Asymptotic approximation of $x^\alpha$ by entire functions 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.
 A: 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. 
A: Start with an entire function $f$ such that $f(x)=1/x + O(1/x^2)$ for $x>0$, $x\rightarrow\infty$. For example $f(z)= (1-e^{-z})/z$. 
Let F be some primitive for $f$: $F(z)=\int_1^z f(s)ds$.
We have $F(x)= ln(x)+C+O(1/x)$,  with C some constant ($ \ C=\int_1^\infty \ (f(x)-{1\over x})\ dx$ ).
Then consider $h(x)=exp(\alpha F(x)-\alpha C)$. 
We get ${h(x)\over x^\alpha} = exp(O(1/x))\rightarrow 1$.
