In this video we learn how to solve limits that involve trigonometric functions. It is all based on using the fundamental trigonometric limit, which is proved using the squeeze theorem. For more lessons: http://www.intuitive-calculus.com/solving-limits.html Watch video: http://www.youtube.com/watch?v=1RqXMJWcRIA

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

8. Example 1
lim
x→0
sin 4x
x

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

20. Example 2
lim
x→0
tan x
x

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