TITLE: Asymptotic approximation of $x^\alpha$ by entire functions
QUESTION [15 upvotes]: Given a non-integral real $\alpha$, is there an entire (see http://en.wikipedia.org/wiki/Entire_function) function $h(x)$ such that $x^{-\alpha}h(x)\longrightarrow 1$
for $x\rightarrow+\infty$ (with $x$ real non-negative)?
Clearly, such a function if it exists is not unique since $h(x)+e^{-x}$ and similar functions
work also.

REPLY [24 votes]: As a matter of fact, real entire functions (that is, entire functions that map the real line into itself, or equivalently, functions represented by a power series centered in  0, with real coefficients and radius of convergence infinite) are dense in $C^0({\mathbb R}, \mathbb{R})$ in the sense of the order, that is:

Theorem (T.Carleman, 1927). For any two continuous real valued
  functions f < g there exists a real
  entire function $\phi$ in between:
$f(x)<\phi(x) < g(x)$ for all $x\in\mathbb{R}$.

So in particular, an entire function may be asymptotic to any continuous real function, and also, it may grow as fast as any continuous function.