Deciding the range of a rational function with undefined points Consider the function,
$$ f(x) = \frac{\sin x}{x}$$
now, we know that x=0 evaluation is undefined but ,
$$ \lim_{x \to 0} f(x) =  \lim_{x \to 0} \frac{\sin x}{x}$$
Now, the limit evaluates to x=1, so is {1} included in the range of this function or not? as in if we define the function from $ \mathbb{R} \to \mathbb{R}$ , is the $ x=0$ point defined or not. As in, the limit exists but the not the function at the point, therefore is 0 included in domain or not?
 A: Function definition designates everything, not reverse. If you consider $f(x) = \frac{\sin x}{x}$ without any note, then, obviously it is not defined in $0$.
But if you consider:
$$f(x)=\begin{cases}
\frac{\sin x}{x}, & x \ne 0 \\
1, & x=0
\end{cases}
$$
then such function is defined on all $\mathbb{R}$.
Formally it have not sense to ask "where is function defined?", because function is given only, when you have domain, co-domain and correspondence between them i.e. set of pairs. But in mathematics under such question implicitly we understand question "where have sense given formula?"
A: A function only takes on values for points where it is defined, regardless of any limits involved. That is, the range of $f$ consists only of those points $f(x)$ where $x$ is from $f$'s domain and $f(x)$ is defined.
This makes no statement as to limits. Thus, $f(x)=\sin(x)/x$ as presented never contains $f(0)$ in its range nor $x=0$ in its domain (as $f(0)$ is undefined) - again, this is regardless of what the limiting value is. (Said limiting value might not even exist in some cases: consider $g(x)=\sin(1/x)$ at $x=0$.)
Your idea has a noteworthy consequence though. We might not be able to define $\sin(x)/x$ at $x=0$, but we can extend the definition to that point. We can define
$$h(x) = \begin{cases}
\dfrac{\sin(x)}{x} & x \ne 0 \\
1 & x = 0 \end{cases}$$
(The limit as $x \to 0$ of $\sin(x)/x$ is $1$.) What this does ensure is that our new $h$ is a continuous and differentiable function, which makes this $h$ a very "natural" extension of $\sin(x)/x$ for this purpose: by extending to the limiting value, we get that $h$ can take on all of the values in the reals, and we also get the properties of continuity and differentiability there.
However, this function $h$ is not the same as the original function $f$. $f$ has a domain of $\Bbb R \setminus \{0\}$, i.e. any real number except zero; $h$ has a domain of $\Bbb R$, i.e. any real number - at least typically in the context you're using these in. (This means that $h(0)=1$ is in the range of $h$, but $f(0)$ is not in the range of $f$ since it is still not defined there.) For two functions to be equal, we typically define them to be so when:

*

*Their domains are equal

*Their codomains are equal

*The values they take on for every point in the domain are equal

This is not the case here, so you can't use this as a means to say "oh, $f(0)=1$ and thus $1$ is in the range of $f$." $f$ is still undefined at $0$, and so $f(0)$ and $1$ are not in $f$'s range.

In short:

*

*The range of a function is limited to the values that function takes on, and those values have to be defined.


*The limiting behavior of a function does not determine whether a value is in the range: only whether the function takes on that value is the concern.


*Therefore, $1$ is not in the range of $\sin(x)/x$ because at no point does that function ever equal $1$ (even if the limit as $x \to 0$ of it is $1$).


*You can extend that function to one where $1$ is in the range of it, but you are dealing with an entirely different function at that point, and this is thus irrelevant (though it has nice consequences in other considerations like continuity/differentiability).
