calcnotes0206.pdf

7/28/2019 CalcNotes0206.pdf

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(Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.1

SECTION 2.6: THE SQUEEZE (SANDWICH) THEOREM

LEARNING OBJ ECTIVES

Understand and be able to rigorously apply the Squeeze (Sandwich) Theoremwhen evaluating limits at a point and long-run limits at ( ) infinity.

PART A: APPLY ING THE SQUEEZE (SANDWICH) THEOREM TOLIMITS AT A POINT

We will formally statethe Squeeze (Sandwich) Theorem in Part B.

Example 1 below is one of many basic examples where we use the Squeeze

(Sandwich) Theorem to show that limx0

f x

( )= 0, where f x

( )is theproduct of a

sine or cosine expressionand amonomial of even degree.

The idea is that something approaching 0 times something bounded(that is, trapped between two real numbers) will approach 0. Informally,

Limit Form 0 bounded( ) 0 .

Example 1 (Applying the Squeeze (Sandwich) Theorem to a Limit at a Point)

Let f x( ) = x2 cos 1x

. Prove that lim

x0f x( ) = 0.

Solution

We firstboundcos1

x

,

which is real for all x 0 .

Multiplyall three parts by x2

so that the middle part becomesf x( ) .

WARNING 1: Wemust observe

that x2 > 0 for all x 0 ,or at least on apuncturedneighborhood ofx= 0 , so that

wecan multiply by x2 withoutreversing inequality symbols.

1 cos1

x

1 x 0( )

x2

x2

cos

1

x

x

2

x 0( )


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As x 0, theleft and rightparts approach 0. Therefore,by the Squeeze (Sandwich)

Theorem, themiddle part,

f x( ) , is forced to approach 0,also. The middle part issqueezed or sandwichedbetween the left and right parts,so itmust approach the samelimit as the other two do.

limx0

x2( ) = 0, and lim

x0x2= 0, so

limx0

x2cos

1

x

= 0 by the Squeeze

Theorem.

Shorthand: As x 0 ,

x2

0

x2 cos1

x

Therefore,0by the Squeeze(Sandwich) Theorem

x2

0

x 0( ).

The graph ofy= x2cos1

x

, together with the squeezing graphs of y= x2

and y= x2, is below.

(The axes are scaled differently.)


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In Example 2 below, f x( ) is theproduct of asine or cosine expressionand amonomial of odd degree.

Example 2 (Handling Complications with Signs)

Let f x( ) = x3sin 1x

3

. Use the Squeeze Theorem to find lim

x0f x( ) .

Solution 1 (Using Absolute Value)

We firstboundsin1

x3

,


WARNING 2: The problemwith multiplying all three parts

by x3 is that x3 < 0whenx< 0. The inequalitysymbols would have to bereversed for x< 0.

Instead, we useabsolute valuehere.We could write

0 sin1

x3

1 x 0( ) ,

but we assume that absolutevalues arenonnegative.

Multiplyboth sides of the

inequality by x3 . We know

x3> 0 x 0( ) .

Theproduct of absolutevalues equals theabsolutevalueof the product.

If, say, a 4, then

4 a 4. Similarly:

1 sin1

x3

1 x 0( )

sin 1

x3

1 x 0( )

x3

sin1

x3

x3 x 0( )

x3sin

1

x3

x3 x 0( )

x3 x3sin1

x3

x3 x 0( )


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Now, apply theSqueeze(Sandwich) Theorem.

limx0

x3( ) = 0, and lim

x0x3= 0, so

limx0

x3sin

1

x3

= 0 by the Squeeze

Theorem.Shorthand: As x 0 ,

x3

0

x3sin1

x3

Therefore, 0by the Squeeze(Sandwich) Theorem

x3

0

x 0( ) .

Solution 2 (Split Into Cases: Analyze Right-Hand and Left-Hand Limits

Separately)

First, we analyze: limx0+

x3sin

1

x3

.

Assume x> 0, since we are taking a limit as x 0+ .

