Trigonometric Limits
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The document discusses fundamental limits involving trigonometric functions as x approaches 0, specifically lim x→0 sin x/x = 1. It provides examples using limits of sin 4x/x, tan x/x, and (1 - cos x)/x, demonstrating techniques like multiplying and dividing to evaluate these limits. The examples utilize basic trigonometric identities and the squeeze theorem to prove these limits.
Trigonometric Limits
- 3. The Fundamental Trig Limit
- 4. The Fundamental Trig Limit This is it:
- 5. The Fundamental Trig Limit This is it: lim x→0 sin x x = 1
- 6. The Fundamental Trig Limit This is it: lim x→0 sin x x = 1 This is proved using the squeeze theorem!
- 7. Example 1
- 9. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4:
- 10. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x =
- 11. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0
- 12. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x
- 13. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x = 4 lim x→0 sin 4x 4x
- 14. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x = 4 lim x→0 & & &&b 1 sin 4x 4x
- 15. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x = 4 lim x→0 sin 4x 4x
- 16. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x = 4 lim x→0 sin 4x 4x = 4.1
- 17. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x = 4 lim x→0 sin 4x 4x = 4.1 = 4
- 18. Example 1 lim x→0 sin 4x x Here the trick is to multiply and divide by 4: lim x→0 sin 4x x = lim x→0 4 sin 4x 4x = 4 lim x→0 sin 4x 4x = 4.1 = 4
- 19. Example 2
- 21. Example 2 lim x→0 tan x x Here we use a basic trig identity:
- 22. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x =
- 23. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0
- 24. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x
- 25. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x = lim x→0 sin x x . 1 cos x
- 26. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x = lim x→0 1 sin x x . 1 cos x
- 27. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x = lim x→0 sin x x . 1 1 cos x
- 28. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x = lim x→0 sin x x . 1 cos x =
- 29. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x = lim x→0 sin x x . 1 cos x = 1
- 30. Example 2 lim x→0 tan x x Here we use a basic trig identity: lim x→0 tan x x = lim x→0 sin x cos x x = lim x→0 sin x x . 1 cos x = 1
- 31. Example 3
- 32. Example 3 lim x→0 1 − cos x x
- 33. Example 3 lim x→0 1 − cos x x In this case a somewhat more elaborate trig identity will be useful:
- 34. Example 3 lim x→0 1 − cos x x In this case a somewhat more elaborate trig identity will be useful: sin x 2 = 1 − cos x 2
- 35. Example 3 lim x→0 1 − cos x x In this case a somewhat more elaborate trig identity will be useful: sin x 2 = 1 − cos x 2 Squaring both sides and solving for 1 − cos x we get:
- 36. Example 3 lim x→0 1 − cos x x In this case a somewhat more elaborate trig identity will be useful: sin x 2 = 1 − cos x 2 Squaring both sides and solving for 1 − cos x we get: 1 − cos x = 2 sin2 x 2
- 37. Example 3 Let’s replace that in our limit:
- 38. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x
- 39. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x =
- 40. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0
- 41. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x
- 42. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2
- 43. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor:
- 44. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = 2 lim x→0 sin x 2 x 2 .2 . sin x 2
- 45. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2
- 46. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 sin x 2 x 2 . sin x 2
- 47. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 U 1 sin x 2 x 2 . sin x 2
- 48. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 sin x 2 x 2 . 0 sin x 2
- 49. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 sin x 2 x 2 . sin x 2
- 50. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 sin x 2 x 2 . sin x 2 = 1.0
- 51. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 sin x 2 x 2 . sin x 2 = 1.0 = 0
- 52. Example 3 Let’s replace that in our limit: lim x→0 1 − cos x x = lim x→0 2 sin2 x 2 x = 2 lim x→0 sin x 2 x . sin x 2 Now the trick is to multiply and divide by 2 in the first factor: = ¡2 lim x→0 sin x 2 x 2 .¡2 . sin x 2 = lim x→0 sin x 2 x 2 . sin x 2 = 1.0 = 0