𝐦 𝒙→𝟑− 𝒇 9 𝑥−3 5 using a table of function values. x 0 1 2 → 𝟑− f(x) = 9 𝑥−3 5 𝐃𝐍𝐄−𝟎. 𝟐𝟖 −𝟗−𝟎. 𝟎𝟑𝟕 𝒍𝒊𝒎 𝒙→𝟑− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑− 𝒇 𝟗 𝒙 − 𝟑 𝟓 = 9 𝟐 − 3 5 = 9 −1 5 = 9 −1 = −9 𝒍𝒊𝒎 𝒙→𝟑− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑− 𝒇 𝟗 𝒙 − 𝟑 𝟓 = 9 𝟏 − 3 5 = 9 −𝟐 5 = 9 −𝟑𝟐 = −𝟎. 𝟐𝟖 𝒍𝒊𝒎 𝒙→𝟑− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑− 𝒇 𝟗 𝒙 − 𝟑 𝟓 = 9 𝟎 − 3 5 = 9 −𝟑 5 = 9 −𝟐𝟒𝟑 = −𝟏 𝟐𝟕 𝐨𝐫 −𝟎. 𝟎𝟑𝟕
- 51. 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇 𝟗 𝒙 − 𝟑 𝟓 𝐥𝐢𝐦 𝒙→𝟑− 𝒇 𝟗 𝒙 − 𝟑 𝟓≠ x 0 1 2 → 𝟑− f(x) = 9 𝑥−3 5 𝐃𝐍𝐄−𝟎. 𝟐𝟖 −𝟗−𝟎. 𝟎𝟑𝟕 𝐥𝐢𝐦 𝒙→𝟑− 𝒇 𝟗 𝒙 − 𝟑 𝟓 𝐥𝐢𝐦 𝒙→𝟑+ 𝒇 𝟗 𝒙 − 𝟑 𝟓 x → 𝟑+ 4 5 6 f(x) = 9 𝑥−3 5 𝐃𝐍𝐄 𝟗 𝟎. 𝟐𝟖 𝟎. 𝟎𝟑𝟕
- 52. For 𝑓 𝑥 = 𝑥2 − 3𝑥 + 4, evaluate:a. lim 𝑥→1 𝑓(𝑥) b. lim 𝑥→1+ 𝑓(𝑥) c. lim 𝑥→1− 𝑓(𝑥) a. lim 𝑥→1 𝑓 𝑥 = lim 𝑥→1 𝑥2 − 3𝑥 + 4 =(1)2 −3(1) + 4 = 𝟏 − 𝟑 + 𝟒
- 53. Get the value of 𝐥𝐢𝐦 𝒙→𝟏+ 𝑥2 − 3𝑥 + 4, using a table of function values. x → 𝟏+ 2 3 4 f(x) = 𝑥2 − 3𝑥 + 4 𝟐 𝟐 𝟒 𝟖 𝒍𝒊𝒎 𝒙→𝟏+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏+ 𝒙 𝟐 − 𝟑𝒙 + 𝟒 = (𝟐) 𝟐−𝟑(𝟐) + 𝟒 = 𝟐= 𝟒 − 𝟔 + 𝟒 𝒍𝒊𝒎 𝒙→𝟏+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏+ 𝒙 𝟐 − 𝟑𝒙 + 𝟒 𝒍𝒊𝒎 𝒙→𝟏+ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏+ 𝒙 𝟐 − 𝟑𝒙 + 𝟒 = (𝟑) 𝟐−𝟑(𝟑) + 𝟒 = 𝟒= 𝟗 − 𝟗 + 𝟒 = (𝟒) 𝟐−𝟑(𝟒) + 𝟒 = 𝟖=16-12+4
- 54. Get the value of 𝐥𝐢𝐦 𝒙→𝟏− 𝑥2 − 3𝑥 + 4, using a table of function values. x -4 -3 -2 → 𝟏− f(x) = 𝑥2 − 3𝑥 + 4 𝟑𝟐 𝟐𝟐 𝟏𝟒 𝟐 𝒍𝒊𝒎 𝒙→𝟏− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏− 𝒙 𝟐 − 𝟑𝒙 + 𝟒 = (−𝟐) 𝟐−𝟑(−𝟐) + 𝟒 =14= 𝟒 + 𝟔 + 𝟒 𝒍𝒊𝒎 𝒙→𝟏− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏− 𝒙 𝟐 − 𝟑𝒙 + 𝟒 𝒍𝒊𝒎 𝒙→𝟏− 𝒇 𝒙 = 𝐥𝐢𝐦 𝒙→𝟏− 𝒙 𝟐 − 𝟑𝒙 + 𝟒 = (−𝟑) 𝟐−𝟑(−𝟑) + 𝟒 = 𝟐𝟐= 𝟗 + 𝟗 + 𝟒 = (−𝟒) 𝟐−𝟑(−𝟒) + 𝟒 = 𝟑𝟐= 𝟏𝟔 + 𝟏𝟐 + 𝟒
- 55. ≠ x -4 -3 -2 → 𝟏− f(x) = 𝑥2 − 3𝑥 + 4 𝟐𝟐𝟐 𝟏𝟒𝟑𝟐 𝐥𝐢𝐦 𝒙→𝟏− 𝒙 𝟐 − 𝟑𝒙 + 𝟒 𝐥𝐢𝐦 𝒙→𝟏+ 𝒙 𝟐 − 𝟑𝒙 + 𝟒 x → 𝟏+ 2 3 4 f(x) = 𝑥2 − 3𝑥 + 4 2 𝟐 𝟒 𝟖 𝐥𝐢𝐦 𝒙→𝟏+ 𝒙 𝟐 − 𝟑𝒙 + 𝟒 𝐥𝐢𝐦 𝒙→𝟏− 𝒙 𝟐 − 𝟑𝒙 + 𝟒
- 57. 1. For 𝑓 𝑥 = 7−x 3− 𝑥+2 , evaluate: a. lim 𝑥→5 𝑓(𝑥) b. lim 𝑥→5+ 𝑓(𝑥) c. lim 𝑥→5− 𝑓(𝑥) 2. For 𝑓 𝑥 = 1 x − 1 11 x−11 , evaluate: a. lim 𝑥→11 𝑓(𝑥) b. lim 𝑥→11+ 𝑓(𝑥) c. lim 𝑥→11− 𝑓(𝑥) 3. For 𝑓 𝑥 = x−7 2−49 𝑥 , evaluate: a. lim 𝑥→0 𝑓(𝑥) b. lim 𝑥→0+ 𝑓(𝑥) c. lim 𝑥→0− 𝑓(𝑥)
