L'Hôpital's Rule

The limits unit established that $0/0$ carries no information, and that clearing it is a matter of algebra, factor, cancel, rationalise. That works when the functions are polynomials or roots, because there is a common factor to remove.

It does not work for $\underset{x\to 0}{lim}\frac{e^x-1}{x}$. Both parts vanish at 0, the form is $0/0$, and no factoring is available: $e^x-1$ has no algebraic factor of $x$ to cancel. The numerical values head somewhere definite, $1.0517$, $1.00502$, $1.00050$ as $x$ closes on 0, but no manipulation reaches it.

L'Hôpital's rule handles exactly these cases, and its statement is almost too simple: differentiate the numerator and the denominator separately, then take the limit again. Here that gives $\frac{e^x}{1}\to 1$, matching the numbers.

The motivation is useful, though it is not the proof. When both quantities vanish, the size of the ratio near the point reflects the relative rates at which each approaches zero, and a derivative measures such a rate. That is why differentiating each part is the right move to try. The theorem itself is proved from the Cauchy mean value theorem, and it requires more than this motivation supplies: the derivatives must exist near the point, $g^{\prime }$ must be nonzero there, and $f^{\prime }/g^{\prime }$ must have a limit.

That also marks its boundary. If the denominator does not vanish, there is no competition to read, and applying the rule answers a different question: $\underset{x\to 0}{lim}\frac{x+1}{x+2}$ is $\frac{1}{2}$, while differentiating top and bottom returns 1. The rule is narrow in scope, and this unit is as much about its limits as its use.

Take the simplest case: $f$ and $g$ both differentiable at $a$, with $f(a)=g(a)=0$ and $g^{\prime }(a)\ne 0$. Then the quotient can be rewritten so that two difference quotients appear:

The first step subtracts $f(a)$ and $g(a)$, which is free precisely because both are zero, that is the indeterminate hypothesis doing its work. The second divides numerator and denominator by $x-a$, which is legitimate because $x\ne a$ throughout a limit.

Now each part is a difference quotient, and as $x\to a$ they tend to $f^{\prime }(a)$ and $g^{\prime }(a)$. By the quotient law, valid since $g^{\prime }(a)\ne 0$,

The limit is decided by the rates at which the two functions leave zero. A numerator vanishing twice as fast as its denominator gives a ratio of 2, whatever the functions are.

Where the hypothesis is load-bearing. If $f(a)\ne 0$, the first step is invalid, subtracting $f(a)$ changes the function. And the conclusion would be wrong: for $\frac{x+1}{x+2}$ at $a=0$ the numerator is 1 and the denominator 2, so the quotient is $\frac{1}{2}$, while $f^{\prime }/g^{\prime }=1/1=1$. Nothing about differentiability failed; the subtraction was simply not free.

Why the general theorem needs more. This argument assumes differentiability at $a$ and a nonzero $g^{\prime }(a)$. The full rule allows $a=\pm \mathrm{\infty }$, allows $f^{\prime }/g^{\prime }$ itself to be indeterminate, and covers $\mathrm{\infty }/\mathrm{\infty }$, where nothing is being subtracted at all. Those cases are proved from the Cauchy mean value theorem rather than this manipulation, but the rate reading survives intact, and it is what makes the statement memorable rather than arbitrary.

A caution the argument makes visible. It uses the derivative's definition, so using the rule to evaluate $\underset{x\to 0}{lim}\frac{\sin x}{x}$ is circular: that limit is what establishes $\frac{d}{dx}\sin x=\cos x$ in the first place. The rule returns the right answer and proves nothing.

L'Hôpital's rule. Let $f$ and $g$ be differentiable on an open interval containing $a$ (except possibly at $a$), with $g^{\prime }(x)\ne 0$ there. Suppose either

If ${\displaystyle \underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}}$ exists or is $\pm \mathrm{\infty }$, then

The statement holds for $a$ real, for one-sided limits, and for $a=\pm \mathrm{\infty }$.

Hypothesis 1 — the form is indeterminate. Without it the conclusion is false, not merely unproved. $\underset{x\to 0}{lim}\frac{x+1}{x+2}=\frac{1}{2}$, while $lim\frac{f^{\prime }}{g^{\prime }}=\frac{1}{1}=1$. The two sides disagree, so the hypothesis cannot be dropped.

Hypothesis 2 — $g^{\prime }$ does not vanish near $a$. The proof forms the quotient $f^{\prime }/g^{\prime }$ and applies the Cauchy mean value theorem, both of which need $g^{\prime }\ne 0$ on the approach. A $g^{\prime }$ vanishing infinitely often near $a$ breaks the argument even when every other condition holds.

