In this section, I'll give proofs of some of the properties of limits. This section is pretty heavy on theory --- more than I'd expect people in a calculus course to know. So unless you're reading this section to learn about analysis, you might skip it, or just look at the statements of the results and the examples.

First, let's recall the definition of a limit.

* Definition.* Let f be a real-valued function
defined on an open interval containing a point , *but possibly not at* c. If , then means: For every , there is a
such that for every x in the domain of f,

Informally, "making x close to c makes close to L". In this section, I'll prove various results for computing limits. But I'll begin with an example which shows that the limit of a function at a point does not have to be defined.

In the next example and in several of the proofs below, I'll need to
use the * Triangle Inequality*. It says that if p
and q are real numbers, then

You often use the Triangle Inequality to combine absolute value terms (going from the left side to the right side) or to break up an absolute value term (going from the right side to the left side).

* Example.* (* A limit that is
undefined*) Let

Prove that

Suppose on the contrary that

This means that for every , there is a such that

(In this case, the "c" of the definition is equal to 0.)

Choose . I'll that there
is * no* number such that if , then

Suppose there * is* such a number . The x's which satisfy the inequality are the points in the interval . Note that there are both positive and
negative numbers in this interval.

Let a be a positive number in . Since , I have , so

Let b be a negative number in . Since , I have , so

Note that

So I can write my two inequalities like this:

Add the two inequalities:

By the Triangle Inequality,

Combining this with , I get

This is a contradiction. Therefore, my assumption that is defined must be incorrect, and the limit is undefined.

* Proposition.* (* The limit of a
constant*) Let and . Then

In other words, the limit of a constant is the constant.

* Proof.* In this case, the function is and the limit is .

Let . Then

Since the conclusion of the statement "If , then " is true, the statement is true regardless of what is. (For the sake of definiteness, I could choose , for example.) This proves that .

* Proposition.* Let . Then

* Proof.* In this case, the function is and the limit is .

Let . Set . Suppose . Since , I have

This proves that .

* Theorem.* (* The limit of a
sum*) Let . Let f and g be functions defined
on an open interval containing c, but possibly not at c. Suppose that

Then

* Proof.* Let . I need to
find a number such that

The idea is that since , I can force to be close to L, and since , I can force to be close to M. Since I want to be within of , I'll split the difference: I'll force to be within of L and force to be within of M.

First, means that I can find a number such that

Likewise, means that I can find a number such that

I'd like to choose so that both of these hold. To do
this, I'll let be the *
smaller* of and . (If and are equal, I choose to be their common
value.) The mathematical notation for this is

Since is the smaller of and , * it must be at
least as small as both*:

Now suppose . Since ,

Therefore,

Since ,

Therefore,

Add the inequalities and :

By the Triangle Inequality,

Combining this with , I get

This proves that .

* Remark.* This result is often written as

But it's important to understand that the equation is true
*provided* that the limits on the right side are defined. If
they are not, then the result might be false. For example, let

In an earlier example, I showed that is undefined. Since , essentially the same proof as in the example shows that is undefined. However,

Hence, the limit-of-a-constant rule shows that

In this case, the equation

The left side is 0, while the right side is undefined.

* Remark.* The rule for sums holds for a sum of
more than 2 terms. Without writing out all the hypotheses, it says

The proof uses * mathematical induction*; I won't
write it out, though it isn't that difficult. I will, however, use
this result in proving the rule for limits of polynomials.

Having just proved a limit rule for sums, it's natural to try to prove a similar rule for products. With the appropriate fine print, it should say that

If you try to write a proof for this, you might find it a bit more challenging than the ones I've done so far. While it's possible to write a direct proof, some of the ones I've seen look a bit magical: They're shorter than the approach I'll take, but it can be hard to see how someone thought of them.

So instead, I'll take a different approach, which is often useful in
writing proofs in math: * If your proof looks too
difficult, try to prove a special case first*. I'll get a bunch
of special cases (which are useful in their own rights), and whose
proofs are fairly straightforward.

I'll begin with the special case where one of the functions in the product is just a constant.

* Theorem.* (* Multiplication by
constants*) Let . Let f be a function
defined on an open interval containing c, but possibly not at c.
Suppose that

Then

* Proof.* First, if , then the limit-of-a-constant rule says

But , so desired equation holds:

Having dealt with the case , I'll assume .

Let . By assumption,

Hence, I may find so that if , then

(Notice that I'm not dividing by 0 on the left side, because .)

With this value of , I have that implies

This proves that

* Remark.* This rule is often written more
concisely as

The multiplication-by-constants rule is a special case of the general rule for products that I'd like to prove, but it's useful in its own right. Here are two easy consequences.

* Corollary.* (* Negatives*)
Let . Let f be a function defined on an open
interval containing c, but possibly not at c. Suppose that

Then

* Proof.* Take in the
multiplication-by-constants rule.

* Corollary.* (* The limit of a
difference*) Let . Let f and g be functions
defined on an open interval containing c, but possibly not at c.
Suppose that

Then

* Proof.* By the preceding corollary, I have

Therefore, by the rule for sums,

Here's another special case of the limit of a product.

