Constructing a function satisfying two limits

Question

Is it possible to construct a smooth function $\rho(x)$ such that,
$$ \rho(x) \stackrel{x\to 0}\sim x^{-3/2} \quad \textrm{and} \quad \rho(x) \stackrel{x\to \infty}\sim x^{-6}$$ My try:
Guess $\rho(x) \sim x^{-3 f(x)}$. So to satisfy the condition, $f(x)$ must follow: $\lim_{x\to0} f(x) =\lim_{x\to \infty} \frac{1}{f(x)}=\frac{1}{2}$. Now my question is how to find out the function $f(x)$ efficiently?

Answer 1

Let's take a look at $\phi(x) = \frac1{\rho(x)}$. We have $$ \phi(x) \stackrel{x\to 0}{\sim}x^{3/2},\quad \phi(x) \stackrel{x\to \infty}{\sim}x^{6} $$ One simple function (perhaps the simplest function) that does this is $x^{3/2} + x^6$, since $x^{3/2}$ dominates for small $x$ and $x^6$ dominates for large $x$. This gives $$ \rho(x) = \frac{1}{x^{3/2} + x^6} $$

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If we want to continue your example, we need a nice function with a given value $\frac12$ at $0$ and a given limit $2$ at $\infty$. There are many ways to do it, for instance by manipulating the $\arctan$ function: $f(x) = \frac12 + \frac{3}{\pi}\arctan(x)$, or by using a fraction: $f(x) = 2 - \frac{3/2}{x+1}$.

Answer 2

The general approach would be to use functions that approaches a constant value at other points of interrest and have the desired behavior and then multiply them together.

The first factor would be for the $0$ limit behavior. This can be achieved by adding one to $x^{-3/2}$ since that would make it behave constant at infinity: $$\rho_0(x) = x^{-3/2}+1$$

The second factor we need to make sure that $x^{-6}$ at infinity converges at $0$. We can do this by adding one in the denominator: $$\rho_\infty(x) = {1\over x^6+1}$$

Then we just multiply them together:

$$\rho(x) = \rho_0(x)\rho_\infty(x) = {x^{-3/2}+1\over x^6+1}$$