Prove that . Begin by letting be given. Find so that if , then , i.e., , i.e., . But this trivial inequality is always true, no matter what value is chosen for . For example, will work. Thus, if , then it follows that . This completes the proof.

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* SOLUTION 2 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" |*x*-10| . Then,

iff

iff

iff

iff

iff .

Now choose . Thus, if , it follows that . This completes the proof.

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* SOLUTION 3 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for"
. Then,

iff

iff

iff

iff

iff

iff .

Now choose . Thus, if , it follows that . This completes the proof.

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* SOLUTION 4 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for"
|*x*-1| . Then,

iff

iff

iff

iff .

We will now ``replace" the term |*x*+1| with an appropriate constant and keep the term |*x*-1| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*-1 < 1 and 0 < *x* < 2 so that 1 < |*x*+1| < 3 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

Now choose (This guarantees that both assumptions made about in the course of this proof are taken into account simultaneously.). Thus, if , it follows that . This completes the proof.

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* SOLUTION 5 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for"
| *x* - (-1) | = | *x* + 1 | . Then,

iff

iff

iff

iff .

We will now ``replace" the term |*x*-1| with an appropriate constant and keep the term |*x*+1| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.). Then implies that -1 < *x*+1 < 1 and -2 < *x* < 0 so that 1 < |*x*-1| < 3 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

Now choose (This guarantees that both assumptions made about in the course of this proof are taken into account simultaneously.). Thus, if , it follows that . This completes the proof.

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* SOLUTION 6 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - 2 | . Then,

iff

iff

iff .

We will now ``replace" the term |3*x*+5| with an appropriate constant and keep the term |*x*-2| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*-2 < 1 and 1 < *x* < 3 so that 8 < |3*x*+5| < 14 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

Now choose (This guarantees that both assumptions made about in the course of this proof are taken into account simultaneously.). Thus, if , it follows that . This completes the proof.

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* SOLUTION 7 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - 3 | . Then,

iff

iff

iff

iff

iff

iff

iff

iff .

We will now ``replace" the term |*x*+3| with an appropriate constant and keep the term |*x*-3| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*-3 < 1 and 2 < *x* < 4 so that 5 < |*x*+3| < 7 and (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff

iff .

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* SOLUTION 8 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - (-6) | = | *x* + 6 | . Then,

iff

iff

iff

iff

iff

iff

iff

iff .

We will now ``replace" the term |2-*x*| with an appropriate constant and keep the term |*x*+6| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*+6 < 1 and -7 < *x* < -5 so that 7 < |2-*x*| < 9 and (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

.

iff .

iff

iff .

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* SOLUTION 9 :*
Prove that .
Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - 3 | . Then,

iff

iff

iff

iff

iff

iff

iff .

We will now ``replace" the term | 4*x*-9 | with an appropriate constant and keep the term |*x*-3| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*-3 < 1 and 2 < *x* < 4 . HOWEVER, THIS RANGE OF X-VALUES IS NOT APPROPRIATE SINCE THE FUNCTION IS NOT DEFINED AT ! Fortunately, this problem can be easily resolved. We simply pick small enough to avoid . For example, assume that . Then implies that and so that 2 < | 4*x*-9 | < 4 and (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

.

iff

iff

iff .

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* SOLUTION 10 :*
Prove that .
Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - 9 | . Then,

iff

iff

(At this point, we need to figure out a way to make | *x*-9 | ``appear'' in our computations. Appropriate use of the conjugate will suffice.)

iff

(Recall that .)

iff

iff .

iff .

We will now ``replace" the term with an appropriate constant and keep the term |*x*-9| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*-9 < 1 and 8 < *x* < 10 so that and (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

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* SOLUTION 11 :*
Prove that .
Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - 4 | . Then,

iff

(At this point, we need to figure out a way to make | *x*-4 | ``appear'' in our computations. Appropriate use of the conjugate will suffice.)

iff

(Recall that .)

iff

iff

iff

iff .

