Supposing that $\lim_{x\downarrow a}f(x) \to \lim_{x\downarrow a}g(x)\to \infty$, and $\lim_{x \downarrow a} \frac{f'(x)}{g'(x)} = L \in \Bbb{R}$, is there a simple way to prove that the limit $\frac{f(x)}{g(x)}$ exists as $x\downarrow a$? (without calculating the limit explicitly)
This is because I want to do an alternate proof of L'Hopital's rule, because the one in my lectures was quite elaborate.
This is the one part of my proof I have left to prove. Currently in my course, I have been taught Cauchy's MVT, and L'Hopital's when $f$ and $g$ both go to zero. I have not learned the Cesaro-Stolz Theorem.
Let $(x_n)$ be a sequence with $x_n > a$ and $x_n \to a$ so that $g(x_n)$ is strictly monotone. We want to show that $$\lim_n \frac{f(x_n)}{g(x_n)}=L$$ From Cauchy's Mean Value Theorem, we have that $$\frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\frac{f'(c_n)}{g'(c_n)}$$ With $c_n$ between $x_n$ and $x_{n+1}$. But $x_n \to a$ and $x_{n+1} \to a$ implies that $c_n \to a$ as well, so taking the limit of both sides we get that $$\exists\lim_n \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\lim_n\frac{f'(c_n)}{g'(c_n)}=\lim_{x\to a+0}\frac{f'(x)}{g'(x)}$$ And Stolz-Cesáro theorem implies that $$\exists \lim_n \frac{f(x_n)}{g(x_n)} = \lim_n \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\lim_{x\to a+0}\frac{f'(x)}{g'(x)}$$