### Subtract 0 and divide by 1

We all know that division by zero is a prohibited operation, and that ratios that reduce to “zero divided by zero” are indeterminate. We probably also recall proving in elementary calculus class that $\displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$

This is an essential step in deriving an expression for the derivative of ${\sin x}$. The limit of ${f(x)/g(x)}$

The limit of a ratio where both the numerator and denominator tend to zero may often be computed by using l’Hôpital’s Rule. If two functions ${f(x)}$ and ${g(x)}$ both tend to zero as ${x\rightarrow a}$, then the ratio tends to ${0/0}$ and is indeterminate. But l’Hôpital’s Rule gives the result that $\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f^\prime(x)}{g^\prime(x)}$

so we just take the limit of the ratio of the derivatives.

Subtract 0. Divide by 1

How is l’Hôpital’s Rule established? This is simplicity itself: we just subtract 0 and divide by 1. We first note that $\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)}$

We have simply subtracted ${f(a)=g(a)=0}$ from top and bottom.

Now we multiply, or divide, by 1: $\displaystyle \lim_{x\rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)} = \lim_{x\rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)} \times \frac{x-a}{x-a}$

But we can easily rearrange this expression to get $\displaystyle \left. \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \middle/ \frac{g(x)-g(a)}{x-a} \right.$

At this point we make the one non-trivial step, using the result that the limit of a quotient equals the quotient of the limits: $\displaystyle \left. \lim_{x\rightarrow a} \left[ \frac{f(x)-f(a)} {x-a} \right] \middle/ \lim_{x\rightarrow a} \left[ \frac{g(x)-g(a)} {x-a} \right] \right.$

Of course, the limits in the numerator and denominator are both derivatives, leading to the desired result: $\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f^\prime(x)}{g^\prime(x)} \,.$

For simplicity, we assume that ${f^\prime(x)}$ and ${g^\prime(x)}$ do not both vanish at ${x=a}$. Although this rule is usually attributed to L’Hôpital, the theorem was actually discovered by the Swiss mathematician Johann Bernoulli.

Guillaume de l’Hôpital and Johann Bernoulli

The Bernoulli brothers, Jakob and Johann, two brilliant Swiss mathematicians, made outstanding advances on the work of Leibniz on calculus and promoted its appreciation throughout Europe. The Marquis de l’Hôpital was an amateur mathematician, studying the subject as a pastime but lacking the creative genius of the Bernoullis.

L’Hôpital was anxious to gain a reputation as a leading mathematician. He asked Johann to tutor him in the emerging and exciting developments in the field. He was wealthy, while Johann was in need of financial support. In a letter to Johann, which came to light only in 1955, l’Hôpital proposed an agreement: he would pay a substantial sum annually to Johann Bernoulli and, in return, Johann would give him exclusive access to any new discoveries that he made.

Billy Nicks Johnnie’s Idea

Johann agreed to these terms as he needed the money. But he was greatly surprised when l’Hôpital published a book containing several of the his theorems. One notable result was the rule for calculating limits of the form 0/0. Bernoulli had made this discovery and had told l’Hôpital about it in 1694.

Following l’Hôpital’s death in 1704, Johann Bernoulli published his own mathematical discoveries but, since l’Hôpital’s book was already out, the question of priority remained murky. It was not until 1955 that the correspondence between Johann and the Marquis was published. This contained the letter of agreement from Billy to John, dated 1694. It also contained a letter from Johnnie to Bill giving the rule for calculating limits of the form 0/0. This letter predated l’Hôpital’s book by two years.