### 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 .