Epsilon-Delta definition of a negative infinite limit as x approaches infinity

Question

I've looked in my notes and searched online for an epsilon-delta styled definition of the limit

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

I've only found the definitions in the lines of

$\forall M < 0$ $\exists N > 0 : f(x) < M$ $\forall x > N$

How do I write an epsilon-delta definition of the limit above? I tried changing M to $\epsilon$ but I haven't known of an $\epsilon$ smaller than zero in real analysis.

Answer 1

The key is that the independent variable becomes larger and larger so we need some stuff like $x>M$ and $f(x) $ becomes smaller and smaller so we need something like $f(x) <N$. And then add quantifier : universal for $N$ as it deals with values of function and existential for $M$ as it is related to the independent variable (the one which occurs below limit symbol).

The desired definition of $\lim\limits_{x\to\infty} f(x) =-\infty$ is as follows:

$\forall N\in\mathbb {R}, \exists M\in\mathbb{R}: \forall x\in\mathbb {R} \, x>M\implies f(x) <N$

Notice that there is no need to specify signs of $M, N$. And if someone insists on using only the symbols $\epsilon, \delta$ just replace $M, N$ in above definition by the symbols $\epsilon, \delta$.

Answer 2

$$\forall M > 0, \exists \delta > 0, : f(x) < - M, \forall x > \frac1\delta$$

Remark: I purposely write $\frac1\delta$ because $\delta$ is usually a quantity that is used to indicate a small number.

Personally, I prefer sticking to the notation of $N$ and $M$. If you want, you can let $M=\frac1\epsilon$.

Answer 3

This is an $\epsilon/\delta$ definition. But it must be adapted to deal with infinite argument and infinite limit value. More precisely, a condition like $|x-x_0|<\epsilon$ must be written $x<-M$, meaning that $x$ is "close to minus infinity" instead of close to $x_0$ (with $M$ as large as you want corresponding to $\epsilon$ as small as you want).

You might as well write $x<-\dfrac1{\epsilon}$, but this is uncommon.