How do epsilon-delta proofs work for limits at negative infinity? How do I write a proof for a limit where $x$ tends to negative infinity, that is
$$\lim_{x\to-\infty} f(x) = L$$
 using the epsilon-delta definition of limits? 
Simply putting $-\infty$ in the definition does not seem to make sense: "$0<|x-(-\infty)|<\delta$"...
 A: The limit of a function $f$ at $-\infty$ is $\ell$ if and only if
$$
\forall\varepsilon\gt0,\ \exists x,\ \forall t,\ t\lt x\implies|f(t)-\ell|\lt\varepsilon.
$$
This can artificially be rewritten in the epsilon-delta frame as
$$
\forall\varepsilon\gt0,\ \exists \delta\gt0,\ \forall t,\ t\lt-1/\delta\implies|f(t)-\ell|\lt\varepsilon.
$$
A: The challenge here is that while the extended real line, $\Bbb R\cup\{-\infty,\infty\}$ with the order topology is metrizable (it's homeomorphic to a closed interval), it doesn't have any metric compatible with the usual one. Thus the metric space interpretation of "$\epsilon$-$\delta$" doesn't really work too well in this context. Furthermore, the extended real line is not a group under real addition, so it's pretty hard to imagine any more literal sort of $\epsilon$-$\delta$ variant working. So really, you have two good options:


*

*Use special definitions of limits approaching $\pm\infty$ that look different from others, or

*Work a bit more explicitly with the order topology, noting that not only open intervals but also open rays must be considered.
A: There are other equivalent definitions, but my preference is as follows:
\begin{align*}
\lim_{x \to - \infty} f(x) & = L
\end{align*}
if and only if
\begin{align*}
(\forall \epsilon > 0) & (\exists N \in \mathbb{N}) & (x > N \Rightarrow |f( - x) - L| < \epsilon)
\end{align*}
That is, for any $\epsilon > 0$, I can choose $N$ such that if $x$ is greater than $N$, then $f(- x)$ is within $\epsilon$ of $L$.
