The Basics of Proofs
7. The Completeness Axiom is sometimes called the Least Upper Bound Principle. The Completeness Axiom comes up frequently in proofs
3.9 The Arithmetic of Limits
The fact that the limits of some functions are easy to prove hides the fact that there are some limits whose validity is considerably more difficult to prove. Fortunately, the limits of most arithmetic combinations of functions work as expected due to the behavior of the arithmetic operations of addition, subtraction, multiplication, and division. In the words of the next chapter, these operations behave well because they are themselves continuous functions of their arguments. That is, for example, the function of two variables f.x; y/ D x C y is a continuous function of x and y.
That continuity allows you to prove the following theorem.
THEOREM: Suppose that f and g are functions both defined on a set with accumulation point a. Let lim
x!af.x/ D L and lim
x!ag.x/ D H. Then 1. lim
x!af.x/ C g.x/ D L C H.
2. lim
x!af.x/ g.x/ D L H.
3. lim
x!af.x/g.x/ D LH.
4. if H ¤0, lim
x!a f.x/
g.x/ D HL.
Consider how to prove each part of the above theorem. In each case you will need to prove the validity of a limit, so the proof can follow the usual proof template for establishing a limit. These proofs differ from limit proofs found earlier in the chapter in that you know less about the functions whose limits you are trying to establish.
On the other hand, you do know that the limits of the functions f and g exist, and that gives you a lot of tools with which to work.
82 3 Limits
3.9.1 Limit of a Sum
So what needs to be done to prove that the limit of the sum of two functions is the sum of their respective limits? As with all limit proofs, the proof will begin with a statement about what is being assumed about two functions f and g. In this case, that would essentially be a restatement of the hypothesis of the theorem that says that the limits of f and g at a are L and H, respectively. The second step of the proof would be to say Let > 0 be given which sets the tolerance to be met by the proof. You know that the end of the proof will need to show that the function in question, f.x/ C g.x/, needs to be within of the proposed limit, L C H. In other words, you will need to establish j
f.x/ C g.x/
.L C H/j < . Clearly, this inequality will depend on properties of the functions f and g. But you know very little about these functions. Actually, knowing very little about the functions makes your job easier. All you know about these functions is that f has L for a limit, and g has H for a limit. This means that your proof can only use these two facts.
Because these two limits exist, you will be able to set up conditions that ensure that jf .x/ Lj and jg.x/ Hj are small. How does this help? It helps because the triangle inequality will allow you to show that the expression j
f.x/ C g.x/
.L C H/j is no bigger than the sum of the two small quantities jf.x/ Lj and jg.x/ Hj. That is, j
f.x/ C g.x/
.L C H/j D j.f .x/ L/ C .g.x/ H/j jf .x/ Lj C jg.x/ Hj. For example, if both jf.x/ Lj and jg.x/ Hj can be made less than2, then their sum will be less than, and the value of j
f.x/ C g.x/
.L C H/j will, in turn, be less than, as desired. How can you arrange for jf .x/ Lj and jg.x/ Hj both to be less than2? You are given that the limits of f and g are L and H, respectively, so, by the definition of limit, you can arrange for each of these quantities to be smaller that any given positive value, such as2, with appropriate choices ofı > 0. The only subtlety here is that the value ofı > 0 needed to assure that jf .x/ Lj is less than 2cannot be assumed to be the same value as theı > 0 needed to assure that jg.x/ Hj is less than2. Thus, two different values ofı should be chosen, and then the minimum of those two will be small enough to guarantee both of the needed inequalities.
Thus, after the proof proposes a given > 0, it can produce a ı1 > 0 small enough so that if x is in the domain of f and0 < jx aj < ı1, then jf.x/ Lj will be less than 2. The existence of thisı1 comes from the definition of lim
x!af.x/ D L.
Similarly, the proof can produce aı2 > 0 coming from the definition of lim
x!ag.x/ D H such that if x is in the domain of g and0 < jx aj < ı2, then jg.x/ Hj will be less than2. The proof then easily follows as described above.
