Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

AP Calculus · Contextual Applications of Differentiation

Using L'Hôpital's Rule for Indeterminate Forms

Two forms qualify directly. Everything else must be reshaped first — and some things never qualify at all.

key idea$\tfrac00\text{ or }\tfrac{\infty}{\infty}\ \Rightarrow\ \lim\frac{f}{g}=\lim\frac{f'}{g'}$

How to choose

L'Hôpital's rule applies to exactly two forms. Everything else must be algebraically reshaped into one of them first — or the rule does not apply at all. Checking the form is the entire skill.

Substituting gives	Indeterminate?	L'Hôpital?	What to do
$\tfrac00$	yes	directly	differentiate top and bottom separately
$\tfrac{\infty}{\infty}$	yes	directly	same
$0\cdot\infty$	yes	after rewriting	move one factor into a denominator
$\infty-\infty$	yes	after rewriting	combine into a single fraction
$1^{\infty}$, $0^{0}$, $\infty^{0}$	yes	after rewriting	take $\ln$, find the limit, then exponentiate
$\tfrac{k}{0}$, $k\neq0$	no	never	one-sided signs → $\pm\infty$ or DNE
a real number	no	never	that number is the limit

The last two rows are where the rule gets misapplied. Applying it to a determinate form produces a confidently wrong answer.

The decision tree

Substitute first. The form you get decides everything.

CHECK

Substitute the value.If you get a real number, that is the limit — stop. If you get $\tfrac{k}{0}$ with $k\neq0$, the rule does not apply; analyse the one-sided signs instead.

FORM A

You get $\tfrac00$ or $\tfrac{\infty}{\infty}$.Apply the rule directly: differentiate numerator and denominator separately — this is not the quotient rule — then substitute again.

FORM B

You get $0\cdot\infty$.Rewrite as a quotient by moving one factor down: $f\cdot g=\dfrac{f}{1/g}$. Choose whichever placement gives a derivative you can handle.

FORM C

You get $\infty-\infty$.Combine over a common denominator, or rationalise, until a single fraction appears. Then check the form again.

FORM D

You get $1^{\infty}$, $0^{0}$ or $\infty^{0}$.Set $y$ equal to the expression, take $\ln$ of both sides, find $\lim\ln y$ with the rule, then exponentiate: the answer is $e^{\lim\ln y}$.

REPEAT

Still indeterminate after one application?Apply the rule again. Each pass must be preceded by re-checking the form — and if it stops being indeterminate, stop.

Worked examples

EX 1

An iterated $\tfrac00$

apply twice

Evaluate $\lim\limits_{x\to0}\dfrac{e^{x}-1-x}{x^{2}}$.

Check the form

$\dfrac{1-1-0}{0}=\dfrac00$ — indeterminate, so the rule applies.

Differentiate top and bottom

$\dfrac{e^{x}-1}{2x}$.

Check again

$\dfrac{1-1}{0}=\dfrac00$ — still indeterminate, so apply the rule a second time.

Differentiate again

$\dfrac{e^{x}}{2}$.

Substitute

$\dfrac{e^{0}}{2}=\dfrac12$. Each pass required re-checking the form first.

EX 2

An $\tfrac{\infty}{\infty}$ form

growth race

Evaluate $\lim\limits_{x\to\infty}\dfrac{\ln x}{x}$.

Check the form

Both $\ln x$ and $x$ grow without bound: $\dfrac{\infty}{\infty}$.

Apply the rule

$\dfrac{1/x}{1}=\dfrac1x$.

Take the limit

$\lim_{x\to\infty}\dfrac1x=0$.

Interpret

$x$ grows much faster than $\ln x$, so the quotient collapses to $0$.

Generalise

The same argument shows $\ln x$ loses to every positive power of $x$.

EX 3

A disguised product

$0\cdot\infty$

Evaluate $\lim\limits_{x\to0^{+}} x\ln x$.

Check the form

$x\to0$ while $\ln x\to-\infty$: the form is $0\cdot(-\infty)$ — indeterminate, but not a quotient.

Rewrite as a quotient

$x\ln x=\dfrac{\ln x}{1/x}$, which now reads $\dfrac{-\infty}{\infty}$.

Apply the rule

$\dfrac{1/x}{-1/x^{2}}$.

Simplify

$=\dfrac{1}{x}\cdot\left(-x^{2}\right)=-x$.

Take the limit

$\lim_{x\to0^{+}}(-x)=0$. Choosing to move $\tfrac1x$ down rather than $\ln x$ is what made the algebra collapse.

EX 4

When the rule must not be used

$\tfrac{k}{0}$

A student applies L'Hôpital to $\lim\limits_{x\to0}\dfrac{\cos x}{x}$ and reports $\lim\dfrac{-\sin x}{1}=0$. Diagnose.

Check the original form

$\dfrac{\cos 0}{0}=\dfrac{1}{0}$ — not indeterminate.

State the consequence

The rule's hypothesis fails, so it may not be applied. The answer $0$ is unjustified.

Find the true behaviour

As $x\to0^{+}$ the quotient $\to+\infty$; as $x\to0^{-}$ it $\to-\infty$.

Give the correct answer

The limit does not exist — nowhere near $0$.

The lesson

L'Hôpital on a determinate form does not merely fail to help; it produces a plausible wrong answer. Always check the form first.

EX 5

An exponential form

$1^{\infty}$

Evaluate $\lim\limits_{x\to\infty}\left(1+\dfrac{3}{x}\right)^{x}$.

Check the form

The base $\to1$ and the exponent $\to\infty$: the form is $1^{\infty}$, which is indeterminate.

Take logarithms

Let $y=\left(1+\tfrac3x\right)^{x}$, so $\ln y=x\ln\left(1+\tfrac3x\right)$.

Rewrite as a quotient

$\ln y=\dfrac{\ln\left(1+3/x\right)}{1/x}$, now $\dfrac00$.

Apply the rule

Differentiating gives $\dfrac{\dfrac{-3/x^{2}}{1+3/x}}{-1/x^{2}}=\dfrac{3}{1+3/x}\to3$.

Exponentiate

$\lim\ln y=3$, so the limit is $e^{3}$. Never forget the last step — the answer is not $3$.

Watch out

Check the form before every application

Applying the rule to $\tfrac{k}{0}$ or to a determinate value gives a wrong answer that looks right — Example 4.

Differentiate top and bottom separately

$\lim\dfrac{f'}{g'}$, not the derivative of the quotient. Using the quotient rule here is a different calculation entirely.

Re-check between repeated applications

Each pass needs its own form check. Continuing past the point where the form becomes determinate introduces errors.

Exponentiate at the end

For $1^{\infty}$, $0^{0}$ and $\infty^{0}$ you found $\lim\ln y$. The limit itself is $e$ to that power.

Course map. Topic 4.7 · LO LIM-4.A — determine limits of functions that result in indeterminate forms (EK LIM-4.A.1, LIM-4.A.2).