Answer
$\lim\limits_{x \to \infty}\frac{x}{x^2-1} = 0$
Work Step by Step
The x^2 term in the denominator increases much faster than the x term in the numerator as x approaches infinity; thus, the denominator will be much larger than the numerator.
We look at the term with the highest degree in both the numerator and denominator, as these are the only values that will matter when x approaches infinity.
$\lim\limits_{x \to \infty}\frac{x}{x^2-1} = \lim\limits_{x \to \infty}\frac{x}{x^2} = \lim\limits_{x \to \infty}\frac{1}{x} = 0$
