When we talk about an annual rate of growth equal to r, we usually mean that:

[A]       P(t+1) = P(t) (1+ r).

If the growth is constant, then, after n years (taking into consideration annual compounding):
[B]       P(t+n) = P(t) (1+ r)ⁿ.

>I remember that from math 101 where we ...
Pay attention!
Now suppose we compound m = 365 times each year, with the rate reduced from an annual r to a daily r / m.
Then the annual growth (after m = 365 days of compounding) is:

[C]       P(t+1) = P(t) (1+ r/m)^m.

>And what if we have continuous compounding ... like m = ∞ ?
If we let m ∞ we'd get:

[D]       P(t+1) = P(t) e^r ... using the fact that (1+ 1/m)^m e as m ∞.

If the growth is constant, then, after n years (taking into consideration annual compounding):
[E]       P(t+n) = P(t) e^{r n}.

>I assume there's a difference between [A] and [D], eh? Between using (1+r) and using er? Yes, there's a difference. However, for small values of r, er ≈ 1+ r. >But if [A] is what's understood by annual growth, then why ...? Mathmanship. Math types gets all excited when exponentials make an appearance. >Huh?

Imagine that the growth changes from year to year ... from r₁ to r₂ to r₃ to ... to r_n.
The final value would be either:
[B1]       P(n) = P(0) (1+ r₁)(1+ r₂)(1+ r₃)...(1+ r_n)
or
[E1]       P(n) = P(0) e ^{r₁+ r₂+r₃+...+r_n}

Which looks more sanitary?

>I'll take [E1] any day!
Yeah ... and notice that, for the [E1] case, we can write:
[E2]       P(n) = P(0) e^{R n} ... where R = (r₁+ r₂+r₃+...+r_n)/n is the Average annual growth rate.

>Aha! Then it looks exactly like [E], right?
Right.

Now, a question of nomenclature.
If somebuddy refers to "continuous" rate of return, it's unclear whether they mean [A] and [D].
If, however, they refer to "continuously compounded" rate of return, then they clearly mean ...

>They mean to use the exponential, eh?
Yes ... possibly sacrificing real-world applicability for the sake of ... of ...

>For the sake of mathmanship!
You got it!

See: Math Stuff