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Good morning, I'm reading L’Hospital’s Rule in my textbook:

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and its proof:

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The author introduces two constants $\alpha_0 < \alpha_1$ in the proof. After some reasoning, we have

  • If (i) holds then $$\frac{f(y)}{g(y)} \le \alpha_1, \quad a<y<x_1$$ and thus $$\limsup_{x \to a} \frac{f(x)}{g(x)} \le \alpha_1$$

  • If (ii) holds then $$\frac{f(x)}{g(x)} < \alpha_1 - \alpha_1 \frac{g(y)}{g(x)} + \frac{f(y)}{g(x)}, \quad a<x<x_2$$ and so $$\limsup_{x \to a}\frac{f(x)}{g(x)} \le \limsup_{x \to a} \left( \alpha_1 - \alpha_1 \frac{g(y)}{g(x)} + \frac{f(y)}{g(x)} \right) = \alpha_1$$

In my understanding, we have $$\limsup_{x \to a}\frac{f(x)}{g(x)} \le \alpha_1$$ in ether case.

Because $\alpha_1$ was chosen arbitrarily close to $\alpha$. Then $$\limsup_{x \to a}\frac{f(x)}{g(x)} \le \alpha$$

With this reasoning, I feel that only $\alpha_1$ is enough for the proof.

My question:

Could you please explain which role $\alpha_0$ plays in the proof? I think I miss something, but could not figure out? Thank you so much for your help!

Akira
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  • Here's another post that might be useful to you https://math.stackexchange.com/questions/3271142/clarification-of-lhopital-proof-pugh – Donlans Donlans Oct 27 '19 at 00:00
  • Hi @DonlansDonlans, I've read that question, but I still don't understand how it resolves my question. Could you please more specific? – Akira Oct 27 '19 at 06:17
  • @Akira Hi Asa, I have three theories. (1) $\alpha_0$ is used later in the proof (the part you didn't show). I view this theory as unlikely. (2) The author did not want to use $h_1(x) < h_2(x)$ for all $x \in (a,c)$ implies $\limsup_{x \to a} h_1(x) \le \limsup_{x \to a} h_2(x)$, which is what you used. He/she might have wanted to shown more steps by saying that since $\alpha_1-\alpha_1 \frac{g(y)}{g(x)}+\frac{f(y)}{g(x)}$ converges to $\alpha_1$, it is less than $\alpha_0$ for $x$ close enough to $a$, so that $\frac{f(x)}{g(x)}$ is less than $\alpha_0$ for $x$ close enough to $a$, and – mathworker21 Oct 27 '19 at 07:40
  • therefore $\limsup_{x \to a} \frac{f(x)}{g(x)} \le \alpha_0$. This (more detailed) argument would not have worked with $\alpha_1$ instead of $\alpha_0$. I would give this theory 80% as being true. (3) The author just was being stupid. I give this as 19% (now you know the probability I view theory (1) as being true). In any event, the essence of what you are saying is completely valid :) – mathworker21 Oct 27 '19 at 07:40
  • First, I would like to thank you so much for your keen support @mathworker21 :) I still don't get what you meant by "This (more detailed) argument would not have worked with $\alpha_1$ instead of $\alpha_0$". Please elaborate more on this point! Btw, your helpful comments deserve to be an answer that I will surely accept and upvote :) – Akira Oct 27 '19 at 08:05
  • @Akira no problem. you ask good questions and are clearly smart and trying to learn the material well. Maybe this example will make things clear. Let $x_n = 1+\frac{1}{n}$ for each $n \ge 1$. We wish to show $\limsup_{n \to \infty} x_n \le 1$. You propose: $\limsup_{n \to \infty} x_n \le \limsup_{n \to \infty} (1+\frac{1}{n}) = 1$. The author proposes: take $\epsilon > 0$; since $\lim_{n \to \infty} 1+\frac{1}{n} = 1$, it's true that there is some $N \ge 1$ so that for all $n \ge N$, $1+\frac{1}{n} < 1+\epsilon$; therefore, $x_n \le 1+\epsilon$ for all $n \ge N$; we conclude – mathworker21 Oct 27 '19 at 08:20
  • $\limsup_{n \to \infty} x_n \le 1$. What I meany by "This (more detailed) argument would not have worked with $\alpha_1$ instead of $\alpha_0$" is, in the analogous example above, that we cannot say "there is some $N \ge 1$ so that $x_n \le 1$ for all $n \ge N$". That is all. – mathworker21 Oct 27 '19 at 08:21
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    Thank you so much for your dedicated support @mathworker21 :)))) I'm now clear. – Akira Oct 27 '19 at 08:29

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