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Behavior near a vertical asymptote

Consider the function f(x)=x2−1x3−xf(x) = \dfrac{x^2-1}{x^3-x}.

Simplify f(x)=x2−1x3−xf(x) = \dfrac{x^2-1}{x^3-x} by factoring, and use the simplified form to evaluate lim⁡x→0x2−1x3−x.\lim_{x \to 0} \frac{x^2-1}{x^3-x}. If the limit is infinite, state whether it is +∞+\infty or −∞-\infty; if it does not exist in either infinite sense, state that as well.

Hint

Factor both numerator and denominator completely first — notice that x3−x=x(x−1)(x+1)x^3 - x = x(x-1)(x+1) and x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1). After canceling common factors, examine what happens to the remaining expression as x→0x \to 0 from the left and from the right.

Step-by-step solution

Factor: x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1) and x3−x=x(x−1)(x+1)x^3-x = x(x-1)(x+1). Canceling the common factor (x−1)(x+1)(x-1)(x+1) (valid since x≠±1x \ne \pm 1 near x=0x=0), we get f(x)=1xf(x) = \dfrac{1}{x} for xx near 00 but x≠0x \ne 0.

Now consider lim⁡x→01x\lim_{x\to 0}\frac{1}{x}. As x→0+x \to 0^+, 1x→+∞\frac{1}{x} \to +\infty, while as x→0−x \to 0^-, 1x→−∞\frac{1}{x} \to -\infty. Since the one-sided limits disagree in sign, the two-sided limit does not exist in either the finite or infinite sense; there is a vertical asymptote at x=0x=0.

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