So I've stumbled upon a proof of L'Hospital rule in my real analysis textbox that I can't seem to understand. Here goes:

$$ \text{Let } f,g:[a, b) \to \mathbb{R} \text{ be functions that satisfy the following conditions: }$$ $$\text{1. } f \text{ and } g \text{ are continuous on the interval } [a, b) $$ $$\text{2. } f \text{ and } g \text{ are differentiable on the interval } (a, b) \text{ where } g'(x) \neq 0 \quad (\forall x\in(a, b)) $$ $$\text{3. } f(a) = g(a) = 0 $$

$$ \text{If } \lim_{x \to a^+}{\frac{f'(x)}{g'(x)}} \text{ exists, then } \lim_{x \to a^+}{\frac{f(x)}{g(x)}} \text{ exists and } \lim_{x \to a^+}{\frac{f'(x)}{g'(x)}} = \lim_{x \to a^+}{\frac{f(x)}{g(x)}} $$


$$ \text{Let } x \in (a, b) \text{ be arbitrary.} $$ $$ \text{Then the functions } f \text{ and } g \text{ satisfy the conditions of Cauchy's theorem on the interval } [a, x] \\ \text{Therefore } \exists \delta = \delta(x) \in (a, x) \text{ such that: }\\ \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(\delta(x))}{g'(\delta(x))} $$

$$ \text{Since } f(a) = g(a) = 0 \text{ we have: } \\ \frac{f(x)}{g(x)} = \frac{f'(\delta(x))}{f'(\delta(x))}$$

$$ \text{Let } \lim_{x \to a^+}{\frac{f'(x)}{g'(x)}} = L \text{. Since } a < \delta(x) < x \text{ , by the squeeze theorem we have } \lim_{x \to a^+}{\delta(x)} = a \\ \text{Since } \delta(x) > a (\forall x \in (a, b)) \text{ we know that } \lim_{x \to a^+}{\frac{f'(\delta(x))}{g'(\delta(x))}} \text{ exists and let: } \\ \lim_{x \to a^+}{\frac{f'(\delta(x))}{g'(\delta(x))}} = L $$

$$ \text{Now, since } \frac{f(x)}{g(x)} = \frac{f'(\delta(x))}{g'(\delta(x))} \text{ , we get } \lim_{x \to a^+}{\frac{f(x)}{g(x)}} = \lim_{x \to a^+}{\frac{f'(\delta(x))}{g'(\delta(x))}} = L = \lim_{x \to a^+}{\frac{f'(x)}{g'(x)}} $$

Now, if I were to pretend to understand how they applied Cauchy's theorem then I would understand the entire proof. What I'm finding hard to understand is the whole notation $$ \delta = \delta(x)$$ What does that mean? Is delta a constant? If so then how can I later evaluate its limit (via squeeze theorem) to be equal to a. If it's not a constant, how come it came from Cauchy's theorem which states that there exists a constant in a given interval for which those rules apply.

I'm so confused about this that I'm not even sure how to properly ask this question.

  • $\begingroup$ youtu.be/kfF40MiS7zA $\endgroup$
    – user29418
    Sep 15 '18 at 9:21
  • 2
    $\begingroup$ $\delta$ is not a constant but rather depends on interval $[a, x] $ and hence depends on $x $. Cauchy's theorem is being applied not on a fixed interval but rather on a varying interval $[a,x]$. $\endgroup$ Sep 15 '18 at 9:23
  • $\begingroup$ It's l'Hôpital's rule, even though l'Hôpital means Hospital in French, the rule was named after the French mathematician Guillaume de l'Hôpital. $\endgroup$ Sep 15 '18 at 9:36
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    $\begingroup$ @Mathematician42: Actually, l'Hôpital is the modern French spelling of what used to be spelled l'Hospital when the guy himself lived. See here, for example. $\endgroup$ Sep 15 '18 at 10:14
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    $\begingroup$ If the $\delta$ from Cauchy's theorem were a constant, it would be a nice project to compute it once and for all. Alas, it cannot be a global constant, but by necessity depends on the function(s) considered and the interval end points. Within this proof of l'Hôpital, $f$ ad $g$ amd $a$ are fixed - but $x$ varies and hence so does $\delta$, which makes it a function of $x$. And this dependency can be expressed notationally by writing it explicitly as a function of $x$, i.e., $\delta(x)$. $\endgroup$ Sep 15 '18 at 10:36

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