So I was watching this video https://www.youtube.com/watch?v=cHvYlrud8j8&ab_channel=AndrewMcCrady and along the proof it says that $(L-\epsilon)(1-\frac{g(c)}{g(\alpha)}) + \frac{f(c)}{g(\alpha)} \lt \frac{f(\alpha)}{g(\alpha)} \lt (L + \epsilon)(1 - \frac{g(c)}{g(\alpha)}) + \frac{f(c)}{g(\alpha)}$ implies that $(L-\epsilon)(1-\delta) - \delta \lt \frac{f(\alpha)}{g(\alpha)} \lt (L + \epsilon) + \delta$ since $0 \lt \frac{g(c)}{g(\alpha)} \lt \delta$ and $\frac{|f(c)|}{g(\alpha)} \lt \delta$.
I understand that $0 \lt \frac{g(c)}{g(\alpha)} \lt \delta$ means that $0 \lt 1 - \delta \lt 1 - \frac{g(c)}{g(\alpha)} \lt 1$, but shouldn't the implication require that $L - \epsilon \gt 0$ for it to work? In this case, it is assuming that $L \gt 0$ is real and that $\epsilon \gt 0$ is arbitrary. So in some cases $L - \epsilon \lt 0$. Can someone please explain? Thank you.