Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

share|improve this question

4 Answers 4

up vote 8 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 $f$).

share|improve this answer

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 $$ \lim_{x\to 0} \frac{x^2 + x}{x} = 1. $$

share|improve this answer

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.

share|improve this answer

$\lim_{x \rightarrow a}$ 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 $a$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.