The rule for indeterminate quotients of the forms zero over zero and infinity over infinity, repeated application, converting products, differences and exponential forms into quotients, the proof of the growth ranking, and the forms that only appear indeterminate.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
L'Hôpital's Rule
Objectives
Five outcomes. The rule is one line, and four of these are about when it applies and how to reach it.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 394-408 — the section these objectives are drawn from
Warm-up
Section 2.3 resolved indeterminate quotients by factoring, rationalising or combining fractions, and Section 4.6 asserted a growth ranking without proving it.
Discussion prompt
The quotient of x to the hundredth by e to the x is infinity over infinity. No algebra factors it. What tool is missing?
Hint: The two parts have very different derivatives.
Answer:
Nothing in Section 2.3 touches this. There is no common factor to cancel, no conjugate to introduce, and no fractions to combine — the algebraic techniques simply do not reach it.
\[ \frac{x^{100}}{e^{x}} \;\to\; \frac{\infty}{\infty} \quad \text{as } x \to \infty \]
What the two parts do have is very different derivatives: differentiating the power lowers its degree while the exponential is unchanged. This section's rule exploits exactly that, and after a hundred differentiations the power is a constant and the answer is obvious.
Concept
When a quotient gives zero over zero or infinity over infinity, the limit of the quotient equals the limit of the quotient of the derivatives — differentiated separately, not by the quotient rule.
L'Hopital's rule — If the quotient of f by g is indeterminate of the form zero over zero or infinity over infinity at a point, and the quotient of their derivatives has a limit there, then the original quotient has the same limit.
\[ \lim\frac{f(x)}{g(x)} = \lim\frac{f'(x)}{g'(x)} \quad \text{when the form is } \tfrac{0}{0} \text{ or } \tfrac{\infty}{\infty} \]
The rule looks like an error the first several times. It is not the quotient rule, which computes the derivative OF a quotient; this computes the LIMIT of a quotient, which is a different question with a different answer.
Figure (svg): The rule stated, with the crucial warning that it differentiates separately
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 394-398
Section
Section 1
Concept
The rule is valid only when direct substitution produces zero over zero or infinity over infinity. Applied to a determinate quotient it produces an answer, and the answer is wrong.
indeterminate form — A combination whose value is not determined by the parts' limits alone, because two effects compete. Zero over zero and infinity over infinity are the two the rule handles directly.
\[ \text{check the form} \;\Longrightarrow\; \text{then differentiate} \]
Checking costs one substitution and prevents the section's characteristic error. The rule gives a confident answer whether or not its hypothesis holds, so nothing about the output signals a misapplication.
Figure (svg): Checking the form before applying the rule, with a case where it does not apply
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 394-399 — the rule and its hypotheses
Picture it
Two quotients, one indeterminate and one not.
Figure (svg): Checking the form before applying the rule, with a case where it does not apply
The left-hand limit is correctly 1; the right-hand one is one half and the rule produces 1. Nothing in the computation flags the error, which is why the check must be made deliberately.
Worked example
Example 4.38. Check the form, then differentiate.
\[ \text{Evaluate } \lim_{x \to 0}\frac{\sin x}{x}. \]
Substitute to check the form
Why: Both parts vanish.
\[ \frac{0}{0},\text{ indeterminate} \]
Differentiate the numerator
Why: Separately.
Differentiate the denominator
Why: Separately.
\[ 1 \]
Form the new quotient and evaluate
Why: Substitute now.
\[ \cos(0) / 1 = 1 \]
State
Why: The limit.
\[ 1 \]
Figure (svg): The solution to Worked example a first application shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0}\frac{\sin x}{x} = 1 \]
Verify: notice the circularity to avoid
Why: Section 2.3 proved this limit by squeezing on the unit circle, and Section 3.5 used it to derive the derivative of sine. So using the derivative of sine to prove the limit would be circular — the result is correct but the argument here is not a proof. That is worth knowing: L'Hopital's rule is a computational tool, and it presupposes the derivatives it uses. For this particular limit the squeeze argument remains the honest one.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 396-397
Sorting
Substitute first and check the form.
Sort into buckets
Sort each limit at the point indicated.
The fourth is worth noticing: it is 0 over 0 so the rule applies, but Section 2.3's factoring settles it faster. The rule being available does not make it the best tool, and for polynomial quotients algebra is usually quicker.
