Does $\lim_{x\to 0} (2x + 5)$ exist if $f(x) = 2x+5$ is defined only over $[0, \infty)$? In the formal definition of Limit, $\delta$ is essentially an interval centered around $x$; so in this case, since the domain of our function only "starts" at $x=0$, does that mean that even though $\lim_{x\to 0} (2x + 5) = 5$ for a non-restricted domain, for our domain of $[0,\infty)$ there is no limit at $x = 0$ because we can't "approach" $x=0$ from values of $x<0$?
If this isn't true and there is in fact a limit at $x=0$ even with our limited domain, why is that the case? 
 A: It still makes sense. The usual notion of a limit only makes sense when we are dealing with the whole of $\mathbb R$. In this more general setting, we need to generalise the notion of a limit. Define $$B_\delta(x)=\{y\in X:|y-x|<\delta\}$$
where $X =[0,\infty)$ and $x\in X$ to be the open ball of radius $\delta$ around $x$. Observe that in the case where $X = \mathbb R$, this is just the usual open interval, but in your example, $B_\delta(0) = [0,\delta)$.
We say that $f(y)\to \ell$ as $y\to x$ if given $\epsilon>0$, there exists some $\delta$ such that $\forall y \in B_\delta(x)$, $|f(y)-\ell|<\epsilon$.
In particular this definition works in any metric space, which is a set $X$ with a measure of distance on it (called a metric); such spaces are genralisations of $\mathbb R$ with the usual notion of distance.
A: The interval in the standard definition
may be centered,
but if the domain of the argument
does not contain
that interval,
then only the part of
the interval in the domain
can be considered.
In your case,
only $x \ge 0$
can be considered.
If the domain of the argument
was
$[2, \infty)$,
the limit would not exist.
