Box 6.0: Proof that Exponentials are Continuous

Why the Limit Always Exists (optional material on exponentials)

In this section, we will show that if b > 0 and the limit shown in equation 6.1-15 exists, then the limit in equation 6.1-16 also must exist. We shall also demonstrate that when r is a rational number, the sequence given in equation 6.1-16 does indeed converge to b^r, which, according to The Stepping Stone Theorem, proves that f(x) = b^x is continuous if you restrict yourself to rational x. By using 6.1-16 to define exponentiation among irrational x, we preserve continuity of f(x) = b^x among all real x, including the irrationals.

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.

Outline of Development

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 is monotonic among rational x's (with the exception of when b = 1). Monotonic means that the function either always increases or always decreases with increasing x. Knowing that exponentials are monotone is enough to prove that the limit in 6.1-6 is always true when r = 0. And finally, knowing that is true for r = 0 is enough show that is it true for all r, both rational and irrational.

Demonstration that Exponentials Functions are Monotonic

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. Now pick any counting number, n. Observe that

   bⁿ < cⁿ                                                        eq. 6.1-17

This seems pretty obvious, but for anybody who is skeptical, you can prove it easily by induction on n.

Point 2: Again pick two real numbers, b and c, and again pick them so that 0 < b < c. Also pick any counting number, n. Observe that

   b^1/n < c^1/n                                                    eq. 6.1-18

This one is not quite as obvious, but follows from point 1. For if it were not true then either b^1/n = c^1/n or b^1/n > c^1/n. They certainly can't be equal because that would mean that if you raised them both to the nth power, you'd have b = c, and we chose b and c not to be equal. Likewise, if b^1/n > c^1/n then by the rule in point 1, you could raise both sides to the nth power and have b > c, and that is not how we chose b and c either. Since neither of the alternatives can be true, 6.1-18 must be true.

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

   b^m/n < c^m/n                                                    eq. 6.1-19a

or equivalently

   (b^1/n)^m < (c^1/n)^m                                              eq. 6.1-19b

which follows immediately from 6.1-17 and 6.1-18. So our little power rule works for all positive rational exponents.

Point 4: Pick b > 1. Pick any counting number, n. Observe that

   1 < bⁿ                                                         eq. 6.1-20

Again this seems obvious, but for the skeptical you can prove it by induction on n. You can immediately extend this by picking another counting number, p, and multiplying both sides of 6.1-20 by b^p.

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

Clearly p < p+n. So let q = p+n. This becomes

   b^p < b^q                                                        eq. 6.1-21b

whenever p < q. This proves that for counting number exponents only, b^k is monotonic increasing whenever b > 1. Using a nearly identical argument, you can also show that for counting number exponents only, b^k is monotonic decreasing whenever 0 < b < 1.

Point 5: Recall from algebra that if you have

   p₁   p₂
      <                                                           eq. 6.1-22a
   q₁   q₂

where the denominators are positive, then it must be true that

   p₁q₂ < p₂q₁                                                    eq. 6.1-22b

The second statement also implies the first. Pick p₁, p₂, q₁, and q₂ as any counting numbers you like, but with the restriction that the inequality in 6.1-22b holds. Pick a real number, b > 1. According to point 4, it must be true that

   b^p₁q₂  <  b^p₂q₁                                                 eq. 6.1-23

If both q₁ and q₂ are counting numbers, then their product is also. That means that if you raise both sides of 6.1-23 to the 1/q₁q₂ power, the rule in point 2 applies, and you get

   b^p₁/q₁  <  b^p₂/q₂                                               eq. 6.1-24

That proves that when b > 1 and the exponents are positive rational numbers, the exponential function, b^x, is monotonic increasing. Using a nearly identical argument you can show that if 0 < b < 1, then the exponential function, b^x, is monotonic decreasing whenever x is a positive rational.

For negative rational exponents, you can show that the same relationships exist. Simply let c = 1/b. Then if b > 1, it must be true that 0 < c < 1, and vice versa. Then use the relationships of

   b^-p/q  =  c^p/q                                                 eq. 6.1-25a

and

   p < q   Û   -p > -q                                            eq. 6.1-25b

In this way you can prove the monotonic properties of exponentials that have negative exponents using what we have shown with positive exponents.

The conclusion is that all exponential functions, b^x, where b > 0 and not equal to 1, are monotonic for all rational exponents. If b > 1, the function is monotonic increasing. If 0 < b < 1, the function is monotonic decreasing.

Proof That Exponentials are Continuous at Zero

We demonstrate this using The Stepping Stone Theorem. We know that b⁰ = 0 for all positive b. If we can show that whenever a sequence, r₁, r₂, r₃, ... , converges to a limit of zero, then b^r₁, b^r₂, b^r₃, ... , must converge to a limit of 1, we can apply The Stepping Stone Theorem and conclude that f(x) = b^x is continuous at x = 0. We shall show it first for rational x, and then demonstrate that it must be true as well for irrational x.

First, let

          1   1   1
   s  =    ,   ,   , ...                                          eq. 6.1-26
         2¹  2²  2³

be our reference sequence. It clearly converges to zero. Also every element of s is rational, so we can use them as exponents according to the algebraic definitions we have already established. We shall use the symbol, s_n, to represent the nth element of the sequence, s.