Worked example
Checkpoint 4.38. A determinate form gives a wrong answer.
\[ \text{What happens if the rule is applied to } \lim_{x \to 0}\frac{x+1}{x+2}? \]
Substitute to find the true value
Why: Both parts are non-zero.
\[ \frac{1}{2} \]
Note the form is determinate
Why: Not 0/0 or infinity over infinity.
Apply it anyway
Why: Differentiate both parts.
\[ \frac{1}{1} \]
Compare
Why: One against one half.
Draw the lesson
Why: Nothing flagged the error.
Figure (svg): Checking the form before applying the rule, with a case where it does not apply
\[ \text{true } \tfrac12; \quad \text{rule misapplied gives } 1 \]
Verify: consider why this is dangerous rather than merely wrong
Why: The misapplied computation is short, clean and produces a plausible number. Nothing about it looks suspect, so the error survives review — unlike an algebraic slip, which usually produces something visibly odd. That is what makes the form check non-negotiable: it is the only step that would have caught this, and it costs one substitution.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 397-398
Trap
\[ \lim_{x \to 0}\frac{\sin x}{x} \]
Differentiate the quotient as a whole
Why: The student applies the quotient rule.
\[ \frac{x\cos x - \sin x}{x^{2}} \;\to\; \tfrac{0}{0} \quad \text{(no progress)} \]
The result is another indeterminate form, and a messier one. The quotient rule answers a different question entirely.
\[ \frac{f'}{g'} = \frac{\cos x}{1} \;\to\; 1 \]
Differentiate the numerator and the denominator SEPARATELY
Why: The rule is about the limit of a quotient, not the derivative of one.
The two theorems are easy to confuse because both involve a quotient and derivatives. The quotient rule computes the derivative OF f over g; L'Hopital's rule computes the LIMIT of f over g using the separate derivatives. They answer different questions and produce different expressions.
Fill the middle
The sine quotient, with the rule applied.
Fill in the blanks
\frac1___ \;\longrightarrow\; \frac______}
Why: The denominator x differentiates to 1. Note this is not the quotient rule: there is no bottom-times-top-prime and no squared denominator, just the two derivatives placed over one another.
Two truths and a lie
All three are about the rule's statement.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The quotient rule computes the derivative of a quotient; L'Hopital's rule computes the limit of one. Applying the quotient rule to the sine quotient gives a messier indeterminate form rather than progress, which shows they are doing different jobs.
Prediction
Commit before reasoning.
Predict first
What happens if the rule is applied to a determinate quotient?
Correct: It produces a plausible answer that is wrong.
\[ \text{true } \tfrac{0+1}{0+2} = \tfrac12 \ne 1 \]
Why: Differentiating the top and bottom of the quotient of x plus 1 by x plus 2 gives 1 over 1, while the true limit is one half. The computation is short and the answer looks reasonable, so nothing signals the error. Unlike an algebraic slip, which usually produces something visibly strange, a misapplied rule fails silently — which is why the one-substitution form check is worth making every time.
Section
Section 2
Concept
The rule may leave another indeterminate form, in which case it can be applied again — but the form must be rechecked each time. When the result becomes determinate, stop.
iterating the rule — Each application requires the current quotient to be indeterminate. The process terminates when substitution gives a determinate value, and continuing past that point produces a wrong answer.
\[ \frac{0}{0} \to \frac{0}{0} \to \cdots \to \text{a value} \]
For a power over an exponential the process is guaranteed to terminate: each application lowers the power's degree by one while the exponential is unchanged, so after n steps the numerator is a constant.
Figure (svg): The growth ranking proved by repeated application of the rule
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 398-402 — repeated application
Picture it
Three applications on a cube over an exponential.
Figure (svg): The growth ranking proved by repeated application of the rule
The numerator's degree falls at every step and the denominator never changes, so the process must terminate. That guarantee is what makes the growth ranking a theorem rather than an observation.
Worked example
Example 4.40. Recheck the form at each step.
\[ \text{Evaluate } \lim_{x \to \infty}\frac{x^{3}}{e^{x}}. \]
Check the form
Why: Both parts grow without bound.
