I evaluate the following limit involving the Airy functions
$$\lim_{l \to 0} \text{Ai}(z+l) \text{Bi}^{'}(z) - \text{Bi}(z+l) \text{Ai}^{'}(z) \qquad (1)$$
by making a Taylor expansion of the Airy functions, writing $\text{Ai}(z+l) \approx \text{Ai}(z) + l \text{Ai}^{'}(z) + \cdots$ and $\text{Bi}^{'}(z+l) \approx \text{Bi}^{'}(z) + l \text{Bi}^{''}(z) + \cdots$. Substituting gives the limit as $$\lim_{l \to 0} W( \text{Ai}(z), \text{Bi}(z) ) + \text{terms of order} \, l.$$ where $W()$ is the Wronskian (equal to $1/\pi$ here) which is expected (for small $l$ you are indeed determining the Wronskian).
The question: Shouldn't I get the same result for $l$ fixed and letting $z \to \infty$?
As $z$ increases the naïve argument appears to be that you are evaluating something between two points that are getting closer and closer to each other. However I don't seem to get the answer expected. For example if I were to look at (1) and make use of the (first order) asymptotic forms for large (positive) $z$ then I would get terms like (ignoring factors like $z^{\pm 1/4}, \pi$ that appear)
$$e^{-2/3 (z+l)^{3/2}} e^{2/3 z^{3/2}} - e^{+2/3 (z+l)^{3/2}} e^{-2/3 z^{3/2}} $$
which is OK for fixed $z$ and letting $l \to 0$ but isn't bounded for $z \to \infty$.