Here's what I'm trying to prove.
Let $f$ be a function such that:
$$\lim_{x \to x_0} f(x) = L, L \neq 0$$
Then, there exist deleted neighbourhoods of $x_0$ such that $f(x)$ has the same sign as $L$.
Proof Attempt:
Let $\epsilon > 0$. Then, we know that:
$$\exists \delta > 0: 0 < |x-x_0| < \delta \implies |f(x) - L| < \epsilon$$.
So:
$$|f(x) - L| < \epsilon$$
$$L - \epsilon < f(x) < L + \epsilon$$
Let $L > 0$ and $\epsilon = L$. Then:
$$0 < f(x) < 2L$$
We are guaranteed the existence of a deleted neighbourhood for this choice of $\epsilon$. So, that proves that there exists one such that $f(x) > 0$.
Now, let $L < 0$ and $\epsilon = -L$. Then:
$$2L < f(x) < 0$$
We are, once again, guaranteed the existence of a deleted neighbourhood for this choice of $\epsilon$. So, that proves that there does exist one such that $f(x) < 0$.
This proves the desired assertion.
Does the proof above work? If it doesn't, how can I fix it?