Day 3 of Free Intuitive Calculus Course: Limits by Factoring
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The document discusses the concept of limits in calculus, focusing on specific examples and how to evaluate them by factoring. It demonstrates the process of resolving indeterminate forms at certain values by simplifying expressions. Additionally, it elaborates on the expansion of polynomial expressions and the application of limits as a method for finding derivatives.
Day 3 of Free Intuitive Calculus Course: Limits by Factoring
- 2. Example 1
- 3. Example 1 Let’s consider the limit:
- 4. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2
- 5. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate.
- 6. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!.
- 7. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor:
- 8. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 =
- 9. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 = lim x→2
- 10. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 = lim x→2 (x − 2)(x + 2) x − 2
- 11. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 = lim x→2 $$$$(x − 2)(x + 2) $$$x − 2
- 12. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 = lim x→2 $$$$(x − 2)(x + 2) $$$x − 2 = lim x→2 (x + 2)
- 13. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 = lim x→2 $$$$(x − 2)(x + 2) $$$x − 2 = lim x→2 (x + 2) = 2 + 2 =
- 14. Example 1 Let’s consider the limit: lim x→2 x2 − 4 x − 2 We note that at x = 2, our function is indeterminate. It equals 0 0!. But we can factor: lim x→2 x2 − 4 x − 2 = lim x→2 $$$$(x − 2)(x + 2) $$$x − 2 = lim x→2 (x + 2) = 2 + 2 = 4
- 15. Example 1
- 16. Example 1 What is the graph of this function?
- 17. Example 1 What is the graph of this function? f (x) = x2 − 2 x + 2
- 18. Example 1 What is the graph of this function? f (x) = x2 − 2 x + 2
- 19. Example 1 What is the graph of this function? f (x) = x2 − 2 x + 2 It is the graph of x + 2, but with a hole!
- 20. Example 2
- 21. Example 2 lim x→1 x3 − 1 x − 1
- 22. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator?
- 23. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 =
- 24. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 = lim x→1
- 25. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 = lim x→1 (x − 1)(x2 + x + 1) x − 1
- 26. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 = lim x→1 $$$$(x − 1)(x2 + x + 1) $$$x − 1
- 27. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 = lim x→1 $$$$(x − 1)(x2 + x + 1) $$$x − 1 = lim x→1 x2 + x + 1
- 28. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 = lim x→1 $$$$(x − 1)(x2 + x + 1) $$$x − 1 = lim x→1 x2 + x + 1 = 12 + 1 + 1
- 29. Example 2 lim x→1 x3 − 1 x − 1 Remember how to factor the numerator? lim x→1 x3 − 1 x − 1 = lim x→1 $$$$(x − 1)(x2 + x + 1) $$$x − 1 = lim x→1 x2 + x + 1 = 12 + 1 + 1 = 3
- 30. Example 3
- 31. Example 3 lim h→0 (a + h)3 − a3 h
- 32. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3:
- 33. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h
- 34. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h =
- 35. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h
- 36. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h = lim h→0
- 37. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h = lim h→0 h 3a2 + 3ah + h2 h
- 38. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h = lim h→0 ¡h 3a2 + 3ah + h2 ¡h
- 39. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h = lim h→0 ¡h 3a2 + 3ah + h2 ¡h = lim h→0 3a2 + 3ah + h2
- 40. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h = lim h→0 ¡h 3a2 + 3ah + h2 ¡h = lim h→0 3a2 + 3ah + h2 = 3a2 + 3a.0 + 02
- 41. Example 3 lim h→0 (a + h)3 − a3 h Here a is a constant. Let’s expand (a + h)3: lim h→0 a3 + 3a2h + 3ah2 + h3 − a3 h = lim h→0 3a2h + 3ah2 + h3 h = lim h→0 ¡h 3a2 + 3ah + h2 ¡h = lim h→0 3a2 + 3ah + h2 = 3a2 + 3a.0 + 02 = 3a2