Find the supremum and infimum of the set $\left \{\frac{x}{x^2 -1} : x\in \Bbb R, x^2 \neq 1\right \}$.

Question

I know how to find the limit to plus and minus infinity of $\frac{x}{x^2 -1}$, which would be $0$. And the limit at $0$ would also be $0$. But I am having a very hard time visualizing this set and even a harder time finding its supremum and infimum.

WolframAlpha says the the image of $\frac{x}{(x^2 - 1)}$ is $\Bbb R$, therefor I think this set is not limited. But I do not know how to prove that the image is in fact $\Bbb R$ and that the supremum and infimum do not exist.

Answer 1

Hint: What does the fact that $$\lim_{x\to 1^-}\frac{x}{x^2-1}=-\infty \quad \text{and} \qquad \lim_{x\to -1^+}\frac{x}{x^2-1}=\infty$$ tell you about $A$? Can it be bounded?

Answer 2

The fact that $x^2$ $\neq$ $1$ means that $x$ $\neq$ $1$ or $-$$1$ but if we get the left and right limits of the function we get $-$$\infty$ and $\infty$ respectively. So we have no supremum or infimum there.