# 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: "$0<|x-(-\infty)|<\delta$"...

• You're going to have to explain what you're talking about a little more clearly. The limit of what as what tends to what? Oct 2, 2013 at 6:17

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 $$\forall\varepsilon\gt0,\ \exists \delta\gt0,\ \forall t,\ t\lt-1/\delta\implies|f(t)-\ell|\lt\varepsilon.$$

• What is the t doing here? Oct 2, 2013 at 6:33
• Tell me.  
– Did
Oct 2, 2013 at 6:34

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:

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

2. Work a bit more explicitly with the order topology, noting that not only open intervals but also open rays must be considered.

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 $L$.