PROOF: Suppose that f and g are functions both defined on a set with accumulation point a. If lim
x!af.x/ D L and limx!ag.x/ D H, then
xlim!af.x/ C g.x/ D L C H.
• Let f and g be functions both defined on a set with accumulation point a with lim
x!af.x/ D L and lim
x!ag.x/ D H.
• Let > 0 be given.
• By the definition of limit, there is aı1> 0 such that if x is in the domain of f and0 < jx aj < ı1, then jf.x/ Lj < 2.
• Similarly, there is aı2 > 0 such that if x is in the domain of g and 0 <
jx aj < ı2, then jg.x/ Hj < 2.
• Letı D min.ı1; ı2/ > 0.
• Then if x is in the domain of f C g with0 < jx aj < ı,
• j
f.x/Cg.x/
.LCH/j D j.f .x/L/C.g.x/H/j jf .x/LjCjg.x/Hj <
2C 2 D .
• This shows that lim
x!af.x/ C g.x/ D L C H.
A proof that the limit of the difference f.x/ g.x/ equals the difference of the individual limits, L H, is very similar to the above proof and is left as an exercise.
3.9.2 Limit of a Product
Proving that the limit of the product f.x/g.x/ equals the product of the individual limits, LH, uses the same techniques as the proof for the limit of a sum but has an added complexity requiring the use of a commonly used trick. The proof of
xlim!af.x/g.x/ D LH follows the usual template for proving the existence of a limit.
Its goal is to establish the inequality jf.x/g.x/ LHj < . Again, you can use the definition of limit to make jf.x/ Lj and jg.x/ Hj as small as you need, but how small these have to be to ensure that jf.x/g.x/LHj is less than is not immediately obvious. The problem is that it is difficult to gauge how close f.x/g.x/ is to LH when you know that f.x/ is close to L, and g.x/ is close to H. The difficulty stems from having to move from f.x/g.x/ to LH, where f .x/ changes to L and g.x/ changes to H at the same time. If only one of these two changes were made, then it might be easier to make the needed estimate. That is, it would be easier to work with an expression like f.x/g.x/ f .x/H than with f .x/g.x/ LH.
Of course, f.x/g.x/ LH is not the same as f .x/g.x/ f .x/H, so one cannot just use f.x/g.x/ f .x/H in place of f .x/g.x/ LH. Sometimes, though, it is worth replacing one expression with another expression that is easier to handle,
84 3 Limits and then adjusting the second expression to make it equivalent to the first. In this case, the change can be accomplished by employing one of the oldest tricks used in mathematical proofs, that of adding and subtracting the same quantity. In particular, you can rewrite jf.x/g.x/ LHj as jf .x/g.x/ f .x/H C f .x/H LHj. The advantage of doing this is that now you can see how the difference between f.x/g.x/ and LH depends on the differences between f.x/ and L and g.x/ and H. Indeed, jf .x/g.x/ LHj D jf .x/g.x/ f .x/H C f .x/H LHj D jf .x/.g.x/ H/ C H.f .x/ L/j jf .x/j jg.x/ Hj C jHj jf .x/ Lj. If each of the two terms, jf .x/j jg.x/ Hj and jHj jf.x/ Lj, can be made smaller than 2, then it will have been shown that jf .x/g.x/ LHj is less than as needed.
So how small does jf.x/ Lj need to be to ensure that jHj jf .x/ Lj is less than 2? Less than 2jHj appears to be small enough, although one needs to handle the embarrassing situation where H D 0. You could handle H D 0 and H ¤ 0 as two separate cases, or you can take care of both cases at once by making jf.x/ Lj less than
2
jHjC1 since jHj C 1 is larger than jHj and can never be 0. Thus, you can select aı1> 0 so that if 0 < jx aj < ı1, then jf.x/ Lj <
2
jHjC1.
