AP Calculus AB and BC

Direct Substitution vs Indeterminate Form

Q: Is 0/0 undefined or indeterminate?

Both words apply, to different things. As arithmetic, \frac{0}{0} is undefined, since no number satisfies it. As a limit form it is called indeterminate, meaning the limit may well exist and its value depends on the particular functions rather than on the form. Undefined describes the arithmetic; indeterminate describes how much the form tells you, which is nothing.

Q: Is 3/0 an indeterminate form?

No, and the difference matters. \frac{3}{0} is undefined but decisive: the values grow without bound, so the graph has a vertical asymptote there and the limit is +\infty, -\infty, or nonexistent. Test the sign of the quotient on each side, combining the sign of the (nonzero) numerator with the sign of the denominator, to decide which of the three to report.

Substitution is always the first move on a limit at a finite point, and what it returns tells you what to do next. A real number is the answer whenever f is continuous at that point. A nonzero number over zero is undefined and signals a vertical asymptote, so the limit is infinite or does not exist.

Direct substitution

Use when: Always, as the opening step on any limit at a finite point, because a real value at the point settles the answer outright for every function continuous there.

Indeterminate form

Use when: Substitution has returned zero over zero or infinity over infinity, which is an instruction to keep working by factoring, rationalising, using a known special limit, or applying l'Hopital's rule.

Side by side

	Direct substitution	Indeterminate form
What it is	A method: evaluate $f(a)$ and see what comes out	A result: an expression such as $\frac{0}{0}$ that no single value fits
Settles the limit	Whenever $f$ is continuous at $a$ and the output is a real number	Never on its own, a second step is always required
Output $\frac{3}{0}$	Undefined, and decisive: there is a vertical asymptote	Not indeterminate, the one-sided values run to $\pm\infty$
Output $\frac{0}{0}$	Substitution has told you nothing yet	Indeterminate, the answer may be any number, infinite, or nonexistent
Next move	If $f$ is continuous at $a$ , none, write the value down	Factor and cancel, rationalise, use $\lim_{x \to 0}\frac{\sin x}{x} = 1$ , or use l'Hopital

Substitution earns its place at the front because polynomials, rational functions, roots, exponentials, logarithms, and trigonometric functions are all continuous wherever they are defined. For those, $\lim_{x \to a} f(x) = f(a)$ by definition of continuity, so plugging in is not a shortcut but the theorem itself. The interesting part is reading the output.

Three common outputs, three conclusions. A real number is the limit provided $f$ is continuous at $a$ ; for a piecewise rule, or any point where the formula changes, check the one-sided limits instead of writing the substituted value down, since a function such as $f(x) = x^2$ for $x \ne 2$ with $f(2) = 7$ substitutes to $7$ while its limit is $4$ . A nonzero number over zero is undefined, which is informative: the magnitude grows without bound, so there is a vertical asymptote and you check each side for the sign. Zero over zero is the one case that reports nothing, because the numerator and denominator are both collapsing and which one collapses faster is exactly what you have not yet determined. Infinity over infinity is the other indeterminate output, arising mainly in limits as $x \to \infty$ , and it is handled the same way as zero over zero.

\lim_{x \to 2}\frac{x^2-4}{x-2} \to \frac{0}{0} \to \lim_{x \to 2}(x+2) = 4 \qquad \lim_{x \to 2}\frac{x^2+4}{x-2} \to \frac{8}{0} \to \text{vertical asymptote}

The mistake: l'Hopital on a form that is not indeterminate

Take $\lim_{x \to 0}\frac{\cos x}{x}$ . Substitution gives $\frac{1}{0}$ , and differentiating top and bottom anyway produces $\lim_{x \to 0}\frac{-\sin x}{1} = 0$ , which is wrong: the true values run to $+\infty$ from the right and $-\infty$ from the left, so the limit does not exist. The rule applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$ only, so name the form before you use it.

Frequently asked questions

Is 0/0 undefined or indeterminate?

Both words apply, to different things. As arithmetic, $\frac{0}{0}$ is undefined, since no number satisfies it. As a limit form it is called indeterminate, meaning the limit may well exist and its value depends on the particular functions rather than on the form. Undefined describes the arithmetic; indeterminate describes how much the form tells you, which is nothing.

What do I do when direct substitution gives 0/0?

Remove the common factor causing the zeros. Factor and cancel for polynomials, multiply by the conjugate for roots, combine the fractions in a compound fraction, recognise a special trigonometric limit, or apply l'Hopital's rule. Then substitute again into the simplified expression.

Is 3/0 an indeterminate form?

No, and the difference matters. $\frac{3}{0}$ is undefined but decisive: the values grow without bound, so the graph has a vertical asymptote there and the limit is $+\infty$ , $-\infty$ , or nonexistent. Test the sign of the quotient on each side, combining the sign of the (nonzero) numerator with the sign of the denominator, to decide which of the three to report.

In the CED: Unit 1: Limits and Continuity