Pick b > 0. Then b^s₁, b^s₂, b^s₃, ... is a sequence as well. We shall first prove using basic algebra that

    lim   b^s_n  =  1                                               eq. 6.1-27
   n  > ¥

Observe that b^s₁ is simply sqrt(b). Further observe that every subsequent term in the sequence is simply the square root of its predecessor. So it is always true that

   (b^s_n+1)²  =  b^s_n                                                eq. 6.1-28

Now we break b^s₁ apart into two components.

    _
   Öb  =  b^s₁  =  1 + h₁                                     eq. 6.1-29a

Notice that all the terms, b^s_n, are positive. And just like the first, we can break each of them up into two components.

    ___
   Öb^s_n  =  b^s_n+1  =  1 + h_n+1                                eq. 6.1-29b

But this means that

   1 + h_n  =  (1 + h_n+1)²  =  1 + 2h_n+1 + h_n+1²                   eq. 6.1-30a

or equivalently

   h_n  =  2h_n+1 + h_n+1²                                           eq. 6.1-30b

Clearly h_n+1² is always positive. That means that h_n+1 is always less than half of h_n. And we must conclude that

    lim   h_n  =  0                                                eq. 6.1-31
   n  > ¥

But h_n is nothing more than the difference between 1 and b^s_n. Since the difference goes to zero in the limit, it must be that

    lim   b^s_n  =  1                                               eq. 6.1-32
   n  > ¥

You can also prove that

    lim   b^-s_n  =  1                                              eq. 6.1-33a
   n  > ¥

letting c = 1/b. Then 6.1-33a becomes

    lim   c^s_n  =  1                                               eq. 6.1-33b
   n  > ¥

which we have already proved.

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 f(a) for any sequence, a_n, that converges to a before you can conclude that f(x) is continuous at a. Proving it for only two special cases just doesn't nail it. So we don't yet know for sure that exponentials are continuous at x = 0.

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 s₁, s₂, s₃ ... and t₁, t₂, t₃ ... be the two sequences (that is the converges-from-above sequence and the converges-from-below sequence respectively) that converge to a and that have the stepping stone property -- that is

    lim   f(s_n)  =   lim   f(t_n)  =  f(a)                         eq. 6.1-34
   n  > ¥          n  > ¥

Let x₁, x₂, x₃ ... be any other sequence that also converges to a limit of a. Now pick any s_j from the converges-from-above sequence and any t_k from the converges-from-below sequence. Clearly a must be between s_j and t_k. That is why we chose the two sequences the way we did.

Since x_n also converges to a, we can always find an n big enough so that

   t_k  <  x_n  <  s_j                                               eq. 6.1-35

is true for that n and all subsequent n's. In other words, you can always go deep enough into the x_n's so that every x_n there and beyond is closer to a than either s_j and t_k.

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

   f(t_k)  <  f(x_n)  <  f(s_j)                                      eq. 6.1-36a

always, or

   f(t_k)  >  f(x_n)  >  f(s_j)                                      eq. 6.1-36b

always, depending upon whether f(x) is monotonic increasing or monotonic decreasing. Either way, f(x_n) is always trapped between f(s_j) and f(t_k) -- always, no matter how close together f(s_j) and f(t_k) are. And since they both converge, we know we can get them as close together as you like -- that is within as small an epsilon of each other as you'd like. Since f(a) is also in between them (again because f(x) is monotonic), that traps f(x_n) within as small an epsilon as you'd like of f(a). And that means that

    lim   f(x_n)  =  f(a)                                          eq. 6.1-37
   n  > ¥

for any x₁, x₂, x₃, ... that converges to a.

Since exponentials are all monotonic, the above applies to them at x = 0. But with exponentials we have still shown only that it works when all the 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 x₁, x₂, x₃, ... is a sequence of reals, then we have

   x₁  =  lim   q_1,n
         n  > ¥


   x₂  =  lim   q_2,n
         n  > ¥


   x₃  =  lim   q_3,n
         n  > ¥
   
   .
   .
   .
                      Table 6.1-1

where all the q's are rational. No matter how you choose s_j and t_k, you can always go down far enough in table 6.1-1 so that eventually you'll find a row that converges between them. Because the exponential is monotonic, that means that for the same row of the table (call it row m), f(q_m,n) will confine itself between f(s_j) and f(t_k) if you make n big enough. And so will all subsequent rows.

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.

Extending the Proof to Show Exponentials are Continuous Everywhere

(this one is easy)

Suppose you have any sequence, x₁, x₂, x₃, ... , that converges to some rational number, q. Then

    lim   x_n  =  q                                                eq. 6.1-37
   n  > ¥

Now subtract q from both sides.

    lim   x_n  - q  =  0                                           eq. 6.1-38
   n  > ¥

Now 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 1.

                            b^x_n
    lim   b^x_n-r  =   lim          =  1                            eq. 6.1-39
   n  > ¥          n  > ¥   b^q

Now simply multipy through by b^q and you have shown that

    lim   b^x_n  =  b^q                                              eq. 6.1-40
   n  > ¥

which means that exponentials of x are continuous among rational values of x.

Finally, if a sequence of rational numbers, q₁, q₂, q₃ ... , confines itself to as small an interval as you like by choosing the subscript large enough, it is also true that b^q₁, b^q₂, b^q₃, ... must confine itself to as small an interval as you like by choosing the subscript large enough. Why? Because the exponential is continuous among the rationals. But that is the same as saying that b^q₁, b^q₂, b^q₃, ... is a Cauchy sequence and must therefore converge to some real number in the limit. So the limit always exists, whether q_n converges to a rational or to an irrational.

Return to main text

email me at hahn@netsrq.com