Apply the rule
Why: Differentiate separately.
\[ 3 x ^{2} / e ^{x},\text{ still } \infty\text{ over } \infty \]
Apply again
Why: Recheck first.
\[ 6 x / e ^{x},\text{ still indeterminate} \]
Apply a third time
Why: Recheck again.
\[ 6 / e ^{x} \]
Now the form is determinate
Why: A constant over something unbounded.
\[ \text{the } \lim\text{ is } 0 \]
Figure (svg): The growth ranking proved by repeated application of the rule
\[ \lim_{x \to \infty}\frac{x^{3}}{e^{x}} = 0 \]
Verify: confirm the process must terminate
Why: Each application differentiates the numerator, lowering its degree by exactly one, while the exponential's derivative is itself and never changes. So after three steps the numerator is the constant 6 and the form is no longer indeterminate. For the hundredth power it would take a hundred steps, which is impractical to write out but guarantees the same conclusion — and that guarantee is what proves the growth ranking of Section 4.6.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 399-400
Fill the middle
The cosine quotient, after one application.
Fill in the blanks
\frac0/0___ \text___ 0 \text___ ___, \text___
Why: Both parts still vanish, so the form remains indeterminate and a second application is licensed. Rechecking before each step is what tells you whether to continue.
Worked example
Checkpoint 4.40. One application too many.
\[ \text{Evaluate } \lim_{x \to 0}\frac{1-\cos x}{x^{2}} \text{ and say when to stop.} \]
Check the form
Why: Both parts vanish.
\[ \frac{0}{0} \]
Apply the rule
Why: Differentiate separately.
\[ \sin(x) / (2 x) \]
Recheck
Why: Still both vanish.
\[ \frac{0}{0}\text{ again} \]
Apply again
Why: Differentiate separately.
\[ \cos(x) / 2 \]
Recheck and stop
Why: Now determinate.
\[ \frac{1}{2} \]
Figure (svg): The solution to Worked example stopping at the right moment shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0}\frac{1-\cos x}{x^{2}} = \tfrac12 \]
Verify: see what a third application would give
Why: Applying the rule to cosine over 2 would give minus sine over 0, which is not a legitimate step at all since the form is no longer indeterminate — and the resulting expression is undefined. The recheck at each stage is what prevents this. Numerically, at x equal to 0.01 the original expression is about 0.4999958, confirming the limit of one half.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 401-402
Error analysis
A student applies the rule repeatedly without rechecking.
Annotate
On: \( \frac{1-\cos x}{x^{2}} \to \frac{\sin x}{2x} \to \frac{\cos x}{2} \to \frac{-\sin x}{0} \)
The form must be rechecked before every application, not just the first. Stopping as soon as substitution gives a value is the rule, and continuing past it turns a correct computation into nonsense.
Ranking
Applying the rule more than once.
Put in order
Why: Step c is the recheck, and it is the one that gets skipped once the process feels mechanical. Without it there is nothing to signal that the iteration should stop, and one application too many produces a meaningless expression.
Sorting
Substitute into the current quotient.
Sort into buckets
Sort each intermediate result.
The last is worth checking carefully: x over x plus 1 at infinity is indeterminate, and one application gives 1 over 1, which is the answer. The degree rule of Section 4.6 gives the same result in one glance, which is usually quicker.
Prediction
Commit before reasoning.
Predict first
Why is a power over an exponential guaranteed to resolve after finitely many applications?
Correct: Because the power's degree drops each time and the exponential does not change.
\[ x^{n} \to nx^{n-1} \to \cdots \to n!, \quad e^{x} \to e^{x} \to \cdots \to e^{x} \]
Why: Differentiating x to the n gives n times x to the n minus 1, so the degree falls by exactly one; differentiating e to the x gives e to the x, unchanged. After n applications the numerator is a constant and the form is no longer indeterminate. That termination guarantee is what turns Section 4.6's growth ranking from a plausible ordering into a proved theorem.
Section
Section 3
Concept
The rule handles only quotients. A product of a vanishing and an unbounded factor is rewritten by moving one factor into the denominator as a reciprocal; a difference of two unbounded quantities is combined over a common denominator.
converting the forms — A product form becomes a quotient by writing one factor as a reciprocal in the denominator. A difference form becomes a quotient by combining over a common denominator or by factoring.
\[ f\cdot g = \frac{f}{1/g}, \qquad f - g = \frac{\cdots}{\cdots} \]
For a product there are two choices of which factor to move, and they are not equally good. One usually leads to a quotient that resolves and the other to one that is worse than the original.