We will now ``replace" the term with an appropriate constant and keep the term |*x*-4| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.) . Then implies that -1 < *x*-4 < 1 and 3 < *x* < 5 so that and (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

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* SOLUTION 12 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - 1 | . Then,

iff

iff

iff

iff

iff

(At this point, we need to figure out a way to make | *x*-1 | ``appear'' in our computations. A simple use of constants will get us started.)

iff

iff

iff

(We need to be able to factor (*x*-1) from the numerator. Apply the conjugate to the term .)

iff

iff

iff

iff

(Now get a common denominator.)

iff

iff

iff

iff

iff .

We will now ``replace" the terms and with appropriate constants and keep the term |*x*-1| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that (This is a valid assumption to make since, in general, once we find a that works, all smaller values of also work.). Then implies that -1 < *x*-1 < 1 and 0 < *x* < 2 so that . In addition, so that and
(Make sure that you understand these steps before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

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* SOLUTION 13 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - *a* | . Then,

iff .

At this point, we need to figure out a way to introduce the term | *x*-*a* | into our computations. The answer lies with the Mean Value Theorem. Consider the function on the interval [*A*, *B*] . Since *f* is continuous on the closed interval [*A*, *B*] and differentiable ( ) on the open interval (*A*, *B*) , according to the Mean Value Theorem there is at least one number *C* , *A* < *C* < *B* , satisfying

,

i.e.,

.

Then

so that

.

This is true for any two real numbers, *A* and *B* . It follows that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

.

Now choose . Thus, if , it follows that . This completes the proof.

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* SOLUTION 14 :*
Prove that . Begin by letting be given. Find (which depends on ) so that if , then . Begin with and ``solve for" | *x* - *a* | . Then,

iff .

At this point, we need to figure out a way to introduce the term | *x*-*a* | into our computations. The answer lies with the Mean Value Theorem. Consider the function on the interval [*A*, *B*] , where *A* and *B* are both positive. Since *f* is continuous on the closed interval [*A*, *B*] and differentiable ( ) on the open interval (*A*, *B*) , according to the Mean Value Theorem there is at least one number *C* , *A* < *C* < *B* , satisfying

,

i.e.,

.

Then, since 0 < *A* < *C* < *B* , it follows that
, so that

.

Thus,

(*) .

This is true for any two positive real numbers *A* and *B* , where *B* > *A* . At this point, we need to consider two cases.

If *x* > *a* , it follows from inequality (*) that (Always make this ``replacement" between your last expression on the left and . This guarantees the logic of the proof.)

iff

iff .

Now choose . Thus, if , it follows that .

If *x* < *a* , then it is reasonable to assume that since we are considering the limit as *x* approaches *a* . Thus, and it follows from inequality (*) that

so that

iff

iff .

Now choose . Thus, if , it follows that . This completes the proof.

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* SOLUTION 15 :*
Let .
Prove that
does not exist .

ASSUME THAT THE LIMIT DOES EXIST. That is, assume that
, where *L* is some real number. It follows that for EACH real number , there exists another real number so that

if , then .

We will proceed to find ONE for which NO works. THIS WILL BE A CONTRADICTION OF OUR ASSUMPTION, making our assumption false, proving that the limit does not exist. (This method is called proof by contradiction.)

By looking at the graph of *f* , which is given above, we see that *x*-*values* chosen ``near'' to *x*=1 but on opposite sides of *x*=1 have corresponding *y*-*values* which are ``about'' one unit apart. Intuitively, this tells us that the limit does not exist and leads us to choose an appropriate leading to the above contradiction.

Consider . Under our assumption that the limit does exist, it follows that there is some number so that if , then . But for ANY choice of

iff (and x not equal to 1)

iff (and x not equal to 1) .

Thus, both and satisfy , which implies that

and .

In addition,

= | -1 |

= 1 .

Now, by the triangle inequality

= 1 .

We have just concluded that

,

an obvious contradiction. It must be that the original limit does not exist.

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Tue May 6 10:38:27 PDT 1997