Epsilon delta for infinite limits

Question

The limit: $$\lim_{x \to \infty} f(x) = -\infty$$ iff $$\big\{ \forall M>0, \exists N >0 \ s.t\ \forall x > N \implies f(x) <-M \big\}$$ This means for all $x \in (N,\infty)$ , $f(x)$ lies in $(-\infty,M)$, however this doesn't account for when $f(x)$ is not defined so it is not $<-M$

Answer 1

As often, this definitions forgets the clause $\forall x\in\text{Dom}(f)$.

Answer 2

The correct definition: let $D \ne \emptyset$ be a subset of $ \mathbb R$ such that $D$ is not bounded from above and let $f:D \to \mathbb R$ a function. Then:

$$\lim_{x \to \infty} f(x) = -\infty$$

$$ \iff$$

for each $M>0$, there is $N \in \mathbb R$ such that $x \in D$ and $x >N$ imply that $f(x)<-M$.

Answer 3

If the limit is $-\infty$, then by definition we are guaranteed $N>0$ such that $$\forall x\in D,\quad x>N \Rightarrow f(x) < -M $$

If you run into trouble with some $f(x)$ being undefined, then you can simply take bigger $N$. If you can't overcome this problem, that simply means the limit is not $-\infty$ as $x\to\infty$.