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
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
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