AP Calculus AB and BC

Limit of 1/(x^2-4) at 2 from the Right

Q: What happens approaching 2 from the left?

Then x - 2 is small and negative while x + 2 is near 4, so the quotient runs to -\infty. The two-sided limit does not exist.

The limit of one over x squared minus four as x approaches two from the right is infinity. Factoring the denominator into x minus two times x plus two shows that only the first factor vanishes, and on the right it is small and positive.

\lim_{x \to 2^+} \frac{1}{x^{2}-4} = \infty

Settled by factoring to isolate the vanishing factor.

Factor before deciding the sign

\frac{1}{x^{2}-4} = \frac{1}{(x-2)(x+2)}

Only $x - 2$ vanishes at $x = 2$ ; the other factor tends to 4, a positive constant. So the sign of the whole expression is the sign of $x - 2$ .

Approaching from the right, $x - 2$ is a small positive number, so the product is small and positive and the reciprocal is large and positive.

The second asymptote

The denominator also vanishes at $x = -2$ , so the graph has two vertical asymptotes. There the roles reverse: $x + 2$ is the vanishing factor and $x - 2$ tends to $-4$ , so approaching $-2$ from the right gives $-\infty$ .

Factoring first is what makes both cases quick. Trying to reason about $x^{2} - 4$ as a single object leaves you guessing at signs.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why factor when the denominator is already simple?

Because the sign near an asymptote is set by the vanishing factor alone. Factoring separates the factor that matters from the one that just contributes a constant.

What happens approaching 2 from the left?

Then $x - 2$ is small and negative while $x + 2$ is near 4, so the quotient runs to $-\infty$ . The two-sided limit does not exist.