- 58. QUIZ #2 a. lim 𝑥→5 𝑓(𝑥) b. lim 𝑥→5+ 𝑓(𝑥) c. lim 𝑥→5− 𝑓(𝑥)
- 60. First, assume that 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 and 𝒍𝒊𝒎 𝒙→𝒂 𝒈 𝒙 exist, and c is any constant. Then, 𝟏. 𝐥𝐢𝐦 𝒙→𝒂 𝒄𝒇 𝒙 = 𝒄 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 In other words, we can “factor” a multiplicative constant out of a limit.
- 61. 𝟏. 𝐥𝐢𝐦 𝒙→𝒂 𝒄𝒇 𝒙 = 𝒄 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝟐 𝟑𝒙 𝟐 = 𝟑 𝐥𝐢𝐦 𝒙→𝟐 𝒙 𝟐 = 𝟑 𝟐 𝟐 = 𝟑(𝟒) = 𝟏𝟐
- 62. 𝟐. 𝐥𝐢𝐦 𝒙→𝒂 )𝒇 𝒙 ± 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 ± 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 So, to take the limit of a sum or difference all we need to do is take the limit of the individual parts and then put them back together with the appropriate sign. This is also not limited to two functions. This fact will work no matter how many functions we’ve got separated by “+” or “-”.
- 63. 𝒇 𝒙 = 𝒙 𝟐 − 𝒙 − 𝟔; 𝒈 𝒙 = 𝒙 𝟐 − 𝟐𝒙 − 𝟑 𝟐. 𝐥𝐢𝐦 𝒙→𝒂 )𝒇 𝒙 ± 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 ± 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 ± 𝒈(𝒙) 𝐥𝐢𝐦 𝒙→3 )𝒇 𝒙 + 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→3 𝒙2 − 𝒙 − 6 + 𝐥𝐢𝐦 𝒙→3 𝒙2 − 2𝒙 − 3 = 𝐥𝐢𝐦 𝒙→3 (𝒙2 −𝒙 − 6) + ( 𝒙2 − 2𝒙 − 3) = 𝐥𝐢𝐦 𝒙→3 𝟐𝒙 𝟐 − 𝟑𝒙 − 𝟗 = 𝟐 𝟑 𝟐 − 𝟑 𝟑 − 𝟗 = 𝟏𝟖 − 𝟗 − 𝟗 𝐥𝐢𝐦 𝒙→3 )𝒇 𝒙 + 𝒈(𝒙 = 𝟎
- 64. 𝒇 𝒙 = 𝒙 𝟐 − 𝒙 − 𝟔; 𝒈 𝒙 = 𝒙 𝟐 − 𝟐𝒙 − 𝟑 𝟐. 𝐥𝐢𝐦 𝒙→𝒂 )𝒇 𝒙 ± 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 ± 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 ± 𝒈(𝒙) 𝐥𝐢𝐦 𝒙→3 )𝒇 𝒙 − 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→3 𝒙2 − 𝒙 − 6 − 𝐥𝐢𝐦 𝒙→3 𝒙2 − 2𝒙 − 3 = 𝐥𝐢𝐦 𝒙→3 (𝒙2 −𝒙 − 6) − ( 𝒙2 − 2𝒙 − 3) = 𝐥𝐢𝐦 𝒙→3 𝒙2 − 𝒙 − 6 − 𝒙 𝟐 + 𝟐𝒙 + 𝟑 = 𝟑 − 𝟑 𝐥𝐢𝐦 𝒙→3 )𝒇 𝒙 + 𝒈(𝒙 = 𝟎 = 𝐥𝐢𝐦 𝒙→3 𝒙 − 𝟑
- 65. 3. 𝐥𝐢𝐦 𝒙→𝒂 )𝒇 𝒙 ∙ 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 ∙ 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 We take the limits of products in the same way that we can take the limit of sums or differences. Just take the limit of the pieces and then put them back together. Also, as with sums or differences, this fact is not limited to just two functions.
