I was watching this video https://www.youtube.com/watch?v=cHvYlrud8j8&ab_channel=AndrewMcCrady and in the proof, he states that when $0 \lt \delta = \min (1, \epsilon, \frac{\epsilon}{|L| + 1}) $,
$(L + \epsilon)(1 - \delta) - \delta \lt \frac{f(\alpha)}{g(\alpha)} \lt (L+\epsilon) + \delta$ implies that $ L - 2 \epsilon \lt \frac{f(\alpha)}{g(\alpha)} \lt L + 2 \epsilon $. What I don't understand is how $ L - 2 \epsilon \lt \frac{f(\alpha)}{g(\alpha)} $, and I've tried proving it myself. Can anyone show me? Thank you.
$\def\e{\epsilon}\def\abajo{\\[0.2cm]}$ By passing terms around, \begin{align*} (L+\e)(1-\delta)-\delta\geq L-2\e&\iff L+\e-\delta(L+\e)-\delta\geq L-2\e \abajo &\iff \delta(L+\e)+\delta\leq 3\e\abajo &\iff \delta(L+1)+\delta(\e-1)+\delta\leq3\e \end{align*} So we need to prove this last inequality. We have $$ \delta(L+1)+\delta(\e-1)+\delta\leq\delta(|L|+1)+\e+\e\leq3\e. $$