Taylor series at boundary of domain

Question

Is it possible to compute the Taylor series expansion of a function at an edge of its domain? If so, what are the conditions for it to hold? For example: Does

$f: \mathbb{R}^+ \to [1\,;+\infty), x \mapsto \exp(\sqrt x)$

have a Taylor series expansion at $x=0$?

Answer 1

No, the function $f$ has not a Taylor expansion at $0$ because $f$ is not differentiable at zero. However $f$ has an expansion (not polynomial) in a right neighbourhood of zero. Since $f$ is the composition of $e^x$, which has a Taylor expansion at $0$, and $\sqrt{x}$, which is continuous at $0$, we have that $$f(x)=\exp(\sqrt x)=\sum_{k=0}^n\frac{(\sqrt x)^k}{k!}+o((\sqrt x)^n).$$ Recall that a Taylor expansion at $0$ is a sum of non-negative integer powers of $x$.

Answer 2

No. The function $f$ is not differentiable at $x=0$, since

$\frac{f(1/n)-f(0)}{1/n}=nf(1/n) \to \infty $ as $n \to \infty$.