Proof of L'Hopital's rule ($x_0 \not\in D$)

Question

Is this a rigorous proof for Hospital rule? (case $\lim_{x \to x_0} f(x) = g(x) = 0$)

We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.

Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: \lim_{n \to \infty} x_n = x_0$.

$\frac{f(x) - f(x_n)}{g(x) - g(x_n)} = \frac{f'(a_n)}{g'(a_n)}$ $\quad$ Cauchy's mean value theorem (with $a_n \in (x,x_n) \,\,\forall \, n$)

If we let $n \to \infty$ we get

$\frac{f(x) - 0}{g(x) - 0)} = \frac{f(x)}{g(x)} = \frac{f'(a_n)}{g'(a_n)}$ $\quad a_n \in (x,x_0)$

$\lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{f'(x_0)}{g'(x_0)}$ $\quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$

Answer 1

The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:

1. You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.
2. You say that you "let $n\to\infty$", and get $\frac{f(x)}{g(x)}=\frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $n\to\infty$" is not particularly rigorous in itself. Two, if you sent $n$ to $\infty$, then how can you get an equality which still depends on $n$?