How to plot and analyze function $f(x)=\lfloor\frac{3-x}{1+x^2}\rfloor$? How to prove $\lim_{x\to -\infty}f(x)=0$ and $\lim_{x\to \infty}f(x)=-1$?
My work:
1) Function is well defined for every $x\in R$.
2) Function is not even nor odd.
3) For $x=0$ we have $f(x)=3$.
4) Function probably has horizontal asymptotes $y=0$ and $y=-1$, but I don't know how to prove that. I could only prove $0\geq\lim_{x\to\pm\infty} f(x)\geq -1$ if limits exist, since we have from the definition of the floor function: $\frac{2-x-x^2}{x^2+1}\leq f(x)\leq \frac{3-x}{x^2+1}$.
5) I can prove that $f(x)$ is smaller than $3+\frac{1}{2}$ for every $x$, but wolframalpha gives better result ($f(x)\leq 3$ and $max f(x)=3$). Also, $min f(x)=-1$ for $x>3$.
Now, I'm stuck at this point.
Any help is welcome. Thanks in advance.
Showing the limits at infinity is straightforward. Simply consider the related rational function $$g(x) = \frac{3-x}{1+x^2}$$ which, by definition of the floor function, satisfies $$f(x) \le g(x) < f(x) + 1$$ for all $x$, and for which $$\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} \frac{\frac{3}{x^2} - \frac{1}{x}}{\frac{1}{x^2} + 1} = 0,$$ and similarly, $$\lim_{x \to \infty} g(x) = 0.$$ However, for $x > 3$, we note $g(x) < 0$, thus $g \nearrow 0$ as $x \to \infty$: that is to say, $g$ approaches $0$ from below, hence $$\lim_{x \to \infty} f(x) = -1.$$ However, for $x < 3$, $g(x) > 0$, and it follows that $g \searrow 0$ as $x \to -\infty$, hence $$\lim_{x \to -\infty} f(x) = 0.$$