delta-epsilon definition of a limit

Question

Suppose $\displaystyle \lim_{n\to \infty}x_n=-\infty$. I want to know the form of delta-epsilon definition.

Can we write it as: there exists $n_0\in \Bbb N$ such that $x_n<-n$ whenever $n\ge n_0$. ?

Answer 1

Your proposal is sufficient for the limit to be $-\infty$, but not necessary. For example, $x_n = -\frac{n}{2}$ certainly goes to $-\infty$, but cannot satisfy your property.

The definition is technically not an $\epsilon$-$\delta$ definition. (And for sequences you don’t have $\epsilon$s anyway.)

The definition is:

Definition. Let $(x_n)$ be a sequence of real numbers. We say that $\displaystyle \lim_{n\to\infty}x_n = \infty$ if and only if for every $M\gt 0$ there exists $N\gt 0$ such that $x_n\gt M$ for all $n\geq N$.

Definition. Let $(x_n)$ be a sequence of real numbers. We say that $\displaystyle \lim_{n\to\infty}x_n = -\infty$ if and only if for every $M\gt 0$ there exists $N\gt 0$ such that $x_n\lt -M$ for all $n\geq N$.

Answer 2

Correct definition: Given any positive number $R$ there exist $n_0$ such that $x_n <-R$ whenever $n \geq n_0$.

What you have written is not correct. For example $x_n =-\frac n 2 \to -\infty$ but your condition is not satisfied.