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 $\exists x \in \left({a \, . \, . \, b}\right): g^{\prime} \left({x}\right) \ne 0$.

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.

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