Limits of functions using the squeeze principle



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LIMITS OF FUNCTIONS USING THE SQUEEZE PRINCIPLE

The following problems involve the algebraic computation of limits using the Squeeze Principle, which is given below.

SQUEEZE PRINCIPLE : Assume that functions f , g , and h satisfy

and


.

Then


.

(NOTE : The quantity A may be a finite number, , or . The quantitiy L may be a finite number, , or .)

The Squeeze Principle is used on limit problems where the usual algebraic methods (factoring, conjugation, algebraic manipulation, etc.) are not effective. However, it requires that you be able to ``squeeze'' your problem in between two other ``simpler'' functions whose limits are easily computable and equal. The use of the Squeeze Principle requires accurate analysis, deft algebra skills, and careful use of inequalities.

SOLUTIONS TO LIMITS USING THE SQUEEZE PRINCIPLE
SOLUTION 1 : First note that

because of the well-known properties of the sine function. Since we are computing the limit as x goes to infinity, it is reasonable to assume that x > 0 . Thus,



.

Since


,

it follows from the Squeeze Principle that



.

SOLUTION 2 : First note that

because of the well-known properties of the cosine function. Now multiply by -1, reversing the inequalities and getting



or


.

Next, add 2 to each component to get



.

Since we are computing the limit as x goes to infinity, it is reasonable to assume that x + 3 > 0. Thus,



.

Since


,

it follows from the Squeeze Principle that



.

SOLUTION 3 : First note that

because of the well-known properties of the cosine function, and therefore



.

Since we are computing the limit as x goes to infinity, it is reasonable to assume that 3 - 2x < 0. Now divide each component by 3 - 2x, reversing the inequalities and getting



,

or


.

Since


,

it follows from the Squeeze Principle that



.

SOLUTION 4 : Note that DOES NOT EXIST since values of oscillate between -1 and +1 as x approaches 0 from the left. However, this does NOT necessarily mean that does not exist ! ? #. Indeed, x3 < 0 and

for x < 0. Multiply each component by x3, reversing the inequalities and getting



or


.

Since


,

it follows from the Squeeze Principle that



.

SOLUTION 5 : First note that

,

so that


and


.

Since we are computing the limit as x goes to infinity, it is reasonable to assume that x+100 > 0. Thus, dividing by x+100 and multiplying by x2, we get



and


.

Then


=

=

=

= .

Similarly,

= .

Thus, it follows from the Squeeze Principle that



= (does not exist).

SOLUTION 6 : First note that

,

so that


,

,

and


.

Then


=

=

=

= 5 .

Similarly,



= 5 .

Thus, it follows from the Squeeze Principle that



= 5 .

SOLUTION 7 : First note that

and


,

so that


and


.

Since we are computing the limit as x goes to negative infinity, it is reasonable to assume that x-3 < 0. Thus, dividing by x-3, we get



or


.

Now divide by x2 + 1 and multiply by x2 , getting



.

Then


=

=

=

=

= 0 .


Similarly,

= 0 .

It follows from the Squeeze Principle that



= 0 .

SOLUTION 8 : Since

=

and


= ,

it follows from the Squeeze Principle that



,

that is,









.

Thus,


.

SOLUTION 9 : a.) First note that (See diagram below.)

area of triangle OAD < area of sector OAC < area of triangle OBC .



The area of triangle OAD is

(base) (height) .

The area of sector OAC is



(area of circle) .

The area of triangle OBC is



(base) (height) .

It follows that



or


.

b.) If , then and , so that dividing by results in



.

Taking reciprocals of these positive quantities gives



or


.

Since


,

it follows from the Squeeze Principle that



.

SOLUTION 10 : Recall that function f is continuous at x=0 if

i.) f(0) is defined ,

ii.) exists ,

and


iii.) .

First note that it is given that

i.) f(0) = 0 .

Use the Squeeze Principle to compute . For we know that



,

so that


.

Since


it follows from the Squeeze Principle that

ii.) .

Finally,

iii.) ,

confirming that function f is continuous at x=0 .






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