0

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.

Zach Hunter
  • 1,828

1 Answers1

1

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 $$\exists \lim_n \frac{f(x_n)}{g(x_n)} = \lim_n \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\lim_{x\to a+0}\frac{f'(x)}{g'(x)}$$

Botond
  • 11,938
  • thank you, however, I am simply trying to show the limit exists, and my course has not taught Stolz-Cesaro – Zach Hunter Mar 04 '20 at 14:03
  • @ZacharyHunter Well, you can prove Stolz-Cesáro before using it. Here is a proof, for example: https://math.stackexchange.com/a/393926/281471 – Botond Mar 04 '20 at 14:09
  • thank you I will look at it – Zach Hunter Mar 04 '20 at 14:10
  • also, why can we restrict ourselves to only considering $(x_n)$ such that $g(x_n)$ is monotonic? – Zach Hunter Mar 04 '20 at 14:13
  • 1
    @ZacharyHunter For the limit of $\frac{f'(x)}{g'(x)}$ to exist, we would need $f'$ to be non-zero around $a$, for example on an interval $(a, b)$ for some $b > a$. Non-zero derivative means strict monotonity on $(a, b)$, so if we pick a decreasing sequence $(x_n)$ from $(a, b)$, we will have that $g(x_n)$ is increasing. – Botond Mar 04 '20 at 16:12