AP Calculus AB · Lesson Review · October 9, 2026

L’Hôpital’s Rule and Indeterminate Forms

Check a quotient’s limiting form, use separate derivatives for valid indeterminate limits, recheck before repeating the rule, and read a needed derivative from a graph.

12-minute edited lesson6 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Indeterminate forms and factoring 1:3802L’Hôpital’s Rule 2:5103Differentiate top and bottom separately 4:5004Repeat only after checking the form 6:4605The familiar sine quotient 7:3506Read a derivative from a graph

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02 · Understand

Learning targets

Recognize 0/0 and signed infinity-over-infinity forms before using the rule.
Differentiate the numerator and denominator separately within a limit.
Recheck the form before each repeated application.
Use a graph’s local slope and explain the conditions needed for the rule.

03 · Connect

Key ideas

The recording compares factoring with derivative ratios, evaluates finite and infinite limits, and reads a derivative from a piecewise-linear graph.

01Check the indeterminate form first: $\frac{0}{0}\quad\text{or}\quad\frac{\pm\infty}{\pm\infty}$ These symbols describe the separate numerator and denominator limits. They do not define a value for the quotient. A nonzero-over-zero form is a different situation.
02This is a limit rule, not the quotient rule: $\lim\frac{f}{g}\overset{\mathrm{LH}}{=}\lim\frac{f'}{g'}$ Differentiate the top and bottom separately only after verifying the hypotheses. The derivative of f/g instead uses (f′g−fg′)/g²; the original quotient and derivative ratio are not generally equal as functions.
03Recheck before repeating: $\frac{x^3-3}{e^x}\longrightarrow\frac{3x^2}{e^x}\longrightarrow\frac{6x}{e^x}\longrightarrow\frac{6}{e^x}$ As x tends to positive infinity, the first three ratios still have infinity-over-infinity form. The last ratio tends directly to zero. Stop when ordinary limit evaluation works.
04The hypotheses matter: $g'(x)\ne0\quad\text{near the limiting point}$ Use differentiable numerator and denominator functions on an appropriate punctured interval (or tail for infinity), with g′ nonzero there. The derivative-ratio limit must exist, either finitely or as signed infinity. One-sided versions use the corresponding side. The target point itself may be excluded.
05Local slopes explain a useful special case: $f(a)=g(a)=0,\ g'(a)\ne0\ \Longrightarrow\ \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(a)}$ If both functions are differentiable at a and vanish there, their small changes resemble their tangent slopes times the same small input change. This explains the special case with nonzero g′(a). It is an added explanation; the full general theorem, including its infinite cases, needs further justification.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example 1

Compare factoring with one application

\lim_{x\to1}\frac{x^2-1}{x^2+x-2}

Direct substitution produces 0/0.

\lim_{x\to1}\frac{(x-1)(x+1)}{(x-1)(x+2)}=\frac{2}{3}

Cancel for x≠1, then evaluate the simpler expression.

\lim_{x\to1}\frac{2x}{2x+1}=\frac{2}{3}

L’Hôpital’s Rule gives the same result; the denominator derivative is nonzero nearby.

Classroom example 2

A polynomial divided by an exponential

\lim_{x\to+\infty}\frac{x^3-3}{e^x}

Both numerator and denominator tend to positive infinity.

\lim_{x\to+\infty}\frac{3x^2}{e^x}=\lim_{x\to+\infty}\frac{6x}{e^x}=\lim_{x\to+\infty}\frac{6}{e^x}=0

Check infinity-over-infinity before each of the first three applications. The final expression can be evaluated directly.

Classroom example 3

Evaluate the familiar sine quotient

\lim_{x\to0}\frac{\sin x}{x}

Work in radians. The form is 0/0.

\lim_{x\to0}\frac{\cos x}{1}=1

This application assumes the sine derivative has already been established independently. It is not an independent foundational proof of that derivative or of the original sine limit.

Classroom example 4

Use the slope of the graph at x = 3

\lim_{x\to3}\frac{f(x)}{x^2-9}

The graph passes through (3,0), so substitution gives 0/0. The graph is linear near x=3 with slope −2.

\lim_{x\to3}\frac{f'(x)}{2x}=\frac{-2}{2(3)}=-\frac{1}{3}

Use the slope of the local line for f′(3). The denominator derivative is nonzero near 3.

f(x)=-2x+6\quad\text{near }x=3

The teacher also suggests an algebraic check: slope −2 and the point (3,0) determine this local piece. No formula for the other pieces is needed.

\lim_{x\to3}\frac{-2(x-3)}{(x-3)(x+3)}=\lim_{x\to3}\frac{-2}{x+3}=-\frac{1}{3}

Cancel only for x≠3. This completes the suggested local algebraic check.

05 · Avoid

Common mistakes

Using the rule for every fraction.

Check for 0/0 or signed infinity-over-infinity. A fraction alone does not authorize the rule.

Calling 0/0 zero or one.

The form is indeterminate: different quotients with that form can have very different limits.

Taking the derivative of the entire quotient.

L’Hôpital’s Rule uses the ratio of separate derivatives. The quotient rule answers a different question.

Differentiating again without checking.

After each application, evaluate the new numerator and denominator limits. Repeat only if the hypotheses still hold.

Ignoring the rule’s hypotheses.

The functions must be differentiable on the relevant punctured interval, with nonzero denominator derivative there and a well-defined derivative-ratio limit. Check each repeated application separately.

06 · Check yourself

Try it first

Use these five fresh problems to check your understanding. Open a solution after you try it.

1Check the form and evaluate:

\lim_{x\to2}\frac{x^2-4}{x-2}

$\frac{0}{0}\ \longrightarrow\ \lim_{x\to2}\frac{2x}{1}=4$

The rule applies. Factoring (x−2)(x+2) gives the same limit.

2Evaluate using a derivative ratio:

\lim_{x\to0}\frac{e^{2x}-1}{x}

$\frac{0}{0}\ \longrightarrow\ \lim_{x\to0}\frac{2e^{2x}}{1}=2$

Use the chain rule when differentiating the numerator.

3Recheck the form and evaluate:

\lim_{x\to0}\frac{1-\cos x}{x^2}

$\lim_{x\to0}\frac{\sin x}{2x}$

The first application leaves another 0/0 form.

$\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2}$

A second valid application works. Trigonometric arguments are in radians.

4Evaluate the limit at positive infinity:

\lim_{x\to+\infty}\frac{\ln x}{x}

$\frac{\infty}{\infty}\ \longrightarrow\ \lim_{x\to+\infty}\frac{1/x}{1}=0$

The logarithm is defined for x>0. Only one application is needed.

5Is L’Hôpital’s Rule applicable? Explain and evaluate:

\lim_{x\to0^+}\frac{x+1}{x}

$\frac{1}{0^+}\quad\text{is not an indeterminate form}$

Do not replace this quotient with its derivative ratio.

$\lim_{x\to0^+}\frac{x+1}{x}=\lim_{x\to0^+}\left(1+\frac{1}{x}\right)=+\infty$

The quotient grows without bound. The incorrect derivative-ratio shortcut would have given 1.

07 · Revisit

Transcript and captions

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