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L'Hospital's rule: let $f$, and $g$ be real functions differentiable on $(a,b)$ except possibly at a point $x_0\in (a,b)$. If $\lim_{x\rightarrow x_0}f(x)=\lim_{x\rightarrow x_0}g(x)=0$, if $g'(x)\neq 0$ for all $x\in (a,b)$ (except possibly at $x_0$), and if $\lim_{x\rightarrow x_0}f'(x)/g'(x)$ exists, then $$\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow x_0}\frac{f'(x)}{g'(x)}$$

The proof (second answer in the link) uses linear approximations. Letting $$\eta (h)= \frac{f(x+h)-f(x)}{h}-f'(x)$$ we have that $\lim_{h\rightarrow 0}\eta(h)=0$, and one can write $f(x+h)=f(x)+f'(x)+\eta(h)$. Defining $\zeta(h)$ similarly for $g$, we have $$\frac{f(x+h)}{g(x+h)}=\frac{f(x)+f'(x)+\eta(h)}{g(x)+g'(x)+\zeta(h)}$$

Taking the limit as $x\rightarrow x_0$ gives $$\frac{f(x_0+h)}{g(x_0+h)}=\frac{hf'(x_0)+\eta(h)}{hg'(x_0)+\zeta(h)}$$

and the result follows by taking the limit as $h\rightarrow 0$.

There are two obvious problems with the proof.

  • First of all $\eta$ and $\zeta$ are functions of two variables ($x$ and $h$), maybe this could be resolved by letting $\eta(h)=\frac{f(x_0+h)}{h}-f'(x_0)$, but this lead us to the second problem:
  • How do we know any of the quotients are well-defined? It might be the case that $g(x_0+h)=0$, for example.

Can these two issues be resolved as to make the proof rigorous?

Arctic Char
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Sam
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  • The proof is already pretty rigorous. We consider $\eta,\zeta$ for fixed $x$, so there's no issue here. Your idea to resolve the "issue" is what's done anyway, essentially. Wether we use $x_0$ and $x$ or $x$ and $x+h$ doesn't matter in the slightest. It's just a change of variables. And the other issue is resolved by the fact that it is assumed as a premise that there is a neighborhood of $x$ in which the quotient is defined. – Vercassivelaunos Nov 08 '20 at 13:16
  • Which premise assumes that there is a neighborhood of $x_0$ for which the quotient $f(x_0+h)/g(x_0+h)$ is defined? – Sam Nov 08 '20 at 13:24
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    One you haven't listed, but which is required to even make sense of the claim. The quotient needs to be defined in order to take its limit. – Vercassivelaunos Nov 08 '20 at 13:27
  • Wait, how would one deal with the fact that $f(x_0)$ need not equal $0$ and $f'(x_0)$ not being defined? – Sam Nov 10 '20 at 00:11
  • It doesn't matter what the values of $f(x_0)$ and $f'(x_0)$ are. Only the limits as $x\to x_0$ matter. – Vercassivelaunos Nov 10 '20 at 05:32

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