When exactly can I use L'Hôpital's Rule?

Only when direct substitution produces the indeterminate quotient 0/0 or ∞/∞, and the functions are differentiable near the point with g'≠ 0. Always check the form first; if you get a determinate value like 3/0 or 4/2, the rule does not apply.

Do I use the quotient rule when applying L'Hôpital's Rule?

No. You differentiate the numerator and the denominator separately and then form a new fraction f'/g'. Using the quotient rule is one of the most common errors and produces a wrong answer.

How do I handle forms like 0·∞ or 1^∞?

Convert them first. For 0·∞, rewrite one factor as a reciprocal to get 0/0 or ∞/∞. For power forms 1^∞, 0^0, ∞^0, take the natural log, evaluate that limit with L'Hôpital, then exponentiate the result with e.

What if applying L'Hôpital's Rule keeps giving an indeterminate form?

You can apply it repeatedly as long as each new limit is still indeterminate. But if it loops or grows more complex, switch strategies — try factoring, dividing by the highest power, or using a known limit like _x→ 0 x/x=1.

AP-CALCBC-4.7

U4.7 L'Hôpital's Rule

Master L'Hôpital's Rule for AP Calc BC: state the hypotheses, apply it to 0/0 and ∞/∞ limits, and convert forms like 0·∞, ∞−∞, and 1^∞.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U4.7 L'Hôpital's Rule, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U4.7 L'Hôpital's Rule free

What this lesson covers

Some limits look impossible: plug in the value and you get

\frac{0}{0}

\frac{\infty}{\infty}

, expressions that could equal almost anything. L'Hôpital's Rule is your power tool for these indeterminate forms, letting you replace a messy ratio of functions with a ratio of their derivatives. On the AP exam, you'll see it applied directly to fractions and, more cleverly, to products and powers that must first be rewritten. This lesson states the rule precisely, shows you exactly when it's legal to use, and walks through the algebraic tricks that convert every indeterminate form into one L'Hôpital's Rule can handle.

Stating the Rule and Its Hypotheses

L'Hôpital's Rule says: suppose

f

and

g

are differentiable near

a

(except possibly at

a

) with

g'(x)\neq 0

near

a

. If

\lim_{x\to a}\frac{f(x)}{g(x)}

produces the indeterminate form

\frac{0}{0}

\frac{\pm\infty}{\pm\infty}

, then

\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}

provided the right-hand limit exists or is

\pm\infty

. The rule also holds for one-sided limits and for

a=\pm\infty

.

The single most important habit: verify the indeterminate form before differentiating. Substitute the target value first. If you get

\frac{0}{0}

\frac{\infty}{\infty}

, proceed. If you get something like

\frac{3}{0}

\frac{5}{2}

, the rule does not apply and using it gives a wrong answer.

A critical misconception: you do NOT use the quotient rule. You differentiate the numerator and denominator separately, then form a new fraction. You may apply the rule repeatedly as long as each new form is still indeterminate. Once substitution yields a determinate value, stop — applying L'Hôpital again would be an error.

Applying It to 0/0 and ∞/∞

The two forms L'Hôpital's Rule handles directly are

\frac{0}{0}

and

\frac{\infty}{\infty}

. Consider

\lim_{x\to 0}\frac{\sin x}{x}

. Substituting gives

\frac{0}{0}

, so differentiate top and bottom:

\lim_{x\to 0}\frac{\cos x}{1}=1

.

For

\frac{\infty}{\infty}

, try

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

. Both grow without bound, so apply the rule:

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

. This illustrates a key fact the AP exam loves: polynomials dominate logarithms, and exponentials dominate polynomials.

Sometimes one pass isn't enough. For

\lim_{x\to 0}\frac{1-\cos x}{x^2}

you get

\frac{0}{0}

, then

\lim_{x\to 0}\frac{\sin x}{2x}

, still

\frac{0}{0}

, then

\lim_{x\to 0}\frac{\cos x}{2}=\frac{1}{2}

Step	Check	Action
Substitute	Get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ?	If yes, continue
Differentiate	$\frac{f'}{g'}$ , not quotient rule	Form new limit
Re-substitute	Determinate now?	If yes, that's the answer
Otherwise	Still indeterminate	Repeat

Converting Other Indeterminate Forms

Many limits arrive disguised. The AP exam expects you to reshape them into

\frac{0}{0}

\frac{\infty}{\infty}

first.

