Find all asymptotes of y=(x³)/(x²-1)

Find all asymptotes of $y=\dfrac{x^3}{x^2-1}$.

Step 1: Vertical asymptotes

The denominator vanishes at $x=\pm1$, where the numerator ($\pm1$) does not. Checking each one-sided limit directly — e.g.\ at $x=-1$, for $x$ slightly greater than $-1$ both $x^3$ and $x^2-1$ are negative, so the ratio is positive, while for $x$ slightly less than $-1$, $x^2-1$ turns positive against a still-negative $x^3$, giving a negative ratio — the four one-sided limits are

\[ \lim_{x\to1^{+}}\frac{x^3}{x^2-1}=+\infty,\quad \lim_{x\to1^{-}}\frac{x^3}{x^2-1}=-\infty,\quad \lim_{x\to-1^{+}}\frac{x^3}{x^2-1}=+\infty,\quad \lim_{x\to-1^{-}}\frac{x^3}{x^2-1}=-\infty. \tag{1} \]

Both $x=1$ and $x=-1$ are vertical asymptotes, with the same $(+\infty$ from the right, $-\infty$ from the left$)$ pattern at each.

Step 2: Oblique asymptote — via polynomial division

\[ \frac{x^3}{x^2-1} = x + \frac{x}{x^2-1}. \tag{2} \]

\[ k = \lim_{x\to\infty}\frac{x^2}{x^2-1} = 1, \qquad b = \lim_{x\to\infty}\!\left(\frac{x^3}{x^2-1}-x\right) = \lim_{x\to\infty}\frac{x}{x^2-1} = 0. \tag{3} \]

Step 3: Verification

\[x=0.99,1.01\text{: }y\approx-48.76,\,51.26\ (\to\mp\infty\text{ as }x\to1^{\mp});\qquad x=-0.99,-1.01\text{: }y\approx48.76,\,-51.26\ (\to\pm\infty\text{ as }x\to-1^{\pm})\ \checkmark\]

\[x=1000\text{: } y-x\approx0.001\to0\ \checkmark\]