4.8 L'Hopital's Rule

Big idea

Some limits refuse to be settled by substitution because both parts of a quotient collapse at once. If the numerator and the denominator both approach zero, the quotient is a contest between two shrinking quantities, and the answer depends on how fast each shrinks. The same contest appears when both parts grow without bound. These are the indeterminate forms, and the name means precisely that the form of the expression does not determine the answer.

The derivative measures exactly the speed of that shrinking, which is why a differentiation rule can resolve the contest. Near a point where $f$ and $g$ both vanish, each is well approximated by its own tangent line through that point, so the quotient is close to the quotient of the two linear pieces. The common factor of the step size cancels and what is left is the ratio of the derivatives. That is the whole idea.

This makes many previously hard limits routine, and it also makes it easy to use the rule where it does not apply. The hypotheses are narrow: you need an honest $0/0$ or an honest quotient of things running to infinity. A limit that is merely awkward, or a quotient whose denominator has a nonzero limit, is not a candidate, and applying the rule there produces a wrong answer with no warning sign.

The other indeterminate forms, products like $0\cdot\infty$, differences like $\infty - \infty$, and powers like $1^{\infty}$, are not directly covered. They become covered after algebra: rewrite the product as a quotient, combine the difference over a common denominator, or take a logarithm. The rewriting step is where the thinking happens.

Decoder

If $f$ and $g$ are differentiable near $a$, if $f(x)/g(x)$ has the form $0/0$ or a quotient of quantities growing without bound as $x \to a$, and if the limit of $f'(x)/g'(x)$ exists, then the original limit equals that one.

Read the operation carefully. You differentiate the numerator and the denominator separately. This is not the quotient rule, and using the quotient rule here is the most common error in the section.

Read the hypothesis just as carefully. The form must be checked before every application, including the second and third application in a chain. A limit that has become determinate, like $2/0^{+}$, must be read off directly; differentiating again would destroy a correct answer. Checking the form is one substitution and it is not optional.

The conclusion is one directional. If the limit of the ratio of derivatives exists, the original equals it. If the ratio of derivatives has no limit, nothing follows about the original; you have simply learned that this tool does not settle the question. The rule also applies as $x \to \infty$ and as $x \to -\infty$, and for one sided limits, with the same hypotheses.

Definitions and results

Indeterminate form $0/0$. As $x \to a$, if $f(x) \to 0$ and $g(x) \to 0$, then $f(x)/g(x)$ has the form $0/0$ and its limit may be any number, may be infinite, or may not exist.

Indeterminate form $\infty/\infty$. If $|f(x)| \to \infty$ and $|g(x)| \to \infty$, the quotient has this form, and again nothing is determined by the form alone.

The rule of L’Hopital. Suppose $f$ and $g$ are differentiable on an open interval containing $a$, except possibly at $a$ itself, with $g'(x) \neq 0$ there. If $f(x)/g(x)$ has the form $0/0$ or $\infty/\infty$ as $x \to a$, then

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

whenever the limit on the right exists or is infinite. The same statement holds with $a$ replaced by $\infty$ or $-\infty$.

Repeating the rule. If the new quotient is again $0/0$ or $\infty/\infty$, apply the rule again. Recheck the form each time, and stop as soon as the form is determinate.

Not for determinate quotients. If the denominator has a nonzero limit, evaluate by substitution. Applying the rule to $\lim_{x\to 0}\frac{x+2}{\cos x}$ replaces the correct answer $2$ with $\lim_{x\to 0} \frac{1}{-\sin x}$, which does not even exist.

The form $0\cdot\infty$. Rewrite the product $f\cdot g$ as $f/(1/g)$ or $g/(1/f)$, choosing whichever gives the simpler derivatives, to land in $0/0$ or $\infty/\infty$.

The form $\infty - \infty$. Combine the two terms over a common denominator, or factor out the dominant term, to produce a quotient.

