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L'Hopital's Rule
Calculus 2 · Chapter 2: Indeterminate Forms & Improper IntegralsSection 2.1 · lesson 8 of 5413 min · free
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In this lesson
Extracting a limit that direct substitution hides.
- Recognise the seven indeterminate forms
- Apply the rule to 0/0 and \infty/\infty, and reapply when needed
- Convert products and differences into quotients first
- Handle 0^{0}, 1^{\infty} and \infty^{0} with logarithms
Transcript
Everything said in the video, section by section (1,850 words). Read along, or search it.
MAT-265 - Calculus 2 · Section 2.1
L'Hopital's Rule
A physics student is studying a falling object with air resistance. Its velocity approaches a terminal value. But at the very first instant — time equals zero — the formula for average speed takes the shape zero over zero. Plug the numbers in and you get nonsense. Zero distance divided by zero time. And yet the object clearly has a well-defined instantaneous speed. It's moving. We just can't read the number off the formula. The same frustrating shape shows up when an economist computes the limiting growth rate of an investment. Or when an engineer evaluates the gain of a filter at a critical frequency. In every one of those cases a limit exists. Direct substitution simply refuses to reveal it. L'Hopital's Rule is the precision tool that extracts these limits — converting a hopeless ratio into a solvable one, through differentiation.
The Problem
A Speed That Has No Obvious Value
Let's be precise about what's going wrong. An indeterminate form is an expression whose limiting value cannot be determined from its surface shape alone. And zero over zero is indeterminate for a specific reason worth understanding. The answer depends entirely on how fast the numerator and denominator each approach zero. It's a competition. And the bare symbols don't reveal who's winning. If the numerator races to zero much faster, the limit is zero. If the denominator does, the limit blows up. If they go at comparable rates, you get something finite in between. L'Hopital's Rule resolves that competition directly — by comparing the rates of change of the two parts. There are seven indeterminate forms in total. Zero over zero. Infinity over infinity. Zero times infinity. Infinity minus infinity. And three exponential ones: zero to the zero, one to the infinity, and infinity to the zero. We'll handle all seven.
The Theorem
Differentiate Top and Bottom - Separately
Here's the statement. Suppose f and g are differentiable near x equals a — except possibly at a itself — and g prime is not zero near a. If the limit of f over g has the form zero over zero, or plus-or-minus infinity over plus-or-minus infinity, then that limit equals the limit of f prime over g prime. Provided the limit on the right exists, or is plus or minus infinity. And a can be a finite number or infinity. One-sided limits are allowed too. Now — the single most important caution is built right into that statement, and it's worth saying out loud. The rule differentiates the numerator and the denominator separately. It is not the quotient rule. There is no g f prime minus f g prime, over g squared, anywhere in this. Just f prime over g prime. Students who confuse these two get consistently wrong answers, and they usually don't notice.
Worked Examples 1 and 2
The Two Basic Forms
Let's see both basic forms. First. The limit as x approaches zero of sine x over x. Direct substitution gives sine zero over zero, which is zero over zero. Indeterminate. So the rule applies. Differentiate top and bottom separately. The derivative of sine x is cosine x. The derivative of x is one. So the limit becomes cosine x over one, which at zero is one. And notice what just happened. That famous limit — the one foundational to the derivative of sine itself — fell out in a single line. Second. The limit as x goes to infinity of log x over x. Both log x and x go to infinity, giving infinity over infinity. Apply the rule. The derivative of log x is one over x. The derivative of x is one. So you get the limit of one over x, which is zero. And that confirms something useful. Any positive power of x eventually outgrows the logarithm. That's the fact behind efficiency comparisons in computer science — why a log-n algorithm beats a linear one, eventually, always.
Conditions
Three Checks, and Knowing When to Go Again
Three conditions have to hold before you may use this rule. And checking them is not optional. First. Verify the form. Substitute, and confirm the limit truly is zero over zero, or infinity over infinity. If substitution gives any determinate value — two over zero, five over three, zero over four — the rule does not apply. And using it will produce a wrong answer. Second. Confirm differentiability. Both functions must be differentiable in a neighborhood of the target point, and g prime must not be zero there. Third. The resulting limit must exist, as a number or as plus or minus infinity. If the limit of f prime over g prime oscillates without settling, the rule is inconclusive. And note carefully — inconclusive does not mean the original limit fails to exist. It just means this tool didn't answer the question. Now, what if one application still leaves you indeterminate? Apply the rule again. And as many times as needed — provided the form stays zero over zero or infinity over infinity at each stage. Take the limit of one minus cosine x, over x squared. Substitution gives zero over zero. Apply once, and you get sine x over two x. Check the new form. Sine zero over zero. Still zero over zero. So apply again. Now you get cosine x over two, which is one half. Two applications were needed. Stopping after the first would have left you with an unevaluated indeterminate form. The form must be re-checked before every single reapplication.