Figure (svg): Converting a product form into a quotient, with the two available choices
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 402-405 — other indeterminate forms
Picture it
A product form rewritten both ways.
Figure (svg): Converting a product form into a quotient, with the two available choices
Moving the logarithm's reciprocal into the denominator produces something worse; moving the x produces a quotient that resolves in one step. Trying the other way when the first stalls is a legitimate tactic.
Worked example
Example 4.42. Choose which factor to move.
\[ \text{Evaluate } \lim_{x \to 0^{+}} x\ln x. \]
Check the form
Why: One factor vanishes, the other is unbounded.
\[ 0 \times\text{ negative } \infty \]
Rewrite as a quotient
Why: Move the x into the denominator as a reciprocal.
\[ \ln x / (\frac{1}{x}) \]
Check the new form
Why: Both parts are unbounded.
Apply the rule
Why: Differentiate separately.
\[ \frac{\frac{1}{x}}{-1 / x ^{2}} \]
Simplify and evaluate
Why: The reciprocals cancel.
\[ -x \to 0 \]
Figure (svg): Converting a product form into a quotient, with the two available choices
\[ \lim_{x \to 0^{+}}x\ln x = 0 \]
Verify: check numerically and see which effect won
Why: At x equal to 0.001 the product is about negative 0.0069, and at 0.000001 about negative 0.0000138 — shrinking toward zero. So the vanishing factor beats the unbounded one, which is the same growth ranking from Section 4.6: x approaches zero faster than the logarithm approaches negative infinity. That the competition has a winner is exactly what makes the form indeterminate before it is resolved.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 403-404
Matching
How each becomes a quotient.
Match the pairs
Why: Only the third needs no preparation. The other three are converted by standard algebra into a quotient, and then the same single rule finishes them — which is why the section is mostly about conversion rather than about the rule.
Worked example
Checkpoint 4.42. Combine over a common denominator.
\[ \text{Evaluate } \lim_{x \to 0^{+}}\left(\frac{1}{x} - \frac{1}{\sin x}\right). \]
Check the form
Why: Both terms grow without bound.
Combine over a common denominator
Why: Standard algebra.
\[ \frac{\sin x - x}{x \sin x} \]
Check the new form
Why: Both parts vanish.
\[ \frac{0}{0} \]
Apply the rule
Why: Differentiate separately.
\[ \frac{\cos x - 1}{\sin x + x \cos x} \]
Recheck and apply again
Why: Still 0/0.
\[ \frac{-\sin x}{2 \cos x - x \sin x} \to 0 \]
Figure (svg): The solution to Worked example a difference form shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0^{+}}\left(\frac1x - \frac{1}{\sin x}\right) = 0 \]
Verify: check numerically and interpret
Why: At x equal to 0.1 the difference is about negative 0.0167, and at 0.01 about negative 0.00167 — closing on zero. The interpretation is that sine of x and x differ by an amount far smaller than either, so their reciprocals differ by less and less. Note the need for two applications, and that each was preceded by a recheck. Combining over a common denominator is what made the rule applicable at all.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 404-405
Trap
\[ \lim_{x \to 0^{+}}x\ln x \]
Differentiate the two factors
Why: The student treats the product like a quotient.
\[ \frac{d}{dx}[x] \cdot \frac{d}{dx}[\ln x] = 1\cdot\frac1x \quad \text{(meaningless)} \]
The rule is stated for quotients only. Differentiating the factors of a product is not any theorem at all.
\[ x\ln x = \frac{\ln x}{1/x} \;\Longrightarrow\; \text{now the rule applies} \]
Convert to a quotient first, then apply the rule
Why: Moving one factor into the denominator as a reciprocal.
Every one of the five remaining indeterminate forms is handled the same way: convert to a quotient, then apply. The rule itself never changes, and the whole difficulty is arranging the expression so that it is usable.
Fill the middle
A product of a vanishing factor and an unbounded one.
Fill in the blanks
x\ln x = \frac1/x___}
Why: Moving x into the denominator as its reciprocal gives an infinity-over-infinity form, which the rule handles. Moving the logarithm instead would give a quotient harder than the original.
Two truths and a lie
All three are about the other forms.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The rule is stated for quotients, and differentiating the two factors of a product is not a theorem of any kind. Every other form must first be converted into a quotient, and that conversion is where most of the work in this section lies.
Prediction
Commit before reasoning.
