Well it isn’t really calculus as there’s no differentiation (I guess you could consider the last step where you take a limit to be calculus), but it’s a bit like differential equations. You can write down the recurrence relation:

  a_(n+2) = a_(n+1) + a_n

Observe that there is a linear solution space (I.e. if you add solutions point wise or multiply each value by the same scalar, you get solutions), and the values a_0 and a_1 are sufficient to determine the sequence. Now guess that a_n = k^n is a solution:

  k^2 = k + 1
  (k - 1/2)^2 = 5/4
  k = (1 +/- sqrt(5))/2
  k = φ or -1/φ, where φ is the golden ratio

Due to linearity, there are a family of solutions a_n = Rφ^n + S(-1/φ)^n for any values of R and S. Because this family provides a solution for any choices of a_0 and a_1, it contains all the solutions.

Because |1/φ|<1, we find that asymptotically a_n ~ Rφ^n as n grows. Therefore the ratio of terms tends to φ in the limit.

You're right, of course. Thanks for the explanation. I was thinking particularly of the comment about the behavior with arbitrary initial numbers. Start with 1, 5000, and it still converges quickly to the golden ratio, as the parent comment mentioned. I enjoy watching the initial constants disappear.