* Lemma.* (* Product of zero
limits*) Let . Let f and g be functions defined
on an open interval containing c, but possibly not at c. Suppose that

Then

* Proof.* Let . I need to
find a number such that

The idea is that I can "control" and , so I'll try to get two inequalities and which multiply to . Since the problem seems to be "symmetric" in f and g, it's natural to use .

Since , I may find a number such that if , then

Since , I may find a number such that if , then

Now let . Then if , I have both and . Thus,

Multiplying the last two inequalities, I get

This proves that .

You can consider the next lemma an example of how you might use the preceding results.

* Lemma.* Let . Let f be a
function defined on an open interval containing c, but possibly not
at c. Suppose that

Then

* Proof.*

Now I'll put together a lot of the previous results to prove the rule for the limit of a product. I actually don't need an proof in this case: Just the earlier rules and some careful algebra.

* Theorem.* (* The limit of a
product*) Let . Let f and g be functions
defined on an open interval containing c, but possibly not at c.
Suppose that

Then

* Proof.* Suppose that

By the last lemma,

I apply the product of zero limits lemma and multiply out the factors in the limit:

(Save this huge expression for a second.)

Now by the rules for multiplication by constants and the limit of a constant,

By the rules for the limit of a sum and a difference,

So again by the rule for the limit of a sum (I'm adding the big expression in the line above, and the big expression two lines above),

But (cancelling 6 terms)

So

* Remark.* I had to be careful in using the rule
for the limit of a sum to ensure that the *component limits*
were defined before applying the rule. That is why I couldn't simply
apply it to the left side of

To apply the sum rule to the left side, I would need to know that exists, but that is part of what I was trying to prove.

You might want to look up the shorter, "magical" proofs of the rule for the Limit of a Product and see if you like them better than this approach.

* Remark.* The rule for products holds for a
product of more than 2 terms. Without writing out all the hypotheses,
it says

The proof uses * mathematical induction*; I won't
write it out, though it isn't that difficult.

My next goal is to prove that if is a polynomial, then

I'll prove it by putting together some preliminary results. Let's start with a really easy one.

* Lemma.*

* Proof.* Let . I have to
find so that if ,
then . Just take . Then

* Proposition.* (* Powers*)
If n is an integer and , then

This proof will use * mathematical induction*.
Explaining induction here would require a separate and fairly lengthy
discussion, so I'll just give the proof and assume that you've seen
induction elsewhere. Or you can just take this result for granted,
since it's not very surprising.

* Proof.* For , the left side is
(by the constants rule)

The right side is . The left and right sides are equal, and the result is true for .

Assume that and the result holds for n:

I will prove it for :

The first and last equalities just used rules for powers. The second equality used the rule for the limit of a product. The third equality used the induction assumption and the previous lemma.

This proves the result for , so the result holds for all by induction.

* Remark.* The rule for powers holds for negative
integer powers. It also holds for rational number powers (with
suitable restrictions --- you can't take the square root of a
negative number, for instance) and even real number powers. I'll
prove some of this below, but the there's an easier way to do all of
these at once The idea is that if r is a real number, I can write

Then I'll need to use limit results on the natural log and exponential functions. That will require a discussion of those functions, which we'll have later.

* Theorem.* (* Polynomials*)
Let , , ... , be real numbers. Consider the
polynomial

Then

In other words, if is a polynomial, then

* Proof.* By the rules for multiplication by
constants and powers, for , ... n, I have

Then by the rule for sums (which I remarked holds for a sum with any number of terms),

* Example.* Compute .

By the rule for polynomials, I can just plug 3 in for x:

We'll see that other functions have the property that you can compute
by "plugging in c for
x". The property is called * continuity*.

You might expect that there would be a rule that says "the limit of a quotient is the quotient of the limits". There is --- though we have to be careful that the component limits exist, and also that we avoid division by 0. As with the rule for products, you can give a proof which looks a little "magical" --- but instead, as I did with the rule for products, I'll derive the rule for quotients from some other rules which are independently useful. There's still a little "magic" in the proof of the lemma for , but it's not too bad if you work backwards "on scratch paper" first.

* Lemma.* Suppose that . Then

* Proof.* (* Scratch work*.)
Before I do the real proof, I do some scratch work so the actual work
doesn't seem too magical. This is going to get a little wordy, so if
you're not interested, you could just skip to the real proof below.

As is common with limit proofs, I work backwards from what I want. According to the definition, I want

Now , so "controls" . I'll do some algebra to try to get a factor of :

I combined the fractions over a common denominator, then broke the result up into three factors. Note that , because the absolute value of a number equals the absolute value of its negative.

The first factor is a constant, so I don't need to worry about it. The third factor is , which I can control using .

In order to get some control over the second factor , I make a preliminary setting of . This isn't a problem, since intuitively I have complete control over . (You'll see how this works out in the real proof.) But how should I set ?

