Asymptotic approximation of the arctangent? That is, I am looking for an algebraic function $f(x)$ that approximates $\arctan x$ for large values of $x$.
The approximation could be reasonably modest -- perhaps something like 
$$\tan (f(x)) = \frac{\pi}{4} + O\left(\frac{1}{x^2}\right).$$

Use the following
$$\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} \Leftrightarrow$$
$$\arctan(x) = \frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$$
The series for $\arctan(x)$ is
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$
Now substitute $\frac1x$ to get the result
$$\arctan(x)=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^3}-\frac{1}{5x^5} +\ldots$$
So for large values of $x$
$$\arctan(x)\sim\frac{\pi}{2}$$
which is logical, because $\displaystyle \lim_{x\to\infty}\arctan(x)=\frac{\pi}{2}$.
