# L'Hôpital's Rule

## Contents

## Theorem

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.

Let:

- $\forall x \in \openint a b: \map {g'} x \ne 0$

where $g'$ denotes the derivative of $g$ with respect to $x$.

Let:

- $\map f a = \map g a = 0$

Then:

- $\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$

provided that the second limit exists.

### Corollary 1

Suppose that $\exists c \in \openint a b: \map f c = \map g c = 0$.

Then:

- $\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$

provided that the second limit exists.

### Corollary 2

Suppose that $\map f x \to \infty$ and $\map g x \to \infty$ as $x \to a^+$.

Then:

- $\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$

provided that the second limit exists.

## Proof 1

Let $l = \displaystyle \lim_{x \mathop \to a^+} \frac{f' \left({x}\right)}{g' \left({x}\right)}$.

Let $\epsilon > 0$.

By the definition of limit, we ought to find a $\delta > 0$ such that:

- $\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l}\right\vert < \epsilon$

Fix $\delta$ such that:

- $\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f' \left({x}\right)} {g' \left({x}\right)} - l}\right\vert < \epsilon$

which is possible by the definition of limit.

Let $x$ be such that $\left\vert{x - a}\right\vert < \delta$.

By the Cauchy Mean Value Theorem with $b = x$:

- $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$

Since $f \left({a}\right) = g \left({a}\right) = 0$, we have:

- $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right)} {g \left({x}\right)}$

Now, as $a < \xi < x$, it follows that $\left\vert{\xi - a}\right\vert < \delta$ as well.

Therefore:

- $\left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l }\right\vert = \left\vert{ \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} - l}\right\vert < \epsilon$

which leads us to the desired conclusion that:

- $\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$

$\blacksquare$

## Proof 2

Take the Cauchy Mean Value Theorem with $b = x$:

- $\displaystyle \exists \xi \in \left({a \,.\,.\, x}\right): \frac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \frac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$

Then if $f \left({a}\right) = g \left({a}\right) = 0$ we have:

- $\displaystyle \exists \xi \in \left({a \,.\,.\, x}\right): \frac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \frac {f \left({x}\right)} {g \left({x}\right)}$

Note that $\xi$ depends on $x$; that is, different values of $x$ may require different values of $\xi$ to make the above statement valid.

It follows from Limit of Function in Interval that $\xi \to a$ as $x \to a$.

Also, $\xi \ne a$ when $x > a$.

So from Hypothesis 2 of Limit of Composite Function, it follows that:

- $\displaystyle \lim_{x \mathop \to a^+} \frac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$

Hence the result.

$\blacksquare$

## Examples

### Example: $\dfrac {\sqrt {1 + x} - 1} x$

- $\displaystyle \lim_{x \mathop \to 0} \dfrac {\sqrt {1 + x} - 1} x = \dfrac 1 2$

## Also known as

Because of variants in the rendition of L'Hôpital's name, this proof is often seen written as **L'Hospital's Rule**.

## Source of Name

This entry was named for Guillaume de l'Hôpital.

## Historical Note

While attributed to Guillaume de l'Hôpital, who included it in his $1696$ work *L'Analyse des Infiniment Petits*, published anonymously, this result was in fact discovered by Johann Bernoulli.

After L'Hôpital's death, Bernoulli claimed that most of the content of *L'Analyse des Infiniment Petits*, including L'Hôpital's Rule, was in fact his own work.

However, it was discovered in $1955$, on the publication of correspondence between L'Hôpital and Bernoulli that there had been an agreement between them to allow L'Hôpital to use Bernoulli's discoveries however he wanted.

Hence L'Hôpital was vindicated, and his name continues to be associated with this result.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.20$: The Bernoulli Brothers - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**de L'Hôpital's rule** - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**L'Hôpital's rule (L'Hospital's rule, de L'Hôpital's rule)** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**de L'Hôpital's rule** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**L'Hôpital's rule (L'Hospital's rule, de L'Hôpital's rule)** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**l'Hôpital's rule** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**l'Hôpital, l'Hôpital's Rule**