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$$ \lim_{x\to a } [ f(x) + g(x) ] = \lim_{x\to a } f(x) + \lim_{x\to a } g(x) $$

$$ \lim_{x\to a } f(x) \ \text{ and } \ \lim_{x\to a } g(x) \ \text{ exist}. $$

I wanted to know that if $f(x)$ and $g(x)$ are defined over an interval $S$, Does $a$ need to be included in $S$?

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up vote 7 down vote accepted

Certainly, $\lim_{x\to a} f(x)$ can indeed exist if $a \notin S$.

For example, consider the function $$f(x) =\frac{x^2-1}{x-1}, \text{ defined }\,\forall x, x\neq 1:$$ $$\begin{align}\lim_{x\to 1} \frac {x^2 - 1}{x - 1} &= \lim_{x\to 1} \frac{(x-1)(x+1)}{x-1}\\ \\ & = \lim_{x\to 1} x+1 = 2\end{align}$$

Note, the limit as $x$ approaches 1exists, even though the function is undefined at $x = 1$ (i.e. even though $1$ is not in the domain of $f$).

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It seems like you want to know whether it makes sense to talk about $$ \lim_{x \to a} f(x) $$ when $a$ is not in the domain of $f$.

It does make sense. Consider for example the limit $$ \lim_{x\to 0} \frac{x^2 + x}{x} = 1. $$

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I wanted to know that if $f(x)$ and $g(x)$ are defined over an interval $S$, Does $a$ need to be included in $S$?

Not necessarily, to put it simply that's because when we use limits, we want to understand the behavior of a function as it approaches a value, and not what the function evaluates to at that particular value. So it doesn't matter at all that the function be defined over that particular value.

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$\lim_{x \rightarrow a}$ makes sense if and only if $a$ is a limit point of $S$, which is true if and only if $a$ is in the closure of $S$ and is not an isolated point of $S$.

Another way of expressing this is that $a$ is a limit point of $S$ if and only if $S$ contains a sequence of points distinct from $a$ but which converge to $a$.

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