Consider the limit $\lim\limits_{x \to -\infty} \frac{1}{x^{m-n}}$.
There are three cases to consider, $m=n$, $m>n$, and $m<n$.
Let's consider the case $m>n$.
$$\forall N<0,\ x < N < 0 \implies \frac{1}{N}<\frac{1}{x}<0$$
If $m-n$ is even, then
$$0<\frac{1}{x^{m-n}}<\frac{1}{N^{m-n}}$$
Therefore, given any $\epsilon>0$, if we choose $N=-\frac{1}{\epsilon^{m-n}}$ then we have
$$x < N < 0 \implies 0<\frac{1}{x^{m-n}}<\epsilon$$
Therefore, we can say that
$$\forall \epsilon>0\ \exists N<0\ \forall x\ x<N<0 \implies 0<\frac{1}{x^{m-n}}<\epsilon\tag{1}$$
My question is, can I assert the following?
$$\forall \epsilon>0\ \exists N<0\ \forall x\ x<N<0 \implies \left | \frac{1}{x^{m-n}} \right | < \epsilon \tag{2}$$
And thus conclude that if $m>n$ then $\lim\limits_{x \to -\infty} \frac{1}{x^{m-n}}=0$? It seems that from $(1)$ to $(2)$ I lost information about $\frac{1}{x^{m-n}}$.
In expression $(2)$ I believe I am saying "for any $\epsilon>0$ there exists an $N$ such that for any $x$ satisfying $x<N<0$ it is also true that $\left | \frac{1}{x^{m-n}} \right | < \epsilon$". But this includes $\frac{1}{x^{m-n}}$ being negative.
I am asking this question because every time I see an expression like $|x|<a$ I conclude that $-a<x<a$, and thus that $x$ can be negative.
For the values of $x$ such that $x<N<0$, it is true that $\left | \frac{1}{x^{m-n}} \right | < \epsilon$, and I guess it is also true that $-\epsilon<\frac{1}{x^{m-n}}<\epsilon$, though this interval is too large and somehow "misleading". Is it fine make a statement such as this?
A more simple case:
Say you know that $a<-5$. Then is it fine to say that $|a|>5$?