Rational Function Approximation of the Maximum Function

Question

Is is possible to create a rational function $R(x)$ that is an approximation of $\max(x,0)$ (that is, as $x \to -\infty, \, R(x) \to 0$ and as $x \to \infty, \, R(x) \to x$)?

Answer 1

No. Given a rational function $R(x) = \frac{f(x)}{g(x)}$, where $f$ and $g$ are polynomials, the behavior is determined by comparing the polynomials' degrees:

$$ \begin{align} \deg f < \deg g &\implies \lim_{x \to -\infty} R(x) = \lim_{x \to +\infty} R(x) = 0 \\ \deg f = \deg g &\implies \lim_{x \to -\infty} R(x) = \lim_{x \to +\infty} R(x) \quad\mbox{is finite and not zero} \\ \deg f > \deg g &\implies \lim_{x \to \pm \infty} R(x) = \pm \infty \end{align} $$

Proof: Suppose the polynomials have coefficients $a_i$ and $b_j$:

$$ \begin{align} f(x) &= a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m \\ g(x) &= b_0 + b_1 x + b_2 x^2 + \cdots + b_n x^n \end{align} $$

where $a_m \neq 0$ and $b_n \neq 0$, so that $m = \deg f$ and $n = \deg g$. If $m \leq n$, we can rewrite $R(x)$ for $x \neq 0$ as:

$$ R(x) = \frac{a_0 + a_1 x + \cdots + a_m x^m}{b_0 + b_1 x + \cdots + b_n x^n} $$ $$ R(x) = \frac{a_0 x^{-m} + a_1 x^{1-m} + \cdots + a_m}{b_0 x^{-m} + b_1^{1-m} + \cdots + b_m + b_{m+1} x + \cdots + b_n x^{n-m}} $$

Then since negative powers of $x$ approach zero when $|x|$ is large,

$$ \lim_{x \to \pm \infty} R(x) = \lim_{x \to \pm \infty} \frac{a_m}{b_m + b_{m+1} x + \cdots + b_n x^{n-m}} $$

If $m < n$, the denominator approaches $\pm \infty$, so $\lim_{x \to \pm \infty} R(x) = 0$. If $m = n$, the denominator is the constant $b_n$, so $\lim_{x \to \pm \infty} R(x) = \frac{a_m}{b_n}$.

If $\deg f > \deg g$, then by the result above,

$$ \lim_{x \to \pm \infty} \frac{1}{R(x)} = \lim_{x \to \pm \infty} \frac{g(x)}{f(x)} = 0 $$

so we must have $\lim_{x \to \pm \infty} R(x) = \pm \infty$.