- 66. 𝒇 𝒙 = 𝒙 𝟐 − 𝒙 − 𝟔; 𝒈 𝒙 = 𝒙 𝟐 − 𝟐𝒙 − 𝟑 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 ∙ 𝒈(𝒙) 𝐥𝐢𝐦 𝒙→3 )𝒇 𝒙 ∙ 𝒈(𝒙 = (𝐥𝐢𝐦 𝒙→3 𝒙2 − 𝒙 − 6)(𝐥𝐢𝐦 𝒙→3 𝒙2 − 2𝒙 − 3) 𝐥𝐢𝐦 𝒙→3 )𝒇 𝒙 ∙ 𝒈(𝒙 = 𝟎 3. 𝐥𝐢𝐦 𝒙→𝒂 )𝒇 𝒙 ∙ 𝒈(𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 ∙ 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→3 (𝒙 𝟐 − 𝒙 − 𝟔)(𝒙 𝟐 − 𝟐𝒙 − 𝟑) = 𝐥𝐢𝐦 𝒙→3 𝒙2 𝒙2 − 2𝒙 − 3 + −𝒙 𝒙2 − 2𝒙 − 3 + −6 𝒙2 − 2𝒙 − 3 = 𝐥𝐢𝐦 𝒙→3 [(𝒙 𝟒 = 𝐥𝐢𝐦 𝒙→3 𝒙4 − 3𝒙3 − 𝟕𝒙2 + 15𝒙 + 18 = 𝟑 𝟒 − 𝟑 𝟑 𝟑 − 𝟕 𝟑 𝟐 + 𝟏𝟓 𝟑 + 𝟏𝟖 = 𝟖𝟏 − 𝟖𝟏 − 𝟔𝟑 + 𝟒𝟓 + 𝟏𝟖 −𝟐𝒙 𝟑 −𝟑𝒙 𝟐 )+(−𝒙 𝟑+𝟐𝒙 𝟐 +𝟑𝒙)+(−𝟔𝒙 𝟐 +𝟏𝟐𝒙 +𝟏𝟖)
- 67. 4. 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 Provided that 𝒍𝒊𝒎 𝒙→𝒂 𝒈 𝒙 ≠ 𝟎 As noted in the statement we only need to worry about the limit in the denominator being zero when we do the limit of a quotient. If it were zero we would end up with a division by zero error and we need to avoid that.
