To understand this, you need to think of the intuition behind the $\epsilon$-$\delta$ definition. We want $\lim_{x\to a}f(x)=L$ if we can make $f(x)$ as close to $L$ as we like by making $x$ sufficiently close to $a$. Worded differently, we might say that:
$\lim_{x\to a}f(x)=L$ if given any neighborhood $U$ of $L$, there is a neighborhood $V$ of $a$ such that elements of $V$ are mapped by $f$ to elements of $U$ (except possibly $a$ itself).
In this context, a "neighborhood" of a point $p$ should be understood to mean "points sufficiently close to $p$". Let's make that precise by defining what we mean by "close". For $\epsilon>0$ (assumed, but not required, to be very small) define
$$B(x,\epsilon):=\{y\,:\,|x-y|<\epsilon\},$$
the ball of radius $\epsilon$ about $x$. For our purposes, we say $U$ is a neighborhood of $x$ if $U=B(x,\epsilon)$ for some $\epsilon>0$. (The usual definition only requires that $U$ contains such a ball.) Assuming $\epsilon>0$ is very small, this agrees with our intuition of what closeness should mean. Now if we go back to our neighborhood "definition" of a limit, you should be able to think about it for a bit and convince yourself that it is equivalent to the usual definition.
How does this relate to the problem with infinity? Given that infinity is not a real number (and things like distance from infinity do not make sense), we must revise what it means to be "close" to infinity. So for $M>0$ (assumed this time to be very large) define
$$B(+\infty,M):=\{y\,:\,y>M\},\quad B(-\infty,M):=\{y\,:\,y<-M\},$$
the neighborhoods of $\pm\infty$. Hopefully you can see why these make sense as definitions; a number should be close to infinity if it is very large (with the correct sign), so a neighborhood of infinity should contain all sufficiently large numbers.
Now we extend our neighborhood definition of limits to include the case where $a$ or $L$ can be $\pm\infty$. It is a similar exercise to before to verify now that the definition is still equivalent to the old one, only now we have in some sense unified somewhat.
To show at least the second problem, assume that $\lim_{x\to 0} f(x) = L$. We will prove that $\lim_{x \to a}f(x-a) = L$.
By definition, we are to show : for all $\epsilon > 0$, there exists $\delta > 0$ such that
\begin{equation}
|y - a |< \delta \implies |f(y-a) - L| < \epsilon \tag{1}
\end{equation}
Fix one such $\epsilon$, say $\epsilon_0$.
We know that $\lim_{x \to 0} f(x) = L$. This means, that for the above $\epsilon_0$, there exists some $\delta_0 > 0$ such that \begin{equation}|x - 0|( = |x|) < \delta_0 \implies |f(x) - L| < \epsilon_0 \tag{2}\end{equation}
By taking $\delta = \delta_0$, we see that substituting $y-a = x$ in $(2)$ gives:
\begin{equation}
|y-a| < \delta_0
\implies |f(y-a) - L |< \epsilon_0 \end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $\lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $\lim_{x \to a} f(x) \neq L$ for some constant $L$.
Best Answer
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 $$ \forall\varepsilon\gt0,\ \exists \delta\gt0,\ \forall t,\ t\lt-1/\delta\implies|f(t)-\ell|\lt\varepsilon. $$