Limits at infinity, in terms of another function Suppose:
$$\lim_{x\to\infty}\; f(x)  +  2x^2 =1$$
Determine if possible:
$$\lim_{x\to\infty}\; \frac{f(x)}{x^2+1}$$
I determined the answer to be $-2$, through substituting for $f(x)$:
$$\frac { 1 - 2x^2 }{x^2 + 1}$$
and I solved for the infinite limit.
however I feel like that substitution is illegal, or is it? 

I think it's okay, but you're skipping a lot of steps that may not work in other problems. To be thorough (and to see exactly what's going on), I'd write it like so:
\begin{align}
\frac{f(x)}{x^2+1} &= \frac{f(x)+2x^2-2x^2}{x^2+1} \\
&= \frac{f(x)+2x^2}{x^2+1} - \frac{2x^2}{x^2+1}
\end{align}
By the laws of limits, we can split the limits across this difference as long as the separated limits converge (which they will):
\begin{align}
&\lim_{x \to \infty} \frac{f(x)+2x^2}{x^2+1} - \lim_{x \to \infty} \frac{2x^2}{x^2+1} \\
&= \lim_{x \to \infty} \frac{1}{x^2+1} - 2 \\
&= 0 - 2 \\
&= \boxed{-2}
\end{align}