Hypothesis 3 — the limit of $f^{\prime }/g^{\prime }$ exists. This is a condition on the conclusion, which is unusual and worth care. If $f^{\prime }/g^{\prime }$ has no limit, the theorem says nothing, and in particular does not say the original limit fails to exist.

The classic witness is $\underset{x\to \mathrm{\infty }}{lim}\frac{x+\sin x}{x}$. The original limit is 1, since $\frac{\sin x}{x}\to 0$. But differentiating gives $\frac{1+\cos x}{1}$, which oscillates between 0 and 2 forever and has no limit. The rule is silent; the limit exists regardless.

The implication runs one way. $lim{f}^{\prime }/g^{\prime }$ existing implies $limf/g$ equals it. The converse fails, as that example shows.

Repeated application. If $f^{\prime }/g^{\prime }$ is again indeterminate and the hypotheses hold for $f^{\prime }$ and $g^{\prime }$, the rule applies again. Each repetition requires a fresh check: $\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}$ gives $\frac{2x}{e^x}$, still $\mathrm{\infty }/\mathrm{\infty }$, then $\frac{2}{e^x}\to 0$. Verifying numerically, $x^2/e^x$ reads $0.1684$ at $x=5$, $0.00454$ at $x=10$ and $8.24\times {10}^{-7}$ at $x=20$.

The same discipline forbids a third application here: once the form is $2/e^x$, the denominator tends to infinity while the numerator is constant, so the quotient is no longer indeterminate and the limit is read directly.

When repetition cannot terminate. For $\underset{x\to \mathrm{\infty }}{lim}\frac{e^x+e^{-x}}{e^x-e^{-x}}$ each application swaps numerator and denominator, returning to the same pair after two steps. The form stays $\mathrm{\infty }/\mathrm{\infty }$ forever. Dividing numerator and denominator by $e^x$ instead gives $\frac{1+e^{-2x}}{1-e^{-2x}}\to 1$, confirmed by the values $1.3130$, $1.00009$, $1.000000004$ at $x=1,5,10$. The rule applies legitimately and simply never finishes, which is a third way it can fail to help.

The canonical definition states the rule and the rewrites. What follows is the full inventory of indeterminate forms and why only two are directly usable.

The two quotient forms — the rule applies directly.

The product form $0\cdot \mathrm{\infty }$ — rewrite as a quotient. There are two ways, and they are not equally useful. For $\underset{x\to 0^{+}}{lim}x\ln x$:

• ${\displaystyle \frac{\ln x}{1/x}}$ gives ${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$, and the rule yields ${\displaystyle \frac{1/x}{-1/x^2}}=-x\to 0$.
• ${\displaystyle \frac{x}{1/\ln x}}$ gives ${\displaystyle \frac{0}{0}}$, and differentiating produces ${\displaystyle \frac{1}{-1/(x{\ln }^{2}x)}}=-x{\ln }^{2}x$, worse than the original.

The working choice is the one whose reciprocal is easy to differentiate. Checking the answer numerically: $x\ln x$ reads $-0.2303$, $-0.04605$, $-0.0069$, $-0.0000138$ at $x=0.1,0.01,0.001,{10}^{-6}$.

The difference form $\mathrm{\infty }-\mathrm{\infty }$ — combine first. Two unbounded quantities may differ by anything at all, so the expression must be turned into a single fraction before the form is even readable. A common denominator, or factoring out the dominant term, usually produces $0/0$.

The three power forms $1^{\mathrm{\infty }}$, $0^0$, ${\mathrm{\infty }}^0$ — take logarithms. For $y=f(x{)}^{g(x)}$, write $\ln y=g(x)\ln f(x)$, which is a $0\cdot \mathrm{\infty }$ product. Resolve that, then exponentiate, and the exponentiation step is not optional: the limit of $\ln y$ is not the limit of $y$.

Why $\frac{0}{\mathrm{\infty }}$ and $\frac{\mathrm{\infty }}{0}$ are absent. Neither is indeterminate. A numerator tending to 0 with a denominator growing without bound gives 0; the reverse gives $\pm \mathrm{\infty }$. Both are settled without any rule, and applying L'Hôpital to them is the same error as applying it to $\frac{x+1}{x+2}$.

