A very quick exercise in L'Hopital's rule

2022 October 10

When some ratio $\frac {f(x)}{g(x)}$ has a limit in an “indeterminate form”, in that $\lim f(x) = 0 = \lim g(x)$ , L’Hopital’s rule (first discovered by Bernoulli) allows for calculating the limit:

$\lim_{x \to c} \frac {f(x)}{g(x)} = \lim_{x \to c} \frac {f'(x)}{g'(x)} \qquad \text{ if } \qquad \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$

under certain conditions. And if at first you don’t succeed, try, try again: when $\frac {f'(x)}{g'(x)}$ is also in indeterminate form, you can differentiate as many times as needed.

As an aside, if we suppose that $f$ and $g$ can be approximated as Taylor series near $c = 0$ , one can easily find that

$\lim_{x \to 0} \frac {a_0 + a_1 x + \cdots}{b_0 + b_1 x + \cdots} = \frac {a_n}{b_n} = \frac {n! a_n}{n! b_n} = \lim_{x \to 0} \frac {D_x^n (a_0 + a_1 x + \cdots)}{D_x^n (b_0 + b_1 x + \cdots)}$

when the leading $n$ terms of $f(x) = a_0 + a_1 x + \cdots$ and $g(x) = b_0 + b_1 x + \cdots$ are both zero, representing the case where $n$ differentiations are needed. Here $D_x$ represents the differentiation-with-respect-to- $x$ operator.

While I have occasionally run into cases where multiple applications of L’Hopital’s rule are needed, the other day¹by which I mean 7 years ago… I am a bit slow at updating this blog I was struck by a situation which necessitated a recursive application of L’Hopital’s rule. Consider the Lambert W function which is defined implicitly by the equation

$W(z) \exp (W(z)) = z,$

that is, $W(z)$ is the inverse of the function $xe^x$ . We will only be interested in the principal real branch, which is defined for $- \frac 1e \leq z < \infty$ . Note that $W(- 1 / e) = -1$ . Also, per Wikipedia, we have

$D_z W(z) = \frac {W(z)}{z(1 + W(z))}.$

Ok, now the calculation I ran into in my research was:

$\lim_{z \to 0} D_z[W((z^2 + z - 1) e^{z - 1})]$

Let $F(z) = (z^2 + z - 1) e^{z - 1}$ and $V(z) = W(F(z))$ so our goal is to calculate $\lim DV$ where the limit goes $z \to 0$ .

Note that $V(0) = W(F(0)) = W(-e^{-1}) = -1$ . Starting by applying the chain rule to the derivative of $W$ , we have

$\begin{aligned} \lim DV = \lim D[W(F(z))] &= \lim \frac {W(F(z))}{F(z) (1 + W(F(z)))} D[F(z)] \\ &= \lim \frac {V}{F(z) (1 + V)} ((2z + 1) e^{z - 1} + (z^2 + z - 1) e^{z - 1}) \\ &...$

From what initially, as I was first calculating it, looked like it would blow up endlessly, instead comes a surprisingly simple and clean result!

Thanks Harry Altman for identifying a mistake in an earlier draft.

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