Predict first
For a product form, which factor should be moved into the denominator?
Correct: Whichever produces a quotient that simplifies; try the other if the first stalls.
\[ \frac{\ln x}{1/x} \text{ resolves}; \quad \frac{x}{1/\ln x} \text{ does not} \]
Why: Both conversions are algebraically valid and only one usually leads anywhere. Moving x into the denominator of x times the logarithm gives a quotient whose derivatives cancel neatly; moving the logarithm gives a reciprocal-of-a-logarithm whose derivative is worse than what it replaced. Recognising a stall and switching is a legitimate and common tactic rather than a sign of error.
Section
Section 4
Concept
The forms one to the infinity, zero to the zero and infinity to the zero all involve a competition in an exponent. Taking logarithms converts each into a product form, which is then converted into a quotient.
the logarithm technique — Set y equal to the expression, take natural logarithms to bring the exponent down, evaluate the resulting limit, and exponentiate that value to recover the original limit.
\[ \lim y = e^{\lim \ln y} \]
The final exponentiation is essential and easy to omit. The rule was applied to the logarithm, so it produced the limit of the logarithm rather than the limit of the original expression.
Figure (svg): The logarithm technique for an exponential indeterminate form
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 405-408 — exponential indeterminate forms
Picture it
The compound-interest limit that defines e.
Figure (svg): The logarithm technique for an exponential indeterminate form
Section 1.5 introduced e as the limit of this expression and could only tabulate it. Here it is evaluated, and the answer is e exactly rather than 2.71828 approximately.
Worked example
Example 4.44. Section 1.5's limit, finally computed.
\[ \text{Evaluate } \lim_{x \to \infty}\left(1+\frac{1}{x}\right)^{x}. \]
Check the form
Why: The base approaches 1 and the exponent grows.
\[ 1\text{ to the } \infty \]
Take logarithms
Why: Set y equal to the expression.
\[ \ln y = x \ln(1 + \frac{1}{x}) \]
Check the new form
Why: A vanishing logarithm times an unbounded x.
\[ 0 \times \infty \]
Rewrite as a quotient and apply the rule
Why: Move x down.
\[ \ln(1 + \frac{1}{x}) / (\frac{1}{x}) \to 1 \]
Exponentiate back
Why: The limit of ln y is 1.
\[ y \to e \]
Figure (svg): The logarithm technique for an exponential indeterminate form
\[ \lim_{x \to \infty}\left(1+\tfrac1x\right)^{x} = e \]
Verify: compare with Section 1.5's table
Why: That section tabulated the expression at n equal to 1, 12, 365 and 8760, obtaining 2.00, 2.613, 2.7146 and 2.7181 — closing on 2.71828 without ever proving what the destination was. This computation proves it is exactly e. Note also why the form is genuinely indeterminate: a base slightly above 1 raised to a huge power could go anywhere, and here the base approaches 1 at exactly the rate that balances the exponent's growth.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 406-407
Fill the middle
An exponential form whose logarithm has been evaluated.
Fill in the blanks
\ln y \to 1 \;\Longrightarrow\; y \to e^1}
Why: The limit of the logarithm is the exponent, so the original expression tends to e to the first power, which is e. Reporting 1 would be reporting the logarithm's limit rather than the function's.
Worked example
Checkpoint 4.44. The same technique.
\[ \text{Evaluate } \lim_{x \to 0^{+}}x^{x}. \]
Check the form
Why: Both base and exponent approach zero.
\[ 0\text{ to the } 0 \]
Take logarithms
Why: The power law brings the exponent down.
\[ \ln y = x \ln x \]
Recognise the limit
Why: From the earlier worked example.
\[ x \ln x \to 0 \]
Exponentiate back
Why: The limit of ln y is 0.
\[ y \to e ^{0} \]
State
Why: The value.
\[ 1 \]
Figure (svg): The solution to Worked example a zero-to-the-zero form shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0^{+}}x^{x} = 1 \]
Verify: check numerically and note the surprise
Why: At x equal to 0.1 the value is about 0.794, at 0.01 about 0.955, and at 0.001 about 0.993 — climbing toward 1. The result is counterintuitive: a shrinking base raised to a shrinking exponent might have gone to 0, or stayed anywhere. The exponent shrinking wins, and the answer is 1. Section 3.9 differentiated this same function, and its minimum at one over e is visible in the numbers dipping before they climb.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 407-408
Error analysis
A student evaluates an exponential indeterminate form.