How small does jg.x/ Hj need to be to ensure that jf .x/j jg.x/ Hj is less than
2? It would be nice to say that jg.x/ Hj < 2jf .x/j suggesting that you setı small enough to ensure jg.x/ Hj < 2jf .x/j , but there is a problem here. The definition of limit requires that the choice ofı come before the choice of x, so you cannot have the value ofı depending on x. What is needed is an upper bound for jf .x/j because, if jf.x/j M, the value of ı can be found to ensure jg.x/ Hj < 2M which will always be small enough to guarantee jf.x/j jg.x/ Hj < 2. You can find such an upper bound for jf.x/j because the limit of f .x/ exists as x approaches a, and so jf .x/j can be restricted to being not much larger than jLj. You could, for example, findı2 > 0 so that if 0 < jx aj < ı2, then jf.x/ Lj < 1. This would ensure that f.x/ is a distance of no more than 1 from L so that jf .x/j < jLj C 1. Then you would only need jg.x/ Hj <
2
jLjC1 to get jf .x/j jg.x/ Hj < 2. This gives you all the pieces necessary to complete the proof as follows.
PROOF: Suppose that f and g are functions both defined on a set with accumulation point a. If lim
x!af.x/ D L and limx!ag.x/ D H, then
xlim!af.x/g.x/ D LH.
• Let f and g be functions both defined on a set with accumulation point a with lim
Finally, the proof that the limit of a quotient is the quotient of the individual limits is much like the proof about the product of limits, although the algebra is more complicated. As in the preceding proof, you can start with the needed inequality which, in this case, is ˇˇˇgf.x/.x/HLˇˇˇ < . Using the trick of adding and subtracting the same quantity, the left side of the inequality can be written as ˇˇˇgf.x/.x/HLˇˇˇ D ˇˇˇf.x/HLg.x/
Both of these terms have a factor of jg.x/j in the denominator. To make the fractions small, you will need to know that jg.x/j does not get too close to zero.
What you do know is that lim
x!ag.x/ D H is not zero because the hypothesis of the theorem will make that assumption. How far away from zero can you require jg.x/j to be? Certainly, this will depend on the value of H. If H is close to zero, then jg.x/j
86 3 Limits will be close to zero as x approaches a. The best you can do is require that jg.x/j be so close to H that it will keep a known distance from zero. For example, you could require that jg.x/ Hj be less than jHj2 . That will ensure that jg.x/j is at least jHj2 which keeps it a known distance away from zero. So, select aı1> 0 such that if x is in the domain of g with0 < jx aj < ı1, then jg.x/ Hj < jHj2 , and jg.x/j will be greater thanjHj2 .
Now for these values of x you will havejf .x/Ljjg.x/j < jf .x/ Lj jHj2 . Thus, it would be sufficient if jf.x/Lj were to be less thanjHj4 which will ensure thatjf .x/Ljjg.x/j < 2. This can be done by choosingı2> 0 small enough so that jf .x/Lj is less thanjHj4 . To make the ˇˇˇL.g.x/H/g.x/H ˇˇˇ term less than 2, you can select aı3 > 0 so that if x is withinı3 of a you will have jg.x/ Hj less than H4jLj2 because that will give ˇˇˇL.g.x/H/g.x/H ˇˇˇ < jg.x/jjHjjLjH24jLj < H2H24
2
D 2. Well, OK, did you catch that the preceding does not work if L D0? To avoid this problem it would be better to make jg.x/ Hj less than H2
4
jLjC1. Putting all of these ideas together gives the following proof.
PROOF: Suppose that f and g are functions both defined on a set with accumulation point a. If lim
x!af.x/ D L and limx!ag.x/ D H with H ¤ 0, then
xlim!a f.x/
g.x/ D HL.
• Let f and g be functions both defined on a set with accumulation point a with lim
3.9.4 Limit of Rational Functions
As a demonstration of the power of these results about the arithmetic of limits, you can now easily prove the following list of results which will allow you to easily calculate limits of polynomials and rational functions of x.
• For any constant c in the real numbers, lim
x!acD c.
• lim
x!axD a.
• For any n in the natural numbers, lim
x!axnD an.
The first two results are very easy to prove, and are left as exercises. The next two results can be proved by using mathematical induction which is often the first technique one considers using when trying to prove statements such as these that depends on a natural number. Here, mathematical induction will be employed to prove statements about the limits of polynomials, and the degree of the polynomial provides a natural number to use as the induction variable.
