The definition of a * limit* involves both
universal and existential quantifiers.

Let f be a function from the real numbers to the real numbers, and let c be a real number. Assume that f is defined on a open interval containing c. The statement means:

For every , there is a , such that if , then .

Think of as a thermostat, as the actual temperature in a room, and L as the ideal temperature. Someone challenges you to make the actual temperature fall within a certain tolerance of the ideal temperature L. You must do that by setting your -thermostat appropriately (so that x is sufficiently close to c).

Moreover, note that it says "*for every* ". It's isn't enough for you to say what you'd
do if you were challenged with or . You
must prove that you can meet the challenge *no matter what
you're challenged with*.

Finally, note the stipulation " ". This implies
that , since gives . Thus, the
conclusion " " *must* hold
only for x's *close to* c, but not necessarily for .
(It *may* hold for , but it doesn't *have to*.)

What does this mean? It's a precise way of saying that the value of
the limit of as x approaches c does not depend on what
does *at* --- over even whether is
defined.

For example, consider the functions whose graphs are shown below.

In both cases,

In the first case, : The value of the function at is different from the value of the limit.

In the second case, is undefined.

The fact that means that f
is *not continuous* at .

* Example.* Use the definition
of the limit to prove that

In this case, , , and . So here is what I need to prove.

Suppose . I must find a such that if , then .

Note that at this point is fixed --- given --- but all you
can assume is that it's some positive number. Since it *is*
given, however, I can use it in finding an appropriate
.

I'll show how to find by working backwards; then I'll write the proof "forwards", the way you should write it.

I want

It looks like I should set .

All of this has been on "scratch paper"; now here's the real proof.

Suppose . Let . If , then

Thus, if and , then . This proves that .

* Example.* Let

Use the definition of the limit to prove that

Let . I must find such that if , then .

Here's my scratch work. First, for ,

It looks like I should take .

For ,

It looks like I should take .

In order to ensure that both the and requirements are satisfied, I'll take to be the smaller of the two: .

Now here's the proof written out correctly.

Suppose . Let , and assume that .

If , then

Now consider the case . Since , and since , I have . Therefore,

(The case is ruled out because .)

Thus, taking guarantees that if , then . This proves that .

* Example.* Use the definition
of the limit to prove that

Let . I want to find such that if , then .

I start out as usual with my scratch work:

Now I have a problem. I can use to control , but what do I do about ?

The idea is this: Since I have complete control over
, I can *assume* . When I finally set
, I can make it smaller if necessary to ensure that this condition is
met.

Now if , then , so , and . In particular, the
*biggest* could be is 5. So now

This inequality suggests that I set --- but then I remember that I needed to assume . I can meet both of these conditions by setting
to *the smaller of* 1 and : that
is, .

That was scratchwork; now here's the real proof.

Let . Set . Suppose .

Since , I have

Therefore, .

Now , so .

Now multiply the inequalities and :

Thus, if and , then . This proves that .

* Example.* Prove that .

Let . I must find such that if , then .

I'll start with some scratchwork.

I can use to control directly. I need to control the
size of . It's important to
think of this as ,
*not* as and !

Assume . Then , so .

For , , so .

For , , so , and .

Since all the number involved are positive, I can multiply the inequalities to obtain

Thus, I'll get if I have , or . Here's the proof.

Let . Set . Suppose .

Since , , and .

First, , so .

Next, , , so , and .

Hence,

In addition, . Therefore,

This proves that .

Copyright 2018 by Bruce Ikenaga