If I were to cover the Fibonacci Sequence in introducing sequences to Calculus students, I would probably avoid some of the more obscure properties. I would focus on what early Calculus students should know: for example, how to prove that a sequence converges, and if it does, then how to find what it converges to.
(Note also the places, not necessarily specified below, in which one needs to remind students about the basic properties of sequences and limits, and when things can be "split up" only when convergence has already been demonstrated.)
In this case, one can show the ratio $\frac{f_{n+1}}{f_n}$ is monotonically increasing and bounded above by $2$.
This is enough to conclude that the ratio converges to some limit as $n$ tends to infinity.
Call this limit $\phi$. Then consider two ways of writing $\frac{f_{n+2}}{f_n}$ as $n \rightarrow \infty$. The first way:
$$\lim_{n \rightarrow \infty} \frac{f_{n+2}}{f_n} = \lim_{n \rightarrow \infty} \frac{f_{n+1} + f_{n}}{f_n} = \lim_{n \rightarrow \infty} \frac{f_{n+1}}{f_n} + \frac{f_{n}}{f_n} = \lim_{n \rightarrow \infty} \frac{f_{n+1}}{f_n} + 1 = \phi + 1.$$
The second way:
$$\lim_{n \rightarrow \infty} \frac{f_{n+2}}{f_n} = \lim_{n \rightarrow \infty} \frac{f_{n+2}}{f_{n+1}} \cdot \frac{f_{n+1}}{f_n} = \lim_{n \rightarrow \infty} \frac{f_{n+2}}{f_{n+1}} \cdot \lim_{n \rightarrow \infty} \frac{f_{n+1}}{f_n} = \phi \cdot \phi = \phi^2.$$
Since these expressions are equal, we find that $\phi^2 = \phi + 1$, whence we can solve for $\phi$ using the quadratic equation. This will give two possibilities: one positive, and one negative. Noting that $\phi > 0$, we find the limit to which our ratio converges.
On the other hand, if you are looking for a somewhat nonstandard problem:
Call a binary string "tripletless" if it never contains three consecutive $0$s or $1$s.
How many tripletless binary strings are there of length $n$?
I worked this problem out some time ago, and found the answer is $2f_{n+1}$.
For example,
Length $1$: $\{0, 1\}$. Total: $2$, i.e., $2f_{2} = 2\cdot1$.
Length $2$: $\{00, 01, 10, 11\}.$ Total: $4$, i.e., $2f_{3} = 2\cdot2$.
Length $3$: $\{001, 010, 011, 100, 101, 110\}.$ Total: $6$, i.e., $2f_{4} = 2\cdot3$.
Length $4$: $\{0010, 0011, 0100, 0101, 0110, 1001, 1010, 1011, 1100, 1101\}.$ Total: $10$, i.e., $2f_{5} = 2\cdot5$.
(If more details on the proof would be helpful then I would be happy to provide them, but I think it is a nice problem to work out. For pedagogically appropriate materials, though, I think the example above is a much better choice.)
Edit (4 April 2016): An irresistible mathematical side-note. The "somewhat nonstandard problem" here essentially asks for a count of
base $2$ strings of length $n$ without $3$ consecutive occurrences of the same digit.
A natural generalization would be to ask for a count of
base $a$ strings of length $b$ without $c$ consecutive occurrences of the same digit.
I asked this very question way back in Apr 2014 on MSE (775863) and it now has a nice answer from Markus Scheuer that uses generating functions and the "Goulden-Jackson Cluster Method" (arXiv).
Edit (27 March 2017): I recently explored with a class of mine the following:
$$1+\frac{1}{1}, 1 + \frac{1}{1 + \frac{1}{1}}, 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}, \ldots $$
where each subsequent term is $1$ plus $1$ over the previous term. These yield
$$\frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \ldots$$
and, more generally, one can show that you get ratios of consecutive Fibonacci numbers. So the next one would be $8/5$, then $13/8$, etc. When you really do this out by hand, e.g., observing that $1 + 1/(8/5)$ is $1 + 5/8 = (8+5)/8 = 13/8$, you not only get to grasp why the Fibonacci numbers are emerging, but by taking the "limit" of the sequence, you have that $x = 1 + 1/x$, which allows you to solve for $x$ and pick the positive root of this disguised quadratic to figure out precisely what the ratio of consecutive Fibonacci terms converges to.