AP Calculus AB and BC

Limit of 1/x^2 as x Approaches 0

Q: Is writing infinity an acceptable answer on the AP exam?

Yes, and it is the expected one. The notation \lim_{x \to 0} \frac{1}{x^2} = \infty is the standard way to record unbounded behaviour, and it also identifies the vertical asymptote at x = 0.

The limit of 1 over x squared as x approaches 0 is infinity. Both sides agree because squaring makes the denominator positive whatever the sign of x, so the values grow without bound from the left and from the right. For 1 over x the two sides disagree, and that two-sided limit does not exist.

\lim_{x \to 0} \frac{1}{x^2} = \infty

Settled by unbounded behaviour at a vertical asymptote.

Checking both sides

For every $x \neq 0$ the square $x^2$ is positive. So as $x$ approaches 0 the denominator shrinks to 0 through positive values only, and its reciprocal grows without bound in the positive direction no matter which side you come in from.

\lim_{x \to 0^-} \frac{1}{x^2} = \infty \qquad \lim_{x \to 0^+} \frac{1}{x^2} = \infty

$x$	$\frac{1}{x^2}$
$\pm 0.1$	100
$\pm 0.01$	10,000
$\pm 0.001$	1,000,000

The one-sided limits agree, so the two-sided limit reports the same unbounded behaviour.

\lim_{x \to 0} \frac{1}{x^2} = \infty

Why substitution fails, and what form this actually is

Substitution gives $\frac{1}{0}$ , which is undefined. It is not the indeterminate $\frac{0}{0}$ : a nonzero numerator over a vanishing denominator always means unbounded values, so the size of the answer is settled before you start. The only open question is the sign.

That makes the work a sign analysis rather than an algebra problem. Look at the denominator, decide its sign on each side of the trouble point, and read off $\infty$ or $-\infty$ . Here the power is even, so both sides come back positive.

The contrast that matters

$\frac{1}{x}$ has an odd power downstairs. From the left the denominator is negative, so $\frac{1}{x} \to -\infty$ ; from the right it is positive, so $\frac{1}{x} \to \infty$ . The one-sided limits disagree, so $\lim_{x \to 0} \frac{1}{x}$ does not exist, not even as an infinite limit.

The mistake students make

Answering that the limit does not exist and stopping there. It is not a real number, true, but $\infty$ records how it fails, and that is what the rubric asks for.
Carrying the $\frac{1}{x}$ answer over out of habit. The sign split there is a property of the odd power, not of reciprocals in general.
Deciding the sign from one side only. Both sides match here, but for $\frac{1}{x(x-1)}$ near 0 the left side runs to $\infty$ and the right side to $-\infty$ , so that two-sided limit does not exist.
Calling $x = 0$ a removable discontinuity and trying to assign a value there. It is an infinite discontinuity, a vertical asymptote, and no assigned value repairs it.

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 does 1/x^2 have an infinite limit while 1/x does not?

The square makes the denominator positive on both sides, so both one-sided limits are $\infty$ and they agree. With $\frac{1}{x}$ the denominator changes sign at 0, so the two sides run to opposite infinities and no two-sided limit exists, finite or infinite.

Is writing infinity an acceptable answer on the AP exam?

Yes, and it is the expected one. The notation $\lim_{x \to 0} \frac{1}{x^2} = \infty$ is the standard way to record unbounded behaviour, and it also identifies the vertical asymptote at $x = 0$ .

Does the same reasoning cover 1/(x-3)^2?

Yes. Any even power in the denominator forces the same positive-from-both-sides behaviour, so $\lim_{x \to 3} \frac{1}{(x-3)^2} = \infty$ by an identical argument, with the vertical asymptote at $x = 3$ .