Why the list is exactly these seven. Each arises where a limit law's hypothesis fails in a way that two competing behaviours leave the outcome open. $\frac{0}{0}$ and $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ are quotients of matched magnitudes; $0\cdot \mathrm{\infty }$ and $\mathrm{\infty }-\mathrm{\infty }$ are their additive and multiplicative cousins; and the three power forms reduce to $0\cdot \mathrm{\infty }$ through the logarithm. Everything else is decided by the laws.

Why differentiating separately is not the quotient rule. The rule replaces $\frac{f}{g}$ by $\frac{f^{\prime }}{g^{\prime }}$. No $g^2$ in the denominator and no subtraction in the numerator. The quotient rule computes the derivative of the quotient, which is a different object and is not what the theorem concerns. Confusing the two is the most common mechanical error in this material.

1. A $0/0$ form: ${\displaystyle \underset{x\to 0}{lim}\frac{e^x-1}{x}}$.

Check the hypotheses. At $x=0$ the numerator is $e^0-1=0$ and the denominator is 0, so the form is $0/0$. Both functions are differentiable near $0$, and the denominator's derivative $\frac{d}{dx}x=1$ is nonzero there. The theorem needs all three: an indeterminate form on its own does not license the rule.

Apply. Differentiate separately: $\frac{d}{dx}(e^x-1)=e^x$ and $\frac{d}{dx}x=1$, so

Check numerically: $1.0517$, $1.00502$, $1.00050$ at $x=0.1,0.01,0.001$, and $0.99950$ at $x=-0.001$.

2. Another $0/0$: ${\displaystyle \underset{x\to 0}{lim}\frac{\sin 3x}{x}}$.

Both parts vanish at 0. Differentiating gives $\frac{3\cos 3x}{1}\to 3$.

Check: $2.9552$, $2.99955$, $2.9999955$ at $x=0.1,0.01,0.001$.

The factor 3 is the point: the numerator leaves zero three times as fast as the denominator, and the limit reads that ratio of rates.

3. An $\mathrm{\infty }/\mathrm{\infty }$ form: ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{\ln x}{x}}$.

Both parts grow without bound. Differentiating gives

Check: $0.2303$, $0.04605$, $0.000921$, $0.0000138$ at $x=10,{10}^2,{10}^4,{10}^6$. The logarithm loses to any positive power, and this is the cleanest demonstration.

4. Two applications: ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}}$.

First pass. Form is $\mathrm{\infty }/\mathrm{\infty }$. Differentiating gives $\frac{2x}{e^x}$.

Re-check. Still $\mathrm{\infty }/\mathrm{\infty }$. This check is mandatory, not a formality.

Second pass. Differentiating again gives $\frac{2}{e^x}$.

Re-check. Now the numerator is constant and the denominator unbounded, so the form is not indeterminate. Stop applying the rule and read the limit: $\frac{2}{e^x}\to 0$.

Check: $x^2/e^x$ reads $0.1684$, $0.00454$, $8.24\times {10}^{-7}$ at $x=5,10,20$.

5. A product, $0\cdot \mathrm{\infty }$: ${\displaystyle \underset{x\to 0^{+}}{lim}x\ln x}$.

The rule applies only to quotients, so rewrite. Choosing $\frac{\ln x}{1/x}$ gives $\frac{-\mathrm{\infty }}{\mathrm{\infty }}$, which the rule covers:

Check: $-0.2303$, $-0.04605$, $-0.0069$, $-0.0000138$ at $x=0.1,0.01,0.001,{10}^{-6}$.

The other rewrite, $\frac{x}{1/\ln x}$, is legitimate and produces $-x{\ln }^{2}x$, harder than the original. Both are valid; only one is useful.

6. Where the rule never finishes: ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{e^x+e^{-x}}{e^x-e^{-x}}}$.

The form is $\mathrm{\infty }/\mathrm{\infty }$, which is necessary but not sufficient. Checking the rest: numerator and denominator are differentiable on the relevant region, and the denominator's derivative $e^x+e^{-x}$ is never zero there. All three conditions hold, so the rule applies. Differentiating gives $\frac{e^x-e^{-x}}{e^x+e^{-x}}$. The original with numerator and denominator swapped. Applying it again returns to the start. The cycle never terminates.

Algebra settles it immediately. Divide numerator and denominator by $e^x$:

Check: $1.3130$, $1.00009$, $1.000000004$ at $x=1,5,10$.

Nothing went wrong. Every application was valid. The rule simply is not a decision procedure, and recognising when to abandon it for algebra is part of using it.

Step 1 — substitute, and name the form. Evaluate the numerator's and denominator's limits separately.

This step is the one that is skipped, and skipping it is what produces wrong answers rather than merely inelegant ones.

