Download - Day 5 of the Intuitive Online Calculus Course: The Squeeze Theorem

The Squeeze Theorem

The Squeeze Theorem

Let’s say f and g are two functions such that one is always belowthe other. Let’s say that g is always above:

The Squeeze Theorem


f (x) ≤ g(x)

The Squeeze Theorem


f (x) ≤ g(x)

Also, let’s say that their limits are equal in some point:

The Squeeze Theorem


f (x) ≤ g(x)

Also, let’s say that their limits are equal in some point:

limx→a

f (x) = limx→a

g(x) = L

The Squeeze Theorem

The Squeeze Theorem

Now, let’s suppose we squeeze a third function between them:

The Squeeze Theorem


Let’s call this third function h:

The Squeeze Theorem


Let’s call this third function h:

f (x) ≤ h(x) ≤ g(x)

The Squeeze Theorem


The Squeeze Theorem


The Squeze Theorem says that:

The Squeeze Theorem


The Squeze Theorem says that:

limx→a

h(x) = L

The Fundamental Trigonometric Limit


We’re going to prove the following limit using the SqueezeTheorem


We’re going to prove the following limit using the SqueezeTheorem

limx→0

sin x

x= 1


∆MOA < Sector MOA < ∆COA


sin x =


sin x =BM

1


sin x =BM

1= BM


sin x =BM

1= BM

∆MOA =


sin x =BM

1= BM

∆MOA =1.BM

2


sin x =BM

1= BM

∆MOA =1.BM

2=

sin x

2


��:

sin x2



sin x

2< Sector MOA < ∆COA


tan x =


tan x =AC

1


tan x =AC

1= AC


tan x =AC

1= AC

∆COA =


tan x =AC

1= AC

∆COA =1.AC

2=


tan x =AC

1= AC

∆COA =1.AC

2=

tan x

2


tan x =AC

1= AC

sin x

2< Sector MOA < ∆COA


tan x =AC

1= AC

sin x

2< Sector MOA <��

�:tan x2

∆COA


tan x =AC

1= AC

sin x

2< Sector MOA <

tan x

2


The area of a circular sector is:



A =θr2

2



A =θr2

2

A simple proof is the following:



A =θr2

2


A(2π) = πr2



A =θr2

2


A(2π) = πr2

If we multiply both sides of this equation by θ2π :



A =θr2

2


A(2π) = πr2


θ

2πA(2π)



A =θr2

2


A(2π) = πr2


θ

2πA(2π) = A(

θ2π

2π)



A =θr2

2


A(2π) = πr2


θ

2πA(2π) = A(

θ��2π

��2π)



A =θr2

2


A(2π) = πr2


θ

2πA(2π) = A(

θ��2π

��2π) = A(θ) =



A =θr2

2


A(2π) = πr2


θ

2πA(2π) = A(

θ2π

2π) = A(θ) =

θπr2

2π



A =θr2

2


A(2π) = πr2


θ

2πA(2π) = A(

θ2π

2π) = A(θ) =

θ�πr2

2�π



A =θr2

2


A(2π) = πr2


θ

2πA(2π) = A(

θ2π

2π) = A(θ) =

θ�πr2

2�π=θr2

2


Now, we have a circle of radius 1 and angle x , so, our area is:



Sector MOA =



Sector MOA =x .12

2=



Sector MOA =x .12

2=

x

2



sin x

2< Sector MOA <

tan x

2



sin x

2<��

��:

x2

Sector MOA <tan x

2



sin x

2<

x

2<

tan x

2



sin x

2<

x

2<

tan x

2

sin x < x < tan x


sin x < x < tan x


sin x < x < tan x

1 <x

sin x<

1

cos x


sin x < x < tan x

1 <x

sin x<

1

cos x

1 >sin x

x> cos x


sin x < x < tan x

1 <x

sin x<

1

cos x

1 >sin x

x>��:

1cos x


sin x < x < tan x

1 <x

sin x<

1

cos x

1 >sin x

x>��:

1cos x

Conclusion:


sin x < x < tan x

1 <x

sin x<

1

cos x

1 >sin x

x>��:

1cos x

Conclusion:

limx→0

sin x

x= 1

Download - Day 5 of the Intuitive Online Calculus Course: The Squeeze Theorem

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