Annotate
On: \( \ln y = x\ln\!\left(1+\tfrac1x\right) \to 1 \;\Longrightarrow\; \lim y = 1 \)
The technique has three stages and the last is the easiest to forget, because by then the hard work feels finished. Writing the answer as e to the computed limit, rather than as the limit itself, makes the omission impossible.
Ranking
An exponential indeterminate form.
Put in order
Why: Step e is the one omitted, and it changes the answer completely — from 1 to e in the standard example. The three exponential forms all funnel through the same route, so learning it once handles all of them.
Sorting
Identify the form first.
Sort into buckets
Sort each limit by the technique required.
The two logarithm cases end up as products, which then become quotients — so every route leads to the same single rule. The variety is entirely in the preparation.
Prediction
Commit before reasoning.
Predict first
Surely one raised to any power is one. Why is the form indeterminate?
Correct: Because the base only approaches 1, and the rate of that approach competes with the exponent's growth.
\[ \left(1+\tfrac{k}{x}\right)^{x} \to e^{k} \quad \text{for any } k \]
Why: If the base were exactly 1 the answer would be 1, but it is 1 plus something shrinking, and raising a number slightly above 1 to a huge power can give anything. Here the balance produces e; changing the expression to 1 plus 2 over x to the power x gives e squared instead. That two influences compete, with the outcome depending on their relative rates, is exactly what makes a form indeterminate.
Section
Section 5
Concept
A form is indeterminate when two effects pull in opposite directions and the outcome depends on their relative rates. When both push the same way, or only one is acting, the value is determined and the rule must not be used.
determinate forms — Combinations whose value follows from the parts' limits alone: a vanishing numerator over an unbounded denominator gives zero, two unbounded quantities added give unbounded, and so on.
\[ \frac{0}{\infty} = 0, \quad \infty + \infty = \infty, \quad 0^{\infty} = 0 \]
The test is whether two influences compete. Zero over infinity has both effects pushing the quotient down, so there is no competition and the answer is zero without any work.
Figure (svg): Forms that look indeterminate and are not, with their actual values
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 398-408 — recognising indeterminate forms
Picture it
Five combinations with definite values.
Figure (svg): Forms that look indeterminate and are not, with their actual values
Each row has both influences pushing the same way, so the outcome is settled before any calculation. Applying the rule to any of them would be a misapplication.
Worked example
Example 4.45. Ask whether anything competes.
\[ \text{Which of } \tfrac{0}{\infty}, \tfrac{\infty}{\infty}, 0\cdot\infty, \infty+\infty \text{ are indeterminate?} \]
Examine zero over infinity
Why: A small numerator and a large denominator.
Conclude for it
Why: No competition.
Examine infinity over infinity
Why: A large numerator and a large denominator.
Examine zero times infinity
Why: One factor shrinks, the other grows.
Examine infinity plus infinity
Why: Both push upward.
Figure (svg): Forms that look indeterminate and are not, with their actual values
\[ \tfrac{\infty}{\infty} \text{ and } 0\cdot\infty \text{ compete}; \tfrac{0}{\infty} \text{ and } \infty+\infty \text{ do not} \]
Verify: test the criterion on a fifth form
Why: Consider infinity to the power zero. The base grows without bound, pushing the value up; the exponent shrinks toward zero, pushing it toward 1. Those compete, so the form IS indeterminate — and indeed it is one of the seven. By contrast zero to the power infinity has a shrinking base raised to a growing power, both pushing toward zero, so it is determinate. The competition test decides both correctly.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 407-408
Sorting
Ask whether two influences compete.
Sort into buckets
Sort each form.
The last is the one most often miscounted: a shrinking base raised to a growing power has both effects driving toward zero, so the value is 0. Compare it with infinity to the power zero, where the two effects genuinely compete and the form IS indeterminate.
Worked example
Checkpoint 4.45. No rule needed.
\[ \text{Evaluate } \lim_{x \to \infty}\frac{\sin x}{x}. \]
Examine the numerator
Why: Sine stays between negative 1 and 1.
Examine the denominator
Why: It grows without bound.
Identify the form
Why: Bounded over unbounded.
Evaluate directly
Why: A bounded quantity over a large one.
\[ 0 \]
Note why the rule would fail
Why: The derivatives give cosine over 1.
