Let’s suppose, for the moment, you’re interested in the function $f(x) = \frac{\pi\sin(x)}{x}$. It’s a perfectly respectable function, defined everywhere except for $x = 0$, where the bottom is 0. The top is also zero there (because $\sin(0) = 0$), so its value is, strictly speaking, *indeterminate* - $\frac{0}{0}$ could, in reality, be anything.

However, if you draw the graph of $y = \frac{\pi\sin(x)}{x}$, you’ll see that near $x=0$, the value gets closer and closer to $\pi$, and it’s the same on either side, so saying $\frac{\pi \sin(0)}{0} “=“ \pi$ seems legit. Unfortunately, we can’t write that, because of the whole dividing by 0 problem, but we can say that as $x \rightarrow 0$ (“$x$ approaches zero”), $\frac{\pi \sin(0)}{0} \rightarrow \pi$. Our answer gets as close to $\pi$ as we could possibly want.

One way to show this is to use something called L’Hôpital’s rule, which I’ve always liked because it has two apostrophes and a circumflex. I’ve recently come to appreciate it more, on the ground that L’Hôpital didn’t discover it, Bernoulli did - and L’Hôpital bought the rights.

Here’s what it says:

If $u(x_0) = 0$ and $v(x_0) = 0$ for some value of $x_0$, and if you can differentiate $u(x)$ and $v(x)$ at $x_0$, then the limit of $\frac{u(x)}{v(x)}$ as $x \rightarrow x_0$ is $\frac{u’(x_0)}{v’(x_0)}$.

I’m not going to go into many of the subtleties of that here, but focus on the main result: if you’ve got a function that gives an indeterminate fraction somewhere, you can find the value that makes sense by dividing the derivatives.

Under no circumstances should you confuse L’Hôpital’s rule with the quotient rule. Different things with different uses. Be careful.

So, in this case, we’ve got $u(x) = \pi \sin(x)$, which gives $u’(x) = \pi \cos(x)$. Meanwhile, $v(x) = x$, so $v(x) = 1$.

Divide them to get $\frac{u’(0)}{v’(0)} = \frac{\pi \times 1}{1} = \pi$, which is what we wanted. Hooray.

The pirate proof is very simple, if you know your Taylor series. You know that $u(x_0 + \epsilon) \simeq u(x_0) + \epsilon u’(x_0) + …$, where $\epsilon$ is small and everything after that is *really* small - so small we can ignore it.

Similarly, $v(x_0 + \epsilon) \simeq v(x_0) + \epsilon v’(x_0) + …$, where again $\epsilon$ is small and everything else is too small to worry about.

So that means $f(x_0 + \epsilon) \simeq \frac{ u(x_0) + \epsilon u’(x_0)}{v(x_0) + \epsilon v’(x_0)}$. However, we also know that $u(x_0) = v(x_0) = 0$, because $f(x)$ is indeterminate. So, that’s all the same as $f(x_0 + \epsilon) = \frac{ \epsilon u’(x_0) + \epsilon^2(\text{stuff})}{ \epsilon v’(x_0) + \epsilon^2 (\text{stuff})}$.

We can divide the top and bottom by $\epsilon$ (which, remember, is small):

$f(x_0 + \epsilon) \simeq \frac{ u’(x_0) + \epsilon(\text{stuff})}{ v’(x_0) + \epsilon(\text{stuff})}$.

Lastly, we see what happens when we let $\epsilon$ approach zero: the “stuff” all becomes zero, so

$f(x_0 + \epsilon) \rightarrow \frac{u’(x_0)}{v’(x_0)}$ as $\epsilon \rightarrow 0$ - or, better still:

$f(x) \rightarrow \frac{u’(x_0)}{v’(x_0)}$ as $x \rightarrow x_0$

## Nathan Briggs

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