In this section, we will show that if
`b > 0``r`
is a rational number, the sequence given in equation 6.1-16
does indeed converge to `b ^{r}`, which,
according to

Why, you may ask, should anybody care? Because in the next section we need to find a way of taking the derivative of an exponential (which will be on the exam). In order to know that taking such a derivative can even be done, we need to know that the definition of exponential functions is useful and that exponentials are continuous (remember that you can't take the derivative of a function that is not continuous). If you plod through this optional material, not only will you gain a better understanding of limits, continuity, and exponentials, but you will also have confidence in your own convictions that the material in the next section is valid.

To prove the limit in 6.1-16, we need to develop our understanding
of exponentials still further. But it all follows from what you
learned about them in algebra. We begin by demonstrating that
the function, `f(x) = b ^{x}`

**Point 0:** Don't let the fancy term, *monotonic*, scare
you. It's just another word for *always*. So when something
is *monotonic increasing*, it is simply *always* increasing.
Likewise if something is *monotonic decreasing* then it is
*always* decreasing. If something is just plain *monotonic*
then it is simply either *always* one or *always* the other.

If you want to accept on faith that exponentials are monotonic and cut right to the chase, then click here, but you'll be missing some very pretty algebra logic.

**Point 1:** Pick two real numbers, `b` and
`c`, but pick them so that
`0 < b < c``n`. Observe that

bThis seems pretty obvious, but for anybody who is skeptical, you can prove it easily by induction on^{n}< c^{n}eq. 6.1-17

**Point 2:** Again pick two real numbers, `b` and
`c`, and again pick them so that
`0 < b < c``n`. Observe that

bThis one is not quite as obvious, but follows from point 1. For if it were not true then either^{1/n}< c^{1/n}eq. 6.1-18

**Point 3:** Pick `b` and `c` in the same way as in the
last two points. Pick any two counting numbers, `m` and `n`.
Observe that

bor equivalently^{m/n}< c^{m/n}eq. 6.1-19a

(bwhich follows immediately from 6.1-17 and 6.1-18. So our little power rule works for all positive rational exponents.^{1/n})^{m}< (c^{1/n})^{m}eq. 6.1-19b

**Point 4:** Pick `b > 1``n`. Observe that

1 < bAgain this seems obvious, but for the skeptical you can prove it by induction on^{n}eq. 6.1-20

bClearly^{p}< b^{p+n}eq. 6.1-21a

bwhenever^{p}< b^{q}eq. 6.1-21b

**Point 5:** Recall from algebra that if you have

pwhere the denominators are positive, then it must be true that_{1}p_{2}~~<~~~~eq. 6.1-22a q~~_{1}q_{2}

pThe second statement also implies the first. Pick_{1}q_{2}< p_{2}q_{1}eq. 6.1-22b

bIf both^{p1q2}< b^{p2q1}eq. 6.1-23

bThat proves that when^{p1/q1}< b^{p2/q2}eq. 6.1-24

For negative rational exponents, you can show that the same relationships
exist. Simply let `c = 1/b``b > 1``0 < c < 1`

band^{-p/q}= c^{p/q}eq. 6.1-25a

p < qIn this way you can prove the monotonic properties of exponentials that have negative exponents using what we have shown with positive exponents.Û-p > -q eq. 6.1-25b

The conclusion is that all exponential functions, `b ^{x}`,
where

We demonstrate this using
*The Stepping Stone Theorem*. We
know that `b ^{0} = 0`

First, let

1 1 1 s =be our reference sequence. It clearly converges to zero. Also every element of~~,~~~~,~~~~, ... eq. 6.1-26 2~~^{1}2^{2}2^{3}

Pick `b > 0``b ^{s1}, b^{s2}, b^{s3}, ... `

lim bObserve that^{sn}= 1 eq. 6.1-27 n~~> ¥~~

(bNow we break^{sn+1})^{2}= b^{sn}eq. 6.1-28

_ Öb = bNotice that all the terms,^{s1}= 1 + h_{1}eq. 6.1-29a

___ ÖbBut this means that^{sn}= b^{sn+1}= 1 + h_{n+1}eq. 6.1-29b

1 + hor equivalently_{n}= (1 + h_{n+1})^{2}= 1 + 2h_{n+1}+ h_{n+1}^{2}eq. 6.1-30a

hClearly_{n}= 2h_{n+1}+ h_{n+1}^{2}eq. 6.1-30b

lim hBut_{n}= 0 eq. 6.1-31 n~~> ¥~~

lim bYou can also prove that^{sn}= 1 eq. 6.1-32 n~~> ¥~~

lim bletting^{-sn}= 1 eq. 6.1-33a n~~> ¥~~

lim cwhich we have already proved.^{sn}= 1 eq. 6.1-33b n~~> ¥~~

If you are still awake, then you ought to be complaining right now that
I have proved the stepping stone premise for only two sequences.
*The Stepping Stone Theorem*
requires that you show that `f(a _{n})`
converges to

But here is the kicker -- and the reason that I went to so much
trouble to show that exponentials are monotonic. If you have
a monotonic function, `f(x)`, then you only need to show
the stepping stone premise for two convergent sequences to show
that is it true for all of them. You just have to make sure that
one of those sequences converges from above and the other from
below, which I have already done for exponentials. You also
have to make sure that *no* element of either sequence is
exactly equal to its limit. That is also true of the two sequences
we developed above.

Here's why monotonic functions are special. Let
`s _{1}, s_{2}, s_{3} ... `

lim f(sLet_{n}) = lim f(t_{n}) = f(a) eq. 6.1-34 n~~> ¥ n~~~~> ¥~~

Since `x _{n}` also converges to

tis true for that_{k}< x_{n}< s_{j}eq. 6.1-35

Because `f(x)` is monotonic, it means that either

f(talways, or_{k}) < f(x_{n}) < f(s_{j}) eq. 6.1-36a

f(talways, depending upon whether_{k}) > f(x_{n}) > f(s_{j}) eq. 6.1-36b

lim f(xfor_{n}) = f(a) eq. 6.1-37 n~~> ¥~~

Since exponentials are all monotonic, the above applies to them at
`x = 0``x _{n}`'s are
rational. But remember that every real number, irrationals included,
can be approached in the limit by a sequence of rationals. So if

xwhere all the_{1}= lim q_{1,n}n~~> ¥ x~~_{2}= lim q_{2,n}n~~> ¥ x~~_{3}= lim q_{3,n}n~~> ¥ . . . Table 6.1-1~~

I realize that this last argument, with the double subscripts and
all, is difficult. If you don't get it at first, you may want to
move on or try again some other time. But it does provide the
final nail in the proof that
all exponentials are continuous at `x = 0`

(this one is easy)

Suppose you have any sequence,
`x _{1}, x_{2}, x_{3}, ... `,

lim xNow subtract_{n}= q eq. 6.1-37 n~~> ¥~~

lim xNow we have a sequence that converges to zero, and we know a thing or two about the exponentials of such sequences from the discussion above. Namely that the exponential of such a sequence has to converge to a limit of_{n}- q = 0 eq. 6.1-38 n~~> ¥~~

bNow simply multipy through by^{xn}lim b^{xn-r}= lim~~= 1 eq. 6.1-39 n~~~~> ¥ n~~~~> ¥ b~~^{q}

lim bwhich means that exponentials of^{xn}= b^{q}eq. 6.1-40 n~~> ¥~~

Finally, if a sequence of rational numbers,
`q _{1}, q_{2}, q_{3} ... `,

email me at *hahn@netsrq.com*