To begin with, try using mathematical induction to prove that lim
x!axn D an for any natural number n. In this mathematical induction argument, the base case is
xlim!axD a, that is, when n D b D 1. The proofs of statements similar to this base case were covered earlier. The induction step in the proof will need to show that if
xlim!axkD akfor some natural number k, then lim
x!axkC1D akC1. But xkC1is just the product xk x, so this result follows immediately from the theorem about the limits of products. That leads to the following proof that uses the template for proofs by mathematical induction.
PROOF: lim
x!axnD anfor any natural number n.
• SET THE CONTEXT: The statement will be proved for all natural numbers n by mathematical induction on n.
• PROVE S.b/: When n D 1, the statement says that lim
x!ax D a which has already been established.
• STATE THE INDUCTION HYPOTHESIS: Assume that for some natural number k, lim
x!axkD ak.
• PERFORM THE INDUCTION STEP: Then since the limit of a product of two functions is the product of the two individual limits, it follows that
xlim!axkC1D lim
x!axk x D .lim
x!axk/.lim
x!ax/ D ak a D akC1. So the statement is true for n D k C1.
• STATE THE CONCLUSION: Therefore, by mathematical induction,
xlim!axnD anis true for all natural numbers n.
88 3 Limits Mathematical induction can again be employed to prove that for every polyno-mial, p.x/, lim
x!ap.x/ D p.a/. As a reminder, a polynomial of degree n is a function, p.x/ D cnxn C cn1xn1 C cn2xn2 C C c1xC c0 where c0; c1; c2; : : : ; cn
are constants with cn ¤ 0. Previously it has been proved that lim
x!acj D cj and
xlim!axj D aj, from which one gets that the limit of a monomial is lim
x!acjxj D cjaj. A polynomial is just a sum of such monomials, so mathematical induction is a convenient tool for showing that this sum of an arbitrary number of monomials has the desired limit.
PROOF: For any constants c0; c1; c2; : : : ; cn and a2 R, the poly-nomial p.x/ D cnxnC cn1xn1C cn2xn2C C c1xC c0 satisfies
xlim!ap.x/ D p.a/.
• SET THE CONTEXT: The statement will be proved by mathematical induction on the degree of the polynomial n.
• PROVE S.b/: lim
x!ac1xCc0 D .lim
x!ac1/.lim
x!ax/C lim
x!ac0 D .c1/.a/Cc0which shows that the statement is true for n D1.
• STATE THE INDUCTION HYPOTHESIS: Assume that for some natural number k, if p.x/ D ckxkC ck1xk1C C c1xC c0, then lim
• STATE THE CONCLUSION: Therefore, by mathematical induction,
xlim!ap.x/ D p.a/ is true for all polynomials p.x/.
Recall that a rational function is just a ratio of polynomials, that is, if p.x/ and q.x/ are polynomials, then pq.x/.x/ is a rational function. It is only a simple step to get the following theorem.
2. Because p and q are polynomials, lim
x!ap.x/ D p.a/ and limx
!aq.x/ D q.a/.
3. Since the limit of the quotient is equal to the quotient of the individual limits when the limit of the denominator is not zero, it follows that lim
x!a p.x/
q.x/ D
xlim!ap.x/
xlim!aq.x/ D pq.a/.a/.
3.9.5 Other Types of Limits
It is time to note that even though all of these limit theorems concerned limits as x approaches a, most can be extended to cover limits as x approaches a from the left, as x approaches a from the right, as x approaches infinity, and as x approaches negative infinity. In particular, most of the theorems apply to the limits of sequences.
Many of these statements can be found in the exercises.
3.9.6 Exercises
Write proofs of each of the following statements.
1. If f and g are defined in an open interval containing a, and if lim
x!af.x/ D L and
xlim!ag.x/ D H, then lim
x!aCf.x/ g.x/ D L H.
2. For any constant c in the real numbers, lim
x!acD c.
This section discusses a few other useful results about limits. They provide an interesting variety of proof strategies to consider.