AP Calculus AB and BC

Limit Solver (Teaching Mode): Which Method and Why

Q: How do I know whether to factor or use the conjugate?

Look for a square root. If the \frac{0}{0} expression is built from polynomials, factor and cancel. If a radical is trapped inside, factoring cannot remove it, so multiply by the conjugate to create a difference of squares, then cancel (Topic 1.6).

Q: Can I just use L'Hopital's Rule on every 0/0 limit?

It works on \frac{0}{0} and \frac{\infty}{\infty}, but it uses derivatives and appears later at Topic 4.7, so early limit questions expect the algebraic methods instead. Always verify the form first: if substitution gives a non-indeterminate form like \frac{5}{0}, the rule does not apply and using it gives a wrong answer.

Which method a limit needs is decided by the form direct substitution returns. A real number is the answer; a nonzero number over 0 is an infinite limit or does not exist; 0/0 is indeterminate, so climb the ladder: factor polynomials, use the conjugate for radicals, match sin x / x trig shapes, and use L'Hopital last.

Pick a limit

\lim_{x \to 2} \left(x^2 + 1\right)

Step 1: try direct substitution. What form do you get?

This tool does more than hand you the value of a limit. For each curated limit it shows which method the limit needs and why, which is the exact skill Topic 1.7 tests: selecting a procedure. Every limit starts the same way, with direct substitution, and the form substitution returns is what picks the method.

Think of the methods as rungs on a ladder. You always step on the bottom rung first (substitute), and you only climb higher when the form you get back is indeterminate. The table below is the whole decision: read what substitution returns on the left, and the method it selects is on the right.

After direct substitution you get	What it means	Method the tool selects
A real number	The function is continuous there (Topic 1.12)	Stop. That number is the limit.
$\frac{0}{0}$ with polynomials	A common factor is hiding	Factor and cancel (Topic 1.6)
$\frac{0}{0}$ with a radical	A root is blocking the cancellation	Multiply by the conjugate (Topic 1.6)
$\frac{0}{0}$ from a $\frac{\sin x}{x}$ shape	A special trig limit	Rewrite to a known trig limit (Topic 1.8)
$\frac{0}{0}$ or $\frac{\infty}{\infty}$ that resists algebra	Still indeterminate	L'Hopital's Rule (Topic 4.7)
A nonzero number over $0$ , like $\frac{5}{0}$	Vertical asymptote	Infinite limit or does not exist (Topic 1.14)

$\frac{0}{0}$ is an indeterminate form, which is a signal to keep working, not an answer. It never means the limit equals $0$ , and it does not by itself mean the limit does not exist. It only tells you a common factor or a special trig shape is hiding.

The sharpest selection is factor versus conjugate, and the expression tells you which. If both the top and bottom are polynomials, factor each one, cancel the shared $(x - c)$ , and substitute into what remains. If a square root is trapped inside the $\frac{0}{0}$ , factoring cannot clear it, so multiply by the conjugate of the radical to turn a difference like $\sqrt{x+4} - 2$ into a difference of squares. Quick rule: polynomials mean factor, a stubborn root means conjugate.

Some $\frac{0}{0}$ limits will not factor and have no radical, because the obstacle is trigonometric. When you see a sine divided by its own argument, or something you can reshape into that, match it to one of the two limits the squeeze theorem gives you (Topic 1.8):

\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0

L'Hopital's Rule is the top rung, and you only reach it when algebra stalls. It applies to exactly two forms, $\frac{0}{0}$ and $\frac{\infty}{\infty}$ , and it uses derivatives, so it lands later in the course at Topic 4.7 (both AB and BC). Verify the indeterminate form first: if substitution already gives a form that is not indeterminate, such as $\frac{5}{0}$ , the rule does not apply. Other forms such as $\infty - \infty$ are not assessed on the AP exam.

Worked example

Choosing the method for a 0/0 limit

Evaluate $\lim_{x \to 7} \dfrac{\sqrt{x+2} - 3}{x - 7}$ .

Start on the bottom rung: direct substitution. Substituting $x = 7$ gives $\dfrac{\sqrt{9} - 3}{7 - 7} = \dfrac{3 - 3}{0} = \dfrac{0}{0}$ . That is indeterminate, so substitution failed and you keep climbing (Topic 1.12).
Select the rung by reading the form. You have $\dfrac{0}{0}$ with a square root sitting in the numerator. A radical blocks ordinary factoring, so the ladder sends you to the conjugate, not to factoring (Topic 1.6).
Multiply the numerator and denominator by the conjugate of the radical, $\sqrt{x+2} + 3$ : $\dfrac{\sqrt{x+2} - 3}{x - 7} \cdot \dfrac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3}$ .
The numerator is a difference of squares: $(\sqrt{x+2})^2 - 3^2 = (x + 2) - 9 = x - 7$ . The expression becomes $\dfrac{x - 7}{(x - 7)(\sqrt{x+2} + 3)}$ .
Cancel the common factor $x - 7$ . This is legal because a limit only looks at $x$ near $7$ , never at $7$ itself, so $x - 7 \ne 0$ for every point that matters. You are left with $\dfrac{1}{\sqrt{x+2} + 3}$ .
Substitution works now, so drop back to the bottom rung on the simplified form: $\dfrac{1}{\sqrt{9} + 3} = \dfrac{1}{3 + 3} = \dfrac{1}{6}$ .

$\dfrac{1}{6}$

Frequently asked questions

Do I always have to try direct substitution first?

Yes. Substitution is the bottom rung and often the entire problem: if the function is continuous at the point, the value you get is the limit (Topic 1.12). You only climb to factoring, conjugates, trig limits, or L'Hopital when substitution returns an indeterminate $\frac{0}{0}$ .

How do I know whether to factor or use the conjugate?

Look for a square root. If the $\frac{0}{0}$ expression is built from polynomials, factor and cancel. If a radical is trapped inside, factoring cannot remove it, so multiply by the conjugate to create a difference of squares, then cancel (Topic 1.6).

Can I just use L'Hopital's Rule on every 0/0 limit?

It works on $\frac{0}{0}$ and $\frac{\infty}{\infty}$ , but it uses derivatives and appears later at Topic 4.7, so early limit questions expect the algebraic methods instead. Always verify the form first: if substitution gives a non-indeterminate form like $\frac{5}{0}$ , the rule does not apply and using it gives a wrong answer.

What if substitution gives a number over zero, like 5/0?

That is not indeterminate. A nonzero number over zero signals a vertical asymptote, so the limit is $+\infty$ , $-\infty$ , or does not exist depending on the sign from each side (Topic 1.14). Only $\frac{0}{0}$ calls for the algebra rungs.