53

Trigonometric Limits

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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

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Page 1: Trigonometric Limits
Page 2: Trigonometric Limits
Page 3: Trigonometric Limits

The Fundamental Trig Limit

Page 4: Trigonometric Limits

The Fundamental Trig Limit

This is it:

Page 5: Trigonometric Limits

The Fundamental Trig Limit

This is it:

limx→0

sin x

x= 1

Page 6: Trigonometric Limits

The Fundamental Trig Limit

This is it:

limx→0

sin x

x= 1

This is proved using the squeeze theorem!

Page 7: Trigonometric Limits

Example 1

Page 8: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Page 9: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

Page 10: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x=

Page 11: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

Page 12: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

Page 13: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

= 4 limx→0

sin 4x

4x

Page 14: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

= 4 limx→0�

���>

1sin 4x

4x

Page 15: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

= 4 limx→0

sin 4x

4x

Page 16: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

= 4 limx→0

sin 4x

4x= 4.1

Page 17: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

= 4 limx→0

sin 4x

4x= 4.1 = 4

Page 18: Trigonometric Limits

Example 1

limx→0

sin 4x

x

Here the trick is to multiply and divide by 4:

limx→0

sin 4x

x= lim

x→0

4 sin 4x

4x

= 4 limx→0

sin 4x

4x= 4.1 = 4

Page 19: Trigonometric Limits

Example 2

Page 20: Trigonometric Limits

Example 2

limx→0

tan x

x

Page 21: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

Page 22: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x=

Page 23: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

Page 24: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

Page 25: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

= limx→0

sin x

x.

1

cos x

Page 26: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

= limx→0�

���

1sin x

x.

1

cos x

Page 27: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

= limx→0

sin x

x.����

11

cos x

Page 28: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

= limx→0

sin x

x.

1

cos x=

Page 29: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

= limx→0

sin x

x.

1

cos x= 1

Page 30: Trigonometric Limits

Example 2

limx→0

tan x

x

Here we use a basic trig identity:

limx→0

tan x

x= lim

x→0

sin xcos x

x

= limx→0

sin x

x.

1

cos x= 1

Page 31: Trigonometric Limits

Example 3

Page 32: Trigonometric Limits

Example 3

limx→0

1 − cos x

x

Page 33: Trigonometric Limits

Example 3

limx→0

1 − cos x

x

In this case a somewhat more elaborate trig identity will be useful:

Page 34: Trigonometric Limits

Example 3

limx→0

1 − cos x

x

In this case a somewhat more elaborate trig identity will be useful:

sinx

2=

√1 − cos x

2

Page 35: Trigonometric Limits

Example 3

limx→0

1 − cos x

x

In this case a somewhat more elaborate trig identity will be useful:

sinx

2=

√1 − cos x

2

Squaring both sides and solving for 1 − cos x we get:

Page 36: Trigonometric Limits

Example 3

limx→0

1 − cos x

x

In this case a somewhat more elaborate trig identity will be useful:

sinx

2=

√1 − cos x

2

Squaring both sides and solving for 1 − cos x we get:

1 − cos x = 2 sin2 x

2

Page 37: Trigonometric Limits

Example 3

Let’s replace that in our limit:

Page 38: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x

Page 39: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x=

Page 40: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

Page 41: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

Page 42: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Page 43: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

Page 44: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= 2 limx→0

sin x2

x2 .2

. sinx

2

Page 45: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

Page 46: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0

sin x2

x2

. sinx

2

Page 47: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0����7

1sin x

2x2

. sinx

2

Page 48: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0

sin x2

x2

.����

0

sinx

2

Page 49: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0

sin x2

x2

. sinx

2

Page 50: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0

sin x2

x2

. sinx

2= 1.0

Page 51: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0

sin x2

x2

. sinx

2= 1.0 = 0

Page 52: Trigonometric Limits

Example 3

Let’s replace that in our limit:

limx→0

1 − cos x

x= lim

x→0

2 sin2 x2

x

= 2 limx→0

sin x2

x. sin

x

2

Now the trick is to multiply and divide by 2 in the first factor:

= �2 limx→0

sin x2

x2 .�2

. sinx

2

= limx→0

sin x2

x2

. sinx

2= 1.0 = 0

Page 53: Trigonometric Limits