L'Hôpital's Rule

Theorem

Let $f$ and $g$ be real functions which are continuous on the closed interval $\left[{a .. b}\right]$ and differentiable on the open interval $\left({a .. b}\right)$.

Suppose that $\forall x \in \left({a .. b}\right): g^{\prime} \left({x}\right) \ne 0$.

Suppose that $f \left({a}\right) = g \left({a}\right) = 0$.

Then:

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

provided that the second limit exists.

Corollary 1

Suppose that instead of $f \left({a}\right) = g \left({a}\right) = 0$, we have that $\exists c \in \left({a .. b}\right): f \left({c}\right) = g \left({c}\right) = 0$.

Then:

$\displaystyle \lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$

provided that the second limit exists.

Corollary 2

Suppose that instead of $f \left({a}\right) = g \left({a}\right) = 0$, we have that $f \left({x}\right) \to \infty$ and $g \left({x}\right) \to \infty$ as $x \to a^+$.

Then:

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

provided that the second limit exists.

Proof

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

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

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

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

Note that $\xi$ depends on $x$, i.e. $\xi$ is a function of $x$.

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:

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

Hence the result.

Proof of Corollary 1

This follows directly from the definition of limit.

If $\displaystyle \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$ exists, it follows that:

$\displaystyle \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)} = \lim_{x \to c^+} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$

That is, if there exists such a limit, it is also a limit from the right.

Proof of Corollary 2

We have that $f \left({x}\right) \to \infty$ and $g \left({x}\right) \to \infty$ as $x \to a^+$.

Thus it follows that $\dfrac 1 {f \left({x}\right)} \to 0$ and $\dfrac 1 {g \left({x}\right)} \to 0$ as $x \to a^+$.

The result follows, after some algebra.

$\blacksquare$

Source of Name

This entry was named for Guillaume de l'Hôpital.

However, this result was in fact discovered by Johann Bernoulli.

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

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 11.8 \ (3)$