For the product form

0\cdot\infty

, rewrite one factor as a reciprocal. Example:

\lim_{x\to 0^+}x\ln x

0\cdot(-\infty)

. Rewrite as

\lim_{x\to 0^+}\frac{\ln x}{1/x}

, now

\frac{-\infty}{\infty}

; L'Hôpital gives

\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+}(-x)=0

.

For

\infty-\infty

, combine into a single fraction using a common denominator, which typically yields

\frac{0}{0}

.

The power forms

1^{\infty}

0^0

, and

\infty^0

all require logarithms. Set

y=f(x)^{g(x)}

, take

\ln y=g(x)\ln f(x)

, find

\lim \ln y

(usually a

0\cdot\infty

product), then exponentiate: the answer is

e^{L}

Form	Rewrite as
$0\cdot\infty$	$\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$
$\infty-\infty$	common denominator $\to \frac{0}{0}$
$1^\infty,\ 0^0,\ \infty^0$	take $\ln$ , then exponentiate

Remember:

2^\infty

\infty+\infty

, and

\frac{5}{0}

are NOT indeterminate — don't waste time on L'Hôpital for those.

How the Exam Tests It

On the multiple-choice section, L'Hôpital problems are usually quick: identify the indeterminate form and differentiate once or twice. Watch for traps where the limit is not actually indeterminate — the correct answer may be found by direct substitution or by recognizing a determinate form like

\frac{4}{0}=\pm\infty

.

Free-response questions often embed a limit inside a larger context, such as evaluating a limit that defines a derivative, or analyzing end behavior. When justification is required, you should explicitly state that the limit has the form

\frac{0}{0}

(or

\frac{\infty}{\infty}

) so that L'Hôpital's Rule applies. Skipping this justification can cost points even if your arithmetic is correct.

A subtle exam favorite: sometimes L'Hôpital loops forever or grows more complicated, signaling you should switch strategies — factor, use a known limit, or divide by the highest power. For instance,

\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x}

cycles under L'Hôpital but is trivially

1

by dividing by

x

. Also beware of applying the rule when the derivative-limit does not exist; if

\lim\frac{f'}{g'}

fails to exist (and isn't

\pm\infty

), the rule is simply inconclusive, not proof that the original limit fails.

Key terms

Indeterminate form.: An expression such as $\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $0\cdot\infty$ , $\infty-\infty$ , $1^\infty$ , $0^0$ , or $\infty^0$ whose value cannot be determined by substitution alone.
L'Hôpital's Rule.: If $\lim\frac{f}{g}$ is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and conditions hold, then $\lim\frac{f}{g}=\lim\frac{f'}{g'}$ .
Determinate form.: An expression whose limit is forced, such as $\frac{5}{0^+}=+\infty$ or $\infty+\infty=\infty$ ; L'Hôpital's Rule does not apply.
0/0 form.: A quotient in which numerator and denominator both approach zero, a primary target of L'Hôpital's Rule.
∞/∞ form.: A quotient in which numerator and denominator both grow without bound; also directly handled by the rule.
Logarithmic conversion.: The technique of taking $\ln$ of a power-form limit $f^g$ , evaluating $\lim g\ln f$ , then exponentiating to get $e^L$ .
Dominance.: The ranking of growth rates ( $\ln x \ll x^n \ll e^x$ ) that predicts the value of many $\frac{\infty}{\infty}$ limits.

Worked example

Evaluate

\lim_{x\to 0^+}(1+2x)^{1/x}

First substitute to identify the form: as

x\to 0^+

, the base

1+2x\to 1

and the exponent

\frac{1}{x}\to+\infty

, giving the indeterminate form

1^\infty

. This is a power form, so use logarithms.

Let

y=(1+2x)^{1/x}

. Take the natural log of both sides:

\ln y=\frac{1}{x}\ln(1+2x)=\frac{\ln(1+2x)}{x}.