Other Forms
Products and Differences Become Quotients
So far every example has been a quotient. That's not an accident — L'Hopital's Rule directly handles only quotients. The remaining indeterminate forms have to be rewritten as one first, using algebra. Start with the product form. Zero times infinity. A product where one factor tends to zero and the other to infinity is indeterminate. Rewrite it as a fraction by moving one factor to the denominator as its reciprocal. You can write f over one-over-g, or g over one-over-f. Choose whichever gives you a cleaner derivative. Try the limit as x approaches zero from the right, of x times log x. x goes to zero while log x goes to minus infinity. So it's zero times minus infinity. Move the x down as one over x. Now you have log x over one over x — which reads as minus infinity over infinity. Apply the rule. The derivative of one over x is minus one over x squared. And the whole thing collapses to the limit of minus x, which is zero. Notice that placing log x in the denominator instead would have produced a much messier derivative. Look ahead one step before you commit. Now the difference form. Infinity minus infinity. This is indeterminate because the result depends on which term grows faster. Combine the two terms into a single fraction — usually over a common denominator — to expose a zero over zero or infinity over infinity form. Then proceed as usual.
Exponential Forms
Take Logarithms First
Three forms remain, and they all have the same shape. A variable base raised to a variable power. Zero to the zero. One to the infinity. Infinity to the zero. These are indeterminate because the base and the exponent pull the result in opposing directions. The universal strategy is to take the natural logarithm. Why does that help? Because the logarithm converts the exponent into a coefficient. And that turns your expression into a zero times infinity product — which you already know how to handle. The procedure has three steps. Let L be the limit. First, take logs: log L equals the limit of g of x times log f of x. Second, evaluate that product limit using the techniques from the previous section. Third, exponentiate to recover L. And that last step is essential. The logarithm gives you log L, not L itself. Students lose marks here constantly. Let's do the most famous one. The limit as x goes to infinity of one plus one over x, all to the x. That's the limit defining continuously compounded interest. The base goes to one and the exponent goes to infinity. One to the infinity. Take logs, rewrite as a quotient, apply the rule, and log L comes out to one. Exponentiate: L equals e to the first, which is e. A dollar compounded continuously for one period at a hundred percent interest grows to exactly e. About two point seven one eight dollars. And one more. The limit of x to the x, as x approaches zero from the right. Base and exponent both go to zero. Take logs and you get the limit of x log x — which we just showed is zero. So L equals e to the zero, which is one. x to the x approaches one. That one surprises almost everybody.
Reference
Every Form, and How to Convert It
Let's put every form in one place, with the conversion that makes L'Hopital applicable. Zero over zero: already a quotient. Apply directly. Infinity over infinity: already a quotient. Apply directly. Zero times infinity: write one factor as a reciprocal to make a quotient. Infinity minus infinity: combine over a common denominator. And zero to the zero, one to the infinity, infinity to the zero: take logs, evaluate, then exponentiate. Five rows. Seven forms. That's the entire section. Now here's something genuinely worth thinking about. The form one to the infinity feels like it should equal one. After all, one raised to any power is one. Yet Example six showed it equals e. Not one. So what resolves that apparent paradox? Consider that the base is never exactly one. It only approaches one. And the gap between the base and one is shrinking at exactly the same time the exponent is growing without bound. Those two effects fight each other. Why does that tension make the form genuinely indeterminate, rather than obviously equal to one?
Avoiding Errors
Two Mistakes, and What to Carry Forward
Two errors account for most mistakes with this rule. The first, and by far the most damaging. Applying the rule to a form that isn't indeterminate. Consider the limit as x approaches zero of cosine x over x. Substitution gives one over zero. That is not indeterminate — the limit diverges. But blindly differentiating gives minus sine x over one, which goes to zero. A completely wrong answer, arrived at confidently. Always confirm the indeterminate form before differentiating. The second error. Differentiating with the quotient rule. L'Hopital requires the numerator and denominator differentiated independently. A student who applies the quotient rule computes the derivative of the whole quotient — an entirely different and irrelevant quantity. There's also a subtler trap. Continuing to apply the rule after the form has become determinate. Once a step yields a definite value, stop and read off the answer. Differentiating further changes the result and introduces error. So — five things to carry forward. Verify the form first. The rule applies only to zero over zero and infinity over infinity. Differentiate separately, not as a quotient. Reapply as needed, re-checking the form each time, and stop the moment it becomes determinate. Convert other forms first — reciprocal quotients for products, common denominators for differences. And for the three exponential forms, take logarithms, evaluate, then exponentiate. Next up is section two point two, Improper Integrals, Type One — where the interval of integration itself runs off to infinity.
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