Proof of L'Hospitals Rule

No matter where I would look it would seem that L'Hospital's Rule has a strange proof-given that they teach it in high school, it seems troublesome that I can't find a solid proof at that level of knowledge. Does anyone have a proof that is fairly basic? Or does proving it simply require higher math?

One can prove it using linear approximation:

Recall the definition of the derivative, $f'(x)$: $$ f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} $$ which we can write as $$ f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h) $$ where $\eta(h)$ is a function such that $\lim_{h\rightarrow 0} \eta(h)=0$ (and is continuous). Then in the above displayed equation multiply through by $h$, which gives (after rearrangment and switching $\eta(h)$ with $-\eta(h)$) $$ f(x+h) = f(x) +f'(x)h+h\cdot \eta(h). $$ Now use L'Hopitals rule for the limit of $f(x)/g(x)$ when $x$ goes to $x_0$, so we have $f(x_0)=g(x_0)=0$. Then by using linear approximation as above in both numerator and denominator (and writing the $\eta$-function for $g$ as $\epsilon$), we get $$ \frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)} $$ and now, in both numerator and denominator, there is a common factor $h$ we can cancel. Taking the limit as $h \rightarrow 0$ now gives the result.

This proof is an upgraded version of Bernoulli's original one. See https://mathoverflow.net/a/51691

A related question is how would you discover L'Hopital's rule. Assuming that $f(x) = g(x) = 0$, one might notice that \begin{align} \frac{f(x + h)}{g(x+h)} &= \frac{\frac{f(x+h) - f(x)}{h}}{\frac{g(x+h)-g(x)}{h}} \\ &\approx \frac{f'(x)}{g'(x)} \end{align} which suggests L'Hopital's rule. This could be turned into a rigorous proof, but we'd need to make some unnecessarily restrictive assumptions.

Here is a version of L'Hopital's rule with a simple proof: Assume $f$ and $g$ are differentiable at $x$ and $g'(x) \neq 0$, and that $f(x) = g(x) = 0$. Then \begin{equation} \lim_{h \to 0} \frac{f(x+h)}{g(x+h)} = \frac{f'(x)}{g'(x)}. \end{equation}

Proving a less restrictive version of L'Hopital's rule requires a less obvious argument.