KindiakMath

The Overrated L’Hôpital’s Rule

I will double down on the title of this post. L’Hôpital’s rule is useful, but gets too much attention from popular science meme pages. Furthermore, I have not seen a meaningful application of L’Hôpital’s rule other than mathematical flexing (that gets very quickly humbled in real analysis).

Nevertheless, perhaps I can suggest a proof of the simplified version of L’Hôpital’s rule, using Cauchy’s mean value theorem. Here’s the motivation behind this result.

Let f, g be real-valued functions continuous on [a, b] and differentiable on (a, b). Suppose g(a) \neq g(b) and g' \neq 0. Using the mean value theorem, find c_1,c_2 \in (a, b) such that

\displaystyle \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c_1)(b-a)}{g'(c_2)(b-a)} = \frac{f'(c_1)}{g'(c_2)}.

Cauchy’s mean value theorem asserts a stronger claim: we can find c such that c_1 = c_2 = c.

Cauchy’s Mean Value Theorem. Let f, g be real-valued functions continuous on [a, b] and differentiable on (a, b). Suppose g(a) \neq g(b) and g' \neq 0 on (a, b). Then there exists c \in (a, b) such that

\displaystyle \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}.

Observe that g(x) = x yields the vanilla mean value theorem.

Proof. Define h : [a, b] \to \mathbb R by

\displaystyle h(x) := f(x) - f(a) - \frac{f(b)-f(a)}{g(b)-g(a)} \cdot (g(x) - g(a)),

where h is continuous, and h' is differentiable on (a, b). In fact,

\displaystyle h'(x) = f'(x) - \frac{f(b)-f(a)}{g(b)-g(a)} \cdot g'(x).

Since h(a) = h(b) = 0, by Rolle’s theorem, there exists c \in (a, b) such that h'(c) = 0:

\displaystyle 0 = h'(c) = f'(c) - \frac{f(b)-f(a)}{g(b)-g(a)} \cdot g'(c).

Algebra yields the desired result.

Now, we will prove a special case of L’Hôpital’s rule for limits of the form 0/0 as x\to 0.

L’Hôpital’s Rule. Let f,g be differentiable at 0 and f(0) = g(0) = 0. Suppose g' \neq 0 for points near 0 and f'(h)/g'(h) \to L as h \to 0. Then f(h)/g(h) \to L as h \to 0.

Proof. For any sufficiently small h, use Cauchy’s mean value theorem to find c_h between 0 and x such that

\displaystyle \frac{f(h)}{g(h)} = \frac{f(h) - f(0)}{g(h) - g(0)} = \frac{f'(c_h)}{g'(c_h)} = L + o(1),

where the last equation holds since c_h \to 0 as h \to 0.

—Joel Kindiak, 31 Oct, 1531H

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Responses

  1. Pragyan Avatar
    Pragyan

    I genuinely want to know how do you define such functions, such as you defined the auxillary function h(x) has some stuff, then magically Rolle it out with h(a) = h(b) = 0… Any particular motivation how could I’ve come up with such a thing by myself? I’m here from your YT vid btw 😉

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    1. joelkindiak Avatar
      joelkindiak

      wow thanks for checking that out; yea I’ll be honest it was mind-blowing for me too my first time learning it 😛 but I guess its like any new idea; I learn it for the very first time from someone else (e.g. lecturer, book, video, etc) and then use it to possibly “control” novel situations (i.e. turn “weird” setups into nice ones) via practice

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