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: "$0<|x-(-\infty)|<\delta$"...
On
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.
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 $$ \forall\varepsilon\gt0,\ \exists \delta\gt0,\ \forall t,\ t\lt-1/\delta\implies|f(t)-\ell|\lt\varepsilon. $$