- 68. 𝒇 𝒙 = 𝒙 𝟐 − 𝒙 − 𝟔; 𝒈 𝒙 = 𝒙 𝟐 − 𝟐𝒙 − 𝟑 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 /𝒈(𝒙) 4. 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝐥𝐢𝐦 𝒙→𝒂 𝒈 𝒙 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐 − 𝒙 − 𝟔 𝒙 𝟐 − 𝟐𝒙 − 𝟑 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 − 𝟑)(𝒙 + 𝟐 𝒙 − 𝟑)(𝒙 + 𝟏 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 + 𝟐 𝒙 + 𝟏 = 𝟑 + 𝟐 𝟑 + 𝟏 𝐥𝐢𝐦 𝒙→3 𝒇 𝒙 𝒈 𝒙 = 5 4 𝒐𝒓 1 1 4 𝒐𝒓 1.25
- 69. 5. 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝒏 = [ 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝒏 Where n is any real number. In this property n can be any real number (positive, negative, integer, fraction, irrational, zero, etc.). That n is an integer this rule can be thought of as an extended case of 3. For example, consider the case of n=2. 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝟐 = [𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 𝟐 = 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙 ∙ 𝒍𝒊𝒎 𝒙→𝒂 𝒇 𝒙
- 70. 𝒇 𝒙 = 𝒙 + 𝟑; Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 𝟐 5. 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝒏 = [ 𝐥𝐢𝐦 𝒙→𝒂 𝒇 𝒙 𝒏 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 + 𝟑 𝟐 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 + 𝟑 ∙ 𝐥𝐢𝐦 𝒙→𝟑 𝒙 + 𝟑 = 𝐥𝐢𝐦 𝒙→𝟑 (𝒙 + 𝟑)(𝒙 + 𝟑 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐 + 𝟔𝒙 + 𝟗 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐 + 𝐥𝐢𝐦 𝒙→𝟑 𝟔𝒙 + 𝐥𝐢𝐦 𝒙→𝟑 𝟗 = 𝟑 𝟐 + 𝟔 𝟑 + 𝟗 = 𝟗 + 𝟏𝟖 + 𝟗 𝐥𝐢𝐦 𝒙→3 𝒇 𝒙 2 = 36
- 71. 𝟔. 𝐥𝐢𝐦 𝒙→𝒂 [ 𝒏 𝒇(𝒙) = 𝒏 𝐥𝐢𝐦 𝒙→𝒂 𝒇(𝒙) This is just a special case of the previous example. 𝐥𝐢𝐦 𝒙→𝒂 [ 𝒏 𝒇(𝒙) = 𝒏 𝐥𝐢𝐦 𝒙→𝒂 𝒇(𝒙) =[lim 𝑥→𝑎 𝒇(𝒙) 𝟏 𝒏
- 72. 𝒇 𝒙 = 𝒙 𝟐 + 𝟔𝒙 + 𝟗; Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 𝟏 𝟐 = 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 𝟔. 𝐥𝐢𝐦 𝒙→𝒂 [ 𝒏 𝒇(𝒙) = 𝒏 𝐥𝐢𝐦 𝒙→𝒂 𝒇(𝒙) = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐 + 𝟔𝒙 + 𝟗 = lim 𝒙→𝟑 𝒙 𝟐 + 𝟔𝒙 + 𝟗 = lim 𝒙→𝟑 𝒙 + 𝟑 𝟐 = lim 𝒙→𝟑 𝒙 + 𝟑 = 𝟑 + 𝟑 𝐥𝐢𝐦 𝒙→𝟑 𝒇 𝒙 𝟏 𝟐 = 𝟔
- 73. 𝟕. 𝐥𝐢𝐦 𝒙→𝒂 𝒄 = 𝒄 In other words, the limit of a constant is just the constant. You should be able to convince yourself of this by drawing the graph of f(x) = c. Example: 𝒇 𝒙 = 𝟗 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇(𝒙) 𝒍𝐢𝐦 𝒙→𝟑 𝟗 = 𝟗
- 74. 𝟖. 𝐥𝐢𝐦 𝒙→𝒂 𝒙 = 𝒂 As with the last one you should be able to convince yourself of this by drawing the graph of f(x) = x. Example: 𝒇 𝒙 = 𝒙 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇(𝒙) 𝒍𝐢𝐦 𝒙→𝟑 𝐱 = 𝟑= 𝐥𝐢𝐦 𝒙→𝟑 𝒙 = 𝐥𝐢𝐦 𝒙→3 3
- 75. 𝟗. 𝐥𝐢𝐦 𝒙→𝒂 𝒙 𝒏 = 𝒂 𝒏 This is really just a special case of property 5 using f(x) = x. Example: 𝒇 𝒙 = 𝒙 𝟑 Given the 𝐥𝐢𝐦 𝒙→𝟑 𝒇(𝒙) 𝒍𝐢𝐦 𝒙→𝟑 𝐱 𝐧 = 𝟗= 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟑 = 𝐥𝐢𝐦 𝒙→3 3 𝟑
- 77. For 𝒇 𝒙 = 𝟏𝟔𝒙 𝟐 − 𝟐𝟒𝒙 + 𝟗 𝒂𝒏𝒅 𝒈 𝒙 = 𝟒𝒙 𝟐 + 𝒙 − 𝟑; evaluate the following limits. 𝒂. lim 𝒙→𝟏 𝒇 𝒙 + 𝒈(𝒙) 𝒃. lim 𝒙→𝟏 𝒇 𝒙 − 𝒈(𝒙) 𝒄. lim 𝒙→𝟐 𝒇 𝒙 𝒈 𝒙 𝒅. lim 𝒙→𝟑 𝒈 𝒙 𝟐 𝒆. lim 𝒙→𝟏 𝒇 𝒙 ∙ 𝒈(𝒙) 𝒇. lim 𝒙→𝟐𝟏 𝒇 𝒙 𝟏 𝟐