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