Why must we ensure that $L - \epsilon > 0$ in this proof of the Limit-Comparison Test?

I'm studying a proof of the Limit-Comparison Test from Youtube. I've taken screenshots of the relevant slides and posted them as reference below (the red annotations on the second slide are mine).

My question is regarding one of the comments/lines during the "Rough Work" slide (second image below).

The instructor says that we need to ensure $$L-\epsilon > 0$$, and proceeds to pick the value for $$\epsilon$$ to fulfill that constraint. This is the only part of the proof I don't understand.

Why do we need to ensure that $$L - \epsilon > 0$$? I've re-watched the video multiple times, and I still don't get it.

Update (refined question):

Thanks to @Surb's comment it's finally starting to make sense.

The Basic Comparison Test (BCT) that we apply in the last step of $$(1)$$ and $$(2)$$ on the last slide requires an $$a \in \mathbb{R}$$, that $$f, g$$ be continuous functions on $$[a, \infty)$$, and for every $$x \geq a \in$$, that $$0 \leq f(x) \leq g(x)$$.

If $$L - \epsilon < 0$$, then $$(L - \epsilon) \cdot g(x) < 0 < f(x)$$. So we cannot apply the BCT, since the chain inequality isn't in the required order.

But is there any problem with letting $$L - \epsilon = 0$$? In that case we have $$0 \leq (L - \epsilon) \cdot g(x) \leq f(x)$$, which is in the required order. We apply the BCT in that case, no? Or is it a problem if $$f(x)$$ happens to be divergent, because $$g(x)$$ is always convergent (since the limit of $$0$$ is always $$0$$)?

• The very last line, if $(L-\varepsilon )g(x)<0<f(x)$ then $\int f$ converges won't implies $\int g$ converges. That's why you need $(L-\varepsilon )g=g\frac{L}{2}>0$. – Surb Aug 13 at 20:22