### SOLUTIONS TO LIMITS OF FUNCTIONS USING THE PRECISE DEFINITION OF LIMIT

SOLUTION 1 :

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.

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,

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iff .

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

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

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iff

iff

iff

iff

iff .

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

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

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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.

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,

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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.

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,

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iff .

We will now ``replace" the term |3x+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 < |3x+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.

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,

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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 .

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.

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,

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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 .

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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.

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,

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iff .

We will now ``replace" the term | 4x-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 < | 4x-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

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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.

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,

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(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 .)

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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 .

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.

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 .)

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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 .

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.

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,

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(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.)

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

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(Now get a common denominator.)

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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 .

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.

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.

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

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iff .

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

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 .

= | -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.