Day 5 of the Intuitive Online Calculus Course: The Squeeze Theorem
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The document discusses the squeeze theorem, which states that if one function is squeezed between two others that converge to the same limit, then it also converges to that limit. It illustrates this concept by proving the fundamental trigonometric limit, lim x→0 sin x/x = 1, using the squeeze theorem. Additionally, it explores relationships between sin x, x, and tan x within the context of circular sectors and their areas.
Day 5 of the Intuitive Online Calculus Course: The Squeeze Theorem
- 3. The Squeeze Theorem Let’s say f and g are two functions such that one is always below the other. Let’s say that g is always above:
- 4. The Squeeze Theorem Let’s say f and g are two functions such that one is always below the other. Let’s say that g is always above: f (x) ≤ g(x)
- 5. The Squeeze Theorem Let’s say f and g are two functions such that one is always below the other. Let’s say that g is always above: f (x) ≤ g(x) Also, let’s say that their limits are equal in some point:
- 6. The Squeeze Theorem Let’s say f and g are two functions such that one is always below the other. Let’s say that g is always above: f (x) ≤ g(x) Also, let’s say that their limits are equal in some point: lim x→a f (x) = lim x→a g(x) = L
- 8. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them:
- 9. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them:
- 10. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them: Let’s call this third function h:
- 11. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them: Let’s call this third function h: f (x) ≤ h(x) ≤ g(x)
- 12. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them:
- 13. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them: The Squeze Theorem says that:
- 14. The Squeeze Theorem Now, let’s suppose we squeeze a third function between them: The Squeze Theorem says that: lim x→a h(x) = L
- 15. The Fundamental Trigonometric Limit
- 16. The Fundamental Trigonometric Limit We’re going to prove the following limit using the Squeeze Theorem
- 17. The Fundamental Trigonometric Limit We’re going to prove the following limit using the Squeeze Theorem lim x→0 sin x x = 1
- 18. The Fundamental Trigonometric Limit
- 19. The Fundamental Trigonometric Limit
- 20. The Fundamental Trigonometric Limit
- 21. The Fundamental Trigonometric Limit ∆MOA < Sector MOA < ∆COA
- 22. The Fundamental Trigonometric Limit sin x =
- 23. The Fundamental Trigonometric Limit sin x = BM 1
- 24. The Fundamental Trigonometric Limit sin x = BM 1 = BM
- 25. The Fundamental Trigonometric Limit sin x = BM 1 = BM ∆MOA =
- 26. The Fundamental Trigonometric Limit sin x = BM 1 = BM ∆MOA = 1.BM 2
- 27. The Fundamental Trigonometric Limit sin x = BM 1 = BM ∆MOA = 1.BM 2 = sin x 2
- 28. The Fundamental Trigonometric Limit ∆MOA < Sector MOA < ∆COA
- 29. The Fundamental Trigonometric Limit $$$$X sin x 2 ∆MOA < Sector MOA < ∆COA
- 30. The Fundamental Trigonometric Limit sin x 2 < Sector MOA < ∆COA
- 31. The Fundamental Trigonometric Limit
- 32. The Fundamental Trigonometric Limit tan x =
- 33. The Fundamental Trigonometric Limit tan x = AC 1
- 34. The Fundamental Trigonometric Limit tan x = AC 1 = AC
- 35. The Fundamental Trigonometric Limit tan x = AC 1 = AC ∆COA =
- 36. The Fundamental Trigonometric Limit tan x = AC 1 = AC ∆COA = 1.AC 2 =
- 37. The Fundamental Trigonometric Limit tan x = AC 1 = AC ∆COA = 1.AC 2 = tan x 2
- 38. The Fundamental Trigonometric Limit tan x = AC 1 = AC sin x 2 < Sector MOA < ∆COA
- 39. The Fundamental Trigonometric Limit tan x = AC 1 = AC sin x 2 < Sector MOA <$$$$X tan x 2 ∆COA
- 40. The Fundamental Trigonometric Limit tan x = AC 1 = AC sin x 2 < Sector MOA < tan x 2
- 41. The Fundamental Trigonometric Limit
- 42. The Fundamental Trigonometric Limit The area of a circular sector is:
- 43. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2
- 44. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following:
- 45. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2
- 46. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π :
- 47. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π)
- 48. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π) = A( θ2π 2π )
- 49. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π) = A( 訨2𠨨2π )
- 50. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π) = A( 訨2𠨨2π ) = A(θ) =
- 51. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π) = A( θ2π 2π ) = A(θ) = θπr2 2π
- 52. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π) = A( θ2π 2π ) = A(θ) = θ&πr2 2&π
- 53. The Fundamental Trigonometric Limit The area of a circular sector is: A = θr2 2 A simple proof is the following: A(2π) = πr2 If we multiply both sides of this equation by θ 2π : θ 2π A(2π) = A( θ2π 2π ) = A(θ) = θ&πr2 2&π = θr2 2
- 54. The Fundamental Trigonometric Limit
- 55. The Fundamental Trigonometric Limit
- 56. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is:
- 57. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: Sector MOA =
- 58. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: Sector MOA = x.12 2 =
- 59. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: Sector MOA = x.12 2 = x 2
- 60. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: sin x 2 < Sector MOA < tan x 2
- 61. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: sin x 2 <$$$$$$$X x 2 Sector MOA < tan x 2
- 62. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: sin x 2 < x 2 < tan x 2
- 63. The Fundamental Trigonometric Limit Now, we have a circle of radius 1 and angle x, so, our area is: sin x 2 < x 2 < tan x 2 sin x < x < tan x
- 64. The Fundamental Trigonometric Limit
- 65. The Fundamental Trigonometric Limit sin x < x < tan x
- 66. The Fundamental Trigonometric Limit sin x < x < tan x 1 < x sin x < 1 cos x
- 67. The Fundamental Trigonometric Limit sin x < x < tan x 1 < x sin x < 1 cos x 1 > sin x x > cos x
- 68. The Fundamental Trigonometric Limit sin x < x < tan x 1 < x sin x < 1 cos x 1 > sin x x >$$$X1 cos x
- 69. The Fundamental Trigonometric Limit sin x < x < tan x 1 < x sin x < 1 cos x 1 > sin x x >$$$X1 cos x Conclusion:
- 70. The Fundamental Trigonometric Limit sin x < x < tan x 1 < x sin x < 1 cos x 1 > sin x x >$$$X1 cos x Conclusion: lim x→0 sin x x = 1