Figure (svg): The solution to Worked example a determinate form evaluated directly shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to \infty}\frac{\sin x}{x} = 0 \]
Verify: see exactly how the rule fails here
Why: Applying the rule gives cosine of x over 1, which oscillates forever and has no limit — so the rule would produce no answer at all, and a student might wrongly conclude the original limit does not exist. In fact a squeeze between minus one over x and one over x settles it immediately. This is a case where the rule is not merely unnecessary but actively misleading, and the form check is what avoids it.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 408-408
Trap
\[ \lim_{x \to \infty}\frac{\sin x}{x} \;\to\; \frac{\cos x}{1} \]
Note that cosine has no limit and conclude the original has none
Why: The student trusts the rule's silence.
\[ \text{'the limit does not exist'} \quad \text{(wrong)} \]
The original limit is 0, by a squeeze. The rule was never applicable, since the form was bounded over unbounded rather than indeterminate.
\[ -\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x} \;\Longrightarrow\; \text{limit } 0 \]
Check the form first, and use a squeeze when a bounded factor is involved
Why: The rule's failure to produce an answer proves nothing about the original limit.
This is a genuine trap because the rule seems to be saying something. It is not: when its hypotheses fail, its output carries no information at all, and Section 2.3's squeeze technique remains the right tool for a bounded numerator.
Fill the middle
A vanishing numerator over an unbounded denominator.
Fill in the blanks
\frac0___ = ___
Why: A small numerator and a large denominator both make the quotient small, so there is no competition and the value is zero. Applying the rule here would be a misapplication.
Two truths and a lie
All three are about determinate forms.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Both quantities are pushing upward, so nothing competes and the sum grows without bound — the value is determined. Infinity MINUS infinity is the indeterminate one, because there the two effects genuinely oppose each other and the outcome depends on which grows faster.
Prediction
Commit before reasoning.
Predict first
How do you decide whether a form is indeterminate?
Correct: Ask whether two influences compete.
\[ \text{competing} \Rightarrow \text{indeterminate}; \quad \text{agreeing} \Rightarrow \text{determined} \]
Why: The list of seven can be reconstructed from that single question rather than memorised. Infinity over infinity competes because a large numerator pushes up and a large denominator pushes down; infinity plus infinity does not, because both push up. Trying the rule is the worst option, since it produces a wrong answer rather than refusing — which is exactly the trap this idea exists to prevent.
Comparison
Fill the blanks. Every route ends at the same rule.
Comparison matrix
| Form | How to prepare it | What it becomes |
|---|---|---|
| 0/0 and inf/inf | nothing | apply the rule directly |
| 0 times inf | write one factor as a reciprocal | a quotient |
| inf minus inf | combine over a common denominator | usually 0/0 |
| 1^inf, 0^0, inf^0 | take logarithms | a product, then a quotient |
The rule itself never changes and applies only to quotients. Everything else in this section is arranging an expression so that it becomes one.
Pattern
Given a limit to evaluate.
Step one is the whole safeguard. The rule produces a plausible wrong answer on a determinate form and no answer at all on a bounded quotient, and neither failure announces itself.
Stewart, Calculus: Early Transcendentals 8e, §4.4 Indeterminate Forms and l'Hospital's Rule §4.4, pp. 304-314
Check
Check the form first.
Check your understanding
Applying the rule to the limit of (x+1)/(x+2) at 0 gives what, and is it right?
Answer: A
Why: The form is determinate, so the rule does not apply; differentiating anyway gives 1 over 1.
Check
Repeated application.
Check your understanding
How many applications does the limit of x^3/e^x at infinity require?
Answer: A
Why: Each application lowers the power's degree by one, so a cube needs three.
Check
The exponential route. Do not stop early.
Check your understanding
For (1 + 1/x)^x at infinity, the limit of ln y is 1. What is the limit of y?
Answer: A
Why: Exponentiating gives e to the first power, which is e.
Real world
A savings account pays an annual rate r compounded n times a year, so a deposit grows by a factor of one plus r over n, raised to the power n, each year. A bank advertises continuous compounding as its best offer.
Discussion prompt
Evaluate the limit as the compounding frequency grows without bound, identify the indeterminate form, and say what it means for a saver choosing between quarterly and continuous compounding.
Hint: This is the same technique as the definition of e, with a rate constant carried through.
