Some troubling counterexamples

An asymptotic expansion that does not come from Taylor's formula -------------------------------------------------------------------

To say that the function f has an asymptotic expansion of order 0 near 0 is equivalent to saying that f (x) = f (0) + o(1), that is, f (x) = f (0) + ε(x), where ε tends to 0 as its argument tends to 0. In other words, this simply says that f is continuous at 0.

To say that the function f has an asymptotic expansion of order 1 is equivalent to saying that f (x) = f (0) + ax + o(x), that is, $\frac{f(x)-f(0)}{x}=f^{\mathrm{\prime }}(0)+\epsilon (x)\backslash frac\{f(x)-f(0)\}\{x\}\; =\; f\text{'}(0)\; +\; \backslash varepsilon\; (x)$ where ε tends to 0 as its argument tends to 0. This is equivalent to saying that f is differentiable at 0 and that f ’(0) = a.

One might be tempted to conjecture that, for every n ≥ 2, the function f has an asymptotic expansion of order n near 0 if and only if f is n times differentiable at 0. One implication is true: if f is n times differentiable at 0, Taylor's formula gives it an asymptotic expansion of order n. The converse, however, is spectacularly false!

Consider the function