I don't want to get too big. But if x is close to 0, then will be large --- for example, . So I want to set so that x doesn't get too close to 0.

controls how close x is to c. And I'm given that . So by forcing x to be close enough to c, I can force x to stay away from 0. There are lots of ways to do this; this picture shows what I will do.

As the picture shows, I'll force x to stay within of c. I can do this by setting .

There are two cases, depending on whether c is positive or negative, but you can see the cases are symmetric. x will lie in an interval around c, and it won't get any closer to 0 than . Thus,

Taking reciprocals,

Now putting this back into the expression above,

I want the left-hand expression to be less than . If the right-hand expression is less than , this will be true:

So how can I make ? Moving the first two terms to the right, I get

But I can control directly using , so I can make this happen if .

Now earlier, I made a preliminary setting of . I seem to have two settings for
. There is a standard trick for getting both of these
at once: Set to the *smaller* of the two. The
notation for this is

Since is the smaller of the two, I get

I arrived at my guess for by working backwards. I have to write the real proof forwards, starting with my guess for . Here it is.

(* Real proof.*) Let . Set . Suppose . Then

Consider the first inequality . This means that x is less than from c. So if c is positive, then

And if c is negative, then , so

In both cases,

Multiply and to get

This proves that .

The next theorem is important in its own right.

* Theorem.* (* Composites*)
Let . Suppose that:

(a) f is a function defined on an open interval containing a, but possibly not at a.

(b) .

(c) g is a function defined on an open interval containing b, but possibly not at b.

(d) .

Then

To write it somewhat roughly,

* Proof.* Let . Since , I can find a number such that if ,
then .

Since , I can find a number such that if , then .

Suppose that . Then . But then

This proves that .

* Example.* Compute .

Let

Then

By the rules for limits of polynomials and composites,

* Theorem.* (* Reciprocals*)
Suppose f is a function defined on an open interval containing a, but
possibly not at a, and

Then

* Proof.* Let . Then

Since , it follows that g is defined on an open interval containing L,

Since , the -lemma implies that

Then the rule for composites implies that

* Theorem.* (* Quotients*)
Suppose f and g are functions defined on an open interval containing
a, but possibly not at a. Suppose that

Then

* Proof.* Note that , and that by
the reciprocal rule

Then by the rule for products,

* Example.* Compute .

By the rules for limits of polynomials and quotients,

The next result is different from the previous results, in that the statement doesn't seem obvious at first glance. However, the conclusion is reasonable if you draw a picture

* Theorem.* (* The Squeezing
Theorem*) Suppose , , and are defined on a open interval I containing c, but
are not necessarily defined at c. Assume that

Then

Here's a picture which makes the result reasonable:

The theorem says that if g is caught between f and h, and if f and h both approach a limit L, then g is "squeezed" to the same limit L.

This result is sometimes called the Sandwich Theorem, the idea being that g is the filling of the sandwich and it's caught between the two slices of bread (f and h).

* Proof.* Let .

Since , I can find so that implies

Since , I can find so that implies

Let . Thus, and .

Thus, if , then

Therefore,

Now means is less than from L, so

And means that is less than from L, so

Hence,

Therefore,

* Example.* Prove that .

Note that as the expression is undefined. So, for instance, you can't use the rule for the limit of a product.

From trigonometry, for all . So

(Note that since , multiplying the inequality by does not cause the inequality to "flip".) Now

By the Squeezing Theorem,

The next result doesn't seem to have a standard name, so I'll call it
* The Neighborhood Theorem*. It says that the
value of depends on the
values of *near* c, not *at* c. I'll
often use this result in computing limits involving *
indeterminate forms*.

In many presentations of calculus, this result isn't stated explicitly. Instead, you'll see it used in the middle of computations like this:

Note that you can only cancel the terms if you know
, i.e. if . The author will
justify this by saying something like: "We can cancel the terms because in taking the limit, we only consider
x's *near* 1, rather than ."

The Neighborhood Theorem applies to this situation in this way: the functions and are equal for all x except . Therefore, the Neighborhood Theorem says that they have the same limit as x approaches 1.

* Theorem.* (* The Neighborhood
Theorem*) Suppose that:

(a) .

(b) for all x in the interval *except* possibly at c.

Then the limits and are either both defined or both undefined. If they are both defined, then they have the same value.

In other words, if two functions are equal in a neighborhood of c, except possibly at c, then they have the same limit at c.

* Proof.* Suppose that and for all x in the interval
*except* possibly at c.

Suppose first that . I will show that .

Let . I must find such that if , then .

Since , the limit definition produces a such that if , then . So take this , and suppose that . By the choice of , I get

But notice that my assumption includes the assumption that . In particular, , since if , then . Since , I have , so

This proves that .

The remaining case is that is undefined. In this case, I must show that is undefined. Suppose not. Then is defined so for some number L. But then the first part of the proof (with the roles of and switched) shows that . This contradicts my assumption that is undefined.

Hence, is undefined.

The functions , , and which are graphed below are equal for all x except . By the Neighborhood Theorem, the three functions have the same limit as x approaches 3:

Copyright 2024 by Bruce Ikenaga