The forms $1^{\infty}$, $0^{0}$ and $\infty^{0}$. Let $y$ be the expression, take logarithms to get $\ln y$ as a product, resolve that limit, then exponentiate. If $\ln y \to L$ then $y \to e^{L}$.

Growth rates. Repeated use of the rule establishes the standard hierarchy: as $x \to \infty$, logarithms grow more slowly than any positive power of $x$, and any positive power of $x$ grows more slowly than any exponential $e^{kx}$ with $k > 0$.

Worked examples

Applying the rule twice

Evaluate $\displaystyle\lim_{x\to 0}\frac{e^{x} - 1 - x}{x^{2}}$.

Substituting $0$ gives $(1 - 1 - 0)/0 = 0/0$, so the rule applies. Differentiate top and bottom separately:

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

Check the form again: the numerator is $0$ and the denominator is $0$, so it is still $0/0$. Differentiate once more:

$$ \lim_{x\to 0}\frac{e^{x}}{2} = \frac{1}{2} $$

This time the form is determinate, so stop. The limit is $1/2$. A numerical check at $x = 0.01$: the numerator is $e^{0.01} - 1.01 \approx 0.0000500$ and the denominator is $0.0001$, giving about $0.50$.

An exponential beating a power

Evaluate $\displaystyle\lim_{x\to\infty}\frac{x^{2}}{e^{x}}$.

Both parts grow without bound, so the form is $\infty/\infty$. One application gives $2x/e^{x}$, still $\infty/\infty$. A second gives

$$ \lim_{x\to\infty}\frac{2}{e^{x}} = 0 $$

The numerator’s degree drops by one each time while the denominator is unchanged, so the exponential wins. The same argument with $x^{n}$ takes $n$ steps and always ends at $0$, which is the precise content of the statement that exponentials outgrow powers.

A product rewritten as a quotient

Evaluate $\displaystyle\lim_{x\to 0^{+}} x\ln x$.

The first factor tends to $0$ and the second to $-\infty$, so the form is $0\cdot\infty$ and the rule does not apply as written. Move the algebraic factor to the denominator as its reciprocal:

$$ x\ln x = \frac{\ln x}{1/x} $$

Now the numerator tends to $-\infty$ and the denominator to $+\infty$, so the form is $\infty/\infty$. Differentiating top and bottom:

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

The limit is $0$. The choice of which factor to invert matters: rewriting as $x/(1/\ln x)$ is also of the form $0/0$, but the derivative of $1/\ln x$ is messier and the resulting quotient is worse than the one you started with.

A power with an indeterminate base and exponent

Evaluate $\displaystyle\lim_{x\to\infty}\left(1 + \frac{3}{x}\right)^{x}$.

The base tends to $1$ and the exponent to $\infty$, the form $1^{\infty}$. Set $y = (1 + 3/x)^{x}$ and take logarithms:

$$ \ln y = x\ln\left(1 + \frac{3}{x}\right) $$

This is $\infty\cdot 0$. Substituting $t = 1/x$, so that $t \to 0^{+}$, turns it into a quotient:

$$ \ln y = \frac{\ln(1 + 3t)}{t} $$

which is $0/0$. Differentiating top and bottom with respect to $t$:

$$ \lim_{t\to 0^{+}}\frac{\dfrac{3}{1+3t}}{1} = 3 $$

So $\ln y \to 3$ and therefore $y \to e^{3}$. Do not stop at $3$; the logarithm was introduced by you and has to be undone. Sanity check with $x = 1000$: $(1.003)^{1000} \approx 20.0$, close to $e^{3} \approx 20.09$.

Practice

Start with direct applications. Confirm the form, differentiate numerator and denominator, and repeat until the form is determinate.

Then the end behavior questions. These include limits at infinity you can settle either by the rule or by comparing growth rates, and it is worth doing several both ways.

Quiz

Five items on recognizing indeterminate forms, applying the rule correctly and repeatedly, rewriting products, differences and powers, and comparing growth rates.