Answer:
The growth factor per year is one plus r over n, all to the power n. As n grows the base approaches 1 and the exponent grows, which is the one to the infinity form — indeterminate, because the base's approach to 1 competes with the exponent's growth.
\[ y = \left(1+\frac{r}{n}\right)^{n}, \qquad \ln y = n\ln\!\left(1+\frac{r}{n}\right) \]
That is a zero-times-infinity form, converted to a quotient and resolved by the rule:
\[ \frac{\ln(1+r/n)}{1/n} \;\longrightarrow\; r \;\Longrightarrow\; y \to e^{r} \]
So continuous compounding multiplies the balance by e to the r each year, which is where Section 1.5's formula came from.
For the saver the practical answer is deflating. At r equal to 5 percent, quarterly compounding gives a factor of 1.050945 and continuous gives 1.051271 — a difference of about three pence on a hundred pounds a year. The limit is mathematically exact and commercially almost irrelevant, because the sequence converges so quickly that quarterly is already within a thousandth of the ceiling.
That gap between mathematical significance and practical significance is worth noticing. The limit tells you where the sequence is going; only computing a few terms tells you whether getting there is worth anything.
Commit first
Answer, then rate your confidence honestly.
Predict first
Which of these is NOT an indeterminate form?
Correct: Zero divided by infinity.
\[ \frac{0}{\infty} = 0 \text{ always}; \quad \frac{0}{0} \text{ can be anything} \]
Why: A vanishing numerator and an unbounded denominator both drive the quotient toward zero, so nothing competes and the value is determined without any work. The other three all involve two influences pulling in opposite directions: two vanishing quantities, two unbounded ones subtracted, and a base approaching 1 against a growing exponent. The competition test reconstructs the whole list of seven and is more reliable than memorising it.
Explain it
They applied the rule to the quotient of x plus 1 by x plus 2 and got 1, and cannot see the problem.
Discussion prompt
In four sentences or fewer, show them what went wrong.
Hint: Have them substitute directly first.
Answer:
Ask them to substitute 0 into the original: they get 1 over 2, which is a perfectly ordinary number. There was never an indeterminate form, so the rule's hypothesis failed and it should not have been used.
The danger is that the rule gave an answer anyway, and a plausible one — nothing about the computation looked wrong. That is why the form check is the first step every time, and it costs a single substitution.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the check, substitute before doing anything else. For conversions, move one factor into the denominator as a reciprocal, or combine over a common denominator. For exponential forms, remember the third stage — exponentiate back. For classification, ask whether two influences compete rather than trying to recall the list of seven. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the rule in a box with a note beside it saying what it is NOT — the quotient rule — and one line showing what the quotient rule would give for the sine quotient. Below, write the seven indeterminate forms in a column with their conversion route beside each, and beneath them list five forms that only look indeterminate with their actual values. In the middle of the page, work the cube over exponential limit through all three applications, marking the form at each stage. Beside it, work x times the logarithm as a product form, showing both possible conversions and which one resolves. In the lower half, work the compound-interest limit completely: the form, taking logarithms, converting to a quotient, applying the rule, and exponentiating back — with the last step circled. At the bottom, work the sine over x limit at infinity by a squeeze, and write what the rule gives and why that proves nothing. In a margin, write the competition test for recognising an indeterminate form.
If your compound-interest answer is 1 rather than e, the exponentiation was omitted — that final stage is the one this section's technique most often loses, and it changes the answer by a factor of e.
Recap
Five things, and the first is a one-substitution check that prevents most of the errors.
| If you see | Then |
|---|---|
| 0/0 or inf/inf | Apply the rule directly |
| A determinate form | Evaluate directly; the rule would mislead |
| 0 times infinity | Move one factor down as a reciprocal |
| infinity minus infinity | Combine over a common denominator |
| An exponent competing with a base | Take logarithms, then exponentiate back |
| A bounded numerator | Use a squeeze, not the rule |
| A power over an exponential | Repeat until the power is a constant |
Section 4.9 turns from evaluating limits to finding roots. Newton's method uses the tangent line of Section 4.2 iteratively, and it solves equations that no algebra can — which the Intermediate Value Theorem of Section 2.4 could only promise had solutions.
OpenStax Calculus Volume 1, §4.8 L'Hôpital's Rule §4.8, pp. 394-408 — everything on these slides traces back here
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