Now evaluate

\lim_{x\to 0^+}\ln y=\lim_{x\to 0^+}\frac{\ln(1+2x)}{x}

. Substituting gives

\frac{\ln 1}{0}=\frac{0}{0}

, so L'Hôpital's Rule applies.

Differentiate numerator and denominator separately. The derivative of

\ln(1+2x)

\frac{2}{1+2x}

, and the derivative of

x

1

\lim_{x\to 0^+}\frac{2/(1+2x)}{1}=\frac{2}{1+0}=2.

\lim_{x\to 0^+}\ln y=2

. But we want

y

, not

\ln y

. Exponentiate to undo the logarithm:

\lim_{x\to 0^+}y=e^{2}.

The answer is

e^2

. Notice the crucial final step: forgetting to exponentiate would leave the answer as

2

, a common and costly mistake.

Practice questions

What is

\lim_{x\to\infty}\frac{3x^2+5}{e^{x}}

$0$
$3$
$\infty$
$\frac{3}{e}$

Answer: $0$

Substituting gives

\frac{\infty}{\infty}

. Apply L'Hôpital:

\frac{6x}{e^x}

, still

\frac{\infty}{\infty}

. Apply again:

\frac{6}{e^x}\to 0

. The exponential dominates the polynomial, so the limit is

0

. This reflects the growth-rate hierarchy

x^n\ll e^x

Evaluate

\lim_{x\to 0^+}\left(\frac{1}{x}-\frac{1}{\sin x}\right)

, showing the steps needed to justify L'Hôpital's Rule.

Answer: $0$

The form is

\infty-\infty

, which is indeterminate, so combine into one fraction:

\frac{\sin x - x}{x\sin x}

. Substituting gives

\frac{0}{0}

, so L'Hôpital applies. Differentiate:

\frac{\cos x-1}{\sin x + x\cos x}

, still

\frac{0}{0}

. Apply again:

\frac{-\sin x}{\cos x + \cos x - x\sin x}=\frac{-\sin x}{2\cos x - x\sin x}

. Substituting

x=0

gives

\frac{0}{2}=0

. So the limit is

0

For which limit does L'Hôpital's Rule NOT apply directly?

$\lim_{x\to 0}\frac{\sin x}{x}$
$\lim_{x\to\infty}\frac{\ln x}{x}$
$\lim_{x\to 0}\frac{\cos x}{x}$
$\lim_{x\to 1}\frac{x^2-1}{x-1}$

Answer: $\lim_{x\to 0}\frac{\cos x}{x}$

Substitution gives

\frac{\cos 0}{0}=\frac{1}{0}

, which is a determinate form (the limit is

\pm\infty

), not

\frac{0}{0}

\frac{\infty}{\infty}

. L'Hôpital's Rule requires an indeterminate quotient. The other three all yield

\frac{0}{0}

\frac{\infty}{\infty}

upon substitution.

FAQ

When exactly can I use L'Hôpital's Rule?: Only when direct substitution produces the indeterminate quotient $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and the functions are differentiable near the point with $g'\neq 0$ . Always check the form first; if you get a determinate value like $\frac{3}{0}$ or $\frac{4}{2}$ , the rule does not apply.
Do I use the quotient rule when applying L'Hôpital's Rule?: No. You differentiate the numerator and the denominator separately and then form a new fraction $\frac{f'}{g'}$ . Using the quotient rule is one of the most common errors and produces a wrong answer.
How do I handle forms like 0·∞ or 1^∞?: Convert them first. For $0\cdot\infty$ , rewrite one factor as a reciprocal to get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . For power forms $1^\infty$ , $0^0$ , $\infty^0$ , take the natural log, evaluate that limit with L'Hôpital, then exponentiate the result with $e$ .
What if applying L'Hôpital's Rule keeps giving an indeterminate form?: You can apply it repeatedly as long as each new limit is still indeterminate. But if it loops or grows more complex, switch strategies — try factoring, dividing by the highest power, or using a known limit like $\lim_{x\to 0}\frac{\sin x}{x}=1$ .

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