Step 2 — convert to a quotient.

• $0\cdot \mathrm{\infty }$: move one factor into the denominator as its reciprocal. Choose the direction whose reciprocal is easier to differentiate, for $x\ln x$ that means $\frac{\ln x}{1/x}$, not $\frac{x}{1/\ln x}$.
• $\mathrm{\infty }-\mathrm{\infty }$: combine over a common denominator, or factor out the dominant term.
• A power form: set $y$ equal to the expression, take $\ln y$, resolve the resulting product, then exponentiate to recover $y$.

Step 3 — try algebra before the rule. Dividing by the highest power, cancelling a common factor or rationalising often settles the limit faster and avoids the cycling case. The rule is for what algebra cannot reach.

Step 4 — differentiate numerator and denominator separately. Not the quotient rule. No $g^2$, no subtraction.

Step 5 — re-check the form, then decide.

• Indeterminate again: repeat from Step 4, provided the hypotheses still hold.
• Determinate: evaluate and stop. Continuing past this point reintroduces the Step 1 error.
• The new quotient has no limit, oscillating, say: the rule is silent. It does not follow that the original limit fails to exist; find it another way.
• The same expression returns after one or two passes: the rule is cycling. Abandon it and use algebra.

Step 6 — check the answer. Evaluate the original expression at points closing in on $a$. The values should approach the claimed limit; if they approach something else, Step 1 was probably skipped.

A standing caution. Do not use the rule for $\underset{x\to 0}{lim}\frac{\sin x}{x}$. It returns 1, but the derivative of $\sin$ is established from that very limit, so the argument is circular. The squeeze theorem is the correct route.

${\displaystyle \frac{0}{0}}$ — ${\displaystyle \underset{x\to 0}{lim}\frac{e^x-1}{x}=1}$. The template case. Differentiating gives $e^x/1\to 1$, and the values $1.0517$, $1.00502$, $1.00050$ confirm it. No algebra clears this one: $e^x-1$ has no factor of $x$ to cancel.

${\displaystyle \frac{0}{0}}$ with a rate ratio — ${\displaystyle \underset{x\to 0}{lim}\frac{\sin 3x}{x}=3}$. Differentiating gives $3\cos 3x\to 3$. The answer is the ratio of the speeds at which the two functions leave zero, and the 3 comes from the chain rule on the numerator.

${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$ — ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{\ln x}{x}=0}$. Differentiating gives $\frac{1/x}{1}\to 0$. The values fall $0.2303$, $0.04605$, $0.000921$, $0.0000138$ at $x=10,{10}^2,{10}^4,{10}^6$. This is the standard statement that logarithmic growth loses to linear growth, and the same argument beats any positive power.

${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$ needing two passes — ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}=0}$. One application gives $\frac{2x}{e^x}$, still indeterminate; a second gives $\frac{2}{e^x}$, which is not, so the limit is read directly. The values $0.1684$, $0.00454$, $8.24\times {10}^{-7}$ at $x=5,10,20$ show the exponential winning comfortably. For $x^n/e^x$ the same argument takes $n$ passes.

$0\cdot \mathrm{\infty }$ — ${\displaystyle \underset{x\to 0^{+}}{lim}x\ln x=0}$. Not a quotient, so it is rewritten as $\frac{\ln x}{1/x}$ before the rule applies. The values $-0.2303$, $-0.04605$, $-0.0069$, $-0.0000138$ climb to 0 from below: $x$ reaches zero faster than $\ln x$ reaches $-\mathrm{\infty }$.

**A quotient that is not indeterminate, ${\displaystyle \underset{x\to 0}{lim}\frac{x+1}{x+2}=\frac{1}{2}}$.** Substitution settles it, and the rule must not be used. Applying it anyway returns $1/1=1$, wrong by a factor of two. It sits beside the others as the case that looks identical in shape and differs entirely in what it permits.

Where the rule is silent — ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x+\sin x}{x}=1}$. The form is $\mathrm{\infty }/\mathrm{\infty }$, and the remaining conditions hold: both parts are differentiable and the denominator's derivative is the constant $1$. So the rule applies legitimately, and differentiating gives $1+\cos x$, which oscillates between 0 and 2 and has no limit. The theorem therefore says nothing. The original limit exists regardless, since $\frac{\sin x}{x}\to 0$ leaves $1+0$.

Two forms are handled directly, one after a rewrite, one only by stopping at the right moment, and two not at all, once because the hypothesis fails and once because the conclusion's condition fails. Reading which case is in front of you is the skill; the differentiation is the easy part.