MATH 136 The Squeezing Theorem Suppose g(x) and h(x) are known functions, with g(x) ≤ h(x), and that g(x) and h(x) both converge to the same limit L at a particular point

€

a .

a

Lg(x )

h(x )

limx→a

g(x ) = L = limx→ a

h( x)

Then let f (x) be any (unknown) function in between g and h : g(x) ≤ f (x) ≤ h(x) . The Squeezing Theorem states that lim

x→af ( x) must exist and must be the same value L .

a

L f ( x)

€

limx→ a

f (x) = L also.

For example, suppose there is some function f (x) such that

2sin πx

2

≤ f (x) ≤ (x −1)2 + 2 , for all

€

x .

What can you conclude from the graphs?

2 sin(π x / 2)

(x − 1)2 + 2f ( x)

1

Because

€

limx→1

2sin πx2

= 2 and lim

x→1( x − 1)2 + 2 = 2 (the functions converge to the

same limit at x = 1), then by the Squeezing Theorem limx→1

f ( x) = 2 also for any function

f (x) such that 2sin πx

2

≤ f (x) ≤ (x −1)2 + 2 .

Special Case

In particular, suppose g(x) ≤ f (x) ≤ h(x) for

€

a < x < b , and g(x) and h(x) have the same one-sided limit L at one of the endpoints. Then by the Squeezing Theorem, the one-sided limit of f (x) exists and is the same value L at that endpoint. Example 1. Suppose

€

3x − 2 < f (x) < 4 − (x − 2)2 , for

€

0 < x < 2. Graph the outer two functions on the interval. Then explain which one-sided limit can be determined and compute this limit. Solution.

€

4 − (x − 2)2

€

3x − 2

€

f (x) in between

We see that

€

limx→ 2–

(3x − 2) = 4 and

€

limx→ 2–

4 − (x − 2)2 = 4 .

Because

€

3x − 2 < f (x) < 4 − (x − 2)2 on

€

0 < x < 2, by the Squeezing Theorem, we must have

€

limx→ 2−

f (x) = 4 also.

Example 2. Suppose

€

−x3 − 4 < f (x) < 3cos(π x) , for

€

−1< x <1. Graph the outer two functions on the interval. Then explain which one-sided limit can be determined and compute this limit. Solution.

€

3cos(π x)

€

−x3 − 4

€

f (x) in between

We see that

€

limx→−1+

− x3 − 4 = −3 and

€

limx→−1+

3cos(π x) = −3.

Because

€

−x3 − 4 < f (x) < 3cos(π x) on

€

−1< x <1, by the Squeezing Theorem, we must have

€

limx→−1+

f (x) = −3 also.

Example 3. For − π

2< θ < 0 and 0 < θ <

π

2, we have

cos θ < sin θθ

< sec θ . What can we conclude?

sec θ

0cos θ

sin θθ

On the left interval, we have lim

θ→ 0−cos θ = 1 = lim

θ →0−sec θ ; thus, lim

θ→ 0−sin θθ

= 1 also.

On the right interval we have limθ→ 0+

cos θ = 1 = limθ →0+

sec θ ; thus, limθ→ 0+

sin θθ

= 1 as well.

We can conclude that limθ→ 0

sin θθ

= 1 .

We note that sin θ

θ is not defined when

€

θ = 0, and direct substitution of

€

θ = 0 yields

a 0/0 indeterminate. But we now know that limθ→ 0

sin θθ

= 1. This limit is one of the most

important results in calculus. It will be used later to derive the derivative of the sine function. To obtain this limit, we used the inequality cos θ < sin θ

θ< sec θ , which we

simply stated after observing the graphs of each function. But this inequality also can be derived geometrically as follows: Let

€

C be a circle with center

€

O and a radius of length 1. Let

€

θ be the measure of an angle in radians, where 0 < θ < π / 2 . Geometrically, we measure

€

θ from the center

€

O between two radii OA and OB .

θO A

B

1

1

Figure 1

θO

B

C

1

Figure 2

The area of the entire circle of radius 1 is π (1)2 = π , and the area of the resulting sector OAB in Figure 1 is θ

2π× π =

θ2

.

On the unit circle, the lengths of triangle sides OC and BC in Figure 2 are given by OC =

€

cosθ and BC =

€

sinθ . Thus the area of the right triangle

€

OCB is 12× cosθ × sin θ .

Because the area of triangle

€

OCB is less than the area of sector

€

OAB , we have

cos θ × sin θ2

<θ2

, which gives sin θθ

<1

cos θ, and thus sin θ

θ< sec θ .

On the exterior right triangle in Figure 3 , we have tan θ =

AD1

; thus, tan θ = AD . The area of this triangle is then

12× 1 × tan θ =

sin θ2 cosθ

.

Because the area of triangle

€

OAD is greater than the area of sector

€

OAB , we have

θ2<sin θ2cos θ

which gives cos θ < sin θθ

.

θO A

D

1

Figure 3

Thus, cos θ < sin θ

θ< sec θ for 0 < θ < π / 2 . Because lim

θ → 0+cosθ = lim

θ → 0+sec θ = 1,

we have limθ → 0+

sinθθ

= 1 by the Squeezing Theorem.

Now consider − π / 2 < θ < 0. Then 0 < −θ < π / 2 . Thus,

cos(−θ ) < sin(−θ )−θ

< sec(−θ ) But cos(−θ ) = cos(θ ) and so sec(−θ ) = sec(θ ). However sin(−θ ) = − sin(θ ) so sin(−θ )−θ

=− sin(θ )−θ

=sin(θ )θ

, so we still have cos θ < sin θθ

< sec θ for 0 < −θ < π / 2 .

Applying the Squeezing Thoerem again gives limθ → 0−

sinθθ

= 1 .

Now because lim

θ → 0−

sinθθ

= 1 and limθ → 0+

sinθθ

= 1 , we have limθ → 0

sin θθ

= 1 .

The following limit also will be needed later to derive the derivative of

€

sin x :

limx→ 0

cos x − 1x

= 0

We note that when first substituting

€

x = 0, we obtain the indeterminate form 0/0. But with some clever algebra we obtain

limx→ 0

cos x − 1x

= limx→0

cos x −1x

×cos x +1cos x +1

= limx→ 0

cos2 x −1x(cos x + 1)

= − limx→ 0

1 − cos2 xx(cos x + 1)

= − limx→ 0

sin2 xx (cos x +1)

= − limx→ 0

sin xx

×sin x

(cos x +1)= − lim

x→ 0

sin xx

× limx→ 0

sin xcos x + 1

= (−1) × 02= 0.