We firstboundsin1

x3

,


Multiplyall three parts by

x3 so that the middle part

becomes f x( ) . We know

x3> 0 for all x> 0.


1 sin1

x3

1 x> 0( )

x3 x3sin1

x3

x3 x> 0( )

limx0+

x3( ) = 0, and lim

x0+x3= 0, so

lim

x0+x3sin

1

x

3

= 0 by the Squeeze

Theorem.

Shorthand: As x 0+ ,

x3

0

x3sin1

x3


x3

0

x> 0( ) .


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Second, we analyze: limx0

x3sin

1

x3

.

Assume x< 0 , since we are taking a limit as x 0 .

We firstboundsin 1x

3

,


Multiplyall three parts by

x3 so that the middle part

becomes f x( ) . We know

x3< 0 for all x< 0, so we

reversethe inequality

symbols.

Reversing the compoundinequality will make iteasier to read.


1 sin 1x

3

1 x< 0( )

x3 x3sin1

x3

x3 x< 0( )

x3 x3sin

1

x3

x3 x< 0( )

limx0

x3= 0, and lim

x0x

3( ) = 0, so

limx

0x3sin

1

x

3

= 0 by the Squeeze

Theorem.

Shorthand: As x 0 ,

x3

0

x3sin1

x3


x3

0

x< 0( ) .

Now, limx0+

x3sin

1

x3

= 0, and lim

x0x3sin

1

x3

= 0, so

limx0

x3sin

1

x3

= 0.


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Example 3 (Limits are Local)

Use limx0

x2= 0 and lim

x0x6= 0 to show that lim

x0x4= 0.

Solution

Let I = 1, 1( ) \ 0{ } . I is apunctured neighborhood of 0.Shorthand: As x 0 ,

x6

0

x4


x2

0

x I( )

WARNING 3: Thedirectionof the inequality symbols may

confuse students. Observe that1

2

4

=

1

16,1

2

2

=

1

4, and

1

16 0, since we are taking a limit as x .

We firstboundsinx.

Divideall three parts by x

x> 0( ) so that the middle

part becomes f x( ) .

1 sinx1 x> 0( )

1

x

sinx

x

1

x

x> 0( )

Now, apply theSqueeze(Sandwich) Theorem. lim

x 1

x

= 0, and lim

x

1

x

= 0, so

limx

sinx

x

= 0 by the Squeeze Theorem.

Shorthand: As x ,

1

x

0

sinx

x


1

x

0

x> 0( ) .

Solution to b)

Assume x< 0, since we are taking a limit as x .

We firstboundsinx.

Divideall three parts by x

so that the middle partbecomes f x( ) . But x< 0,so we must reversethe inequality symbols.

Reversing the compoundinequality will make iteasier to read.

1 sinx1 x< 0( )

1

x

sinx

x

1

x

x< 0

( )

1

x

sinx

x

1

x

x< 0( )


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(Section 2.6: The Squeeze (Sandwich) Theorem) 2.6.9.

Now, apply theSqueeze(Sandwich) Theorem. lim

x

1

x

= 0, and limx

1

x

= 0, so

limx

sinx

x

= 0 by the Squeeze Theorem.

Shorthand: As x ,

1

x

0

sinx

x


1

x

0

x< 0( )

The graph ofy=sinx

x, together with the squeezing graphs of y=

1

xand

y= 1x

, is below. We can now justify theHA at y = 0 (thex-axis).

(The axes are scaled differently.)

Variation for Long-Run Limits to the Right

Let B x( ) f x( ) T x( ) on somex-interval of the form c,( ), c .

Iflim

xB x

( )= L

andlim

xT x

( )= L L

( ), thenlim

xf x

( )= L

. In Example 4a, we used c= 0. We need the compound inequality to holdforever after some point c.

Variation for Long-Run Limits to the Left

Let B x( ) f x( ) T x( ) on somex-interval of the form ,c( ) , c .

If limx

B x( ) = L and limx

T x( ) = L L( ) , then limx

f x( ) = L .

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