Silly question: suppose that $f$ has domain $(a,b)$, then does $f$ have a limit as $x$ tends to either $a$ or $b$?

Question

For example, suppose that $f(x) = x^2$ with domain $(-1,1)$. Then we can take the limit as $x \to -1$ or $x \to 1$?

The function itself is not even defined there. That's my biggest concern.

Answer 1

No need for the function to be defined there. Asserting that $\lim_{x\to a}f(x)=l$ means (assuming that $a$ is a limit point of the domain $D_f$ of $f$) that, for each $\varepsilon>0$, there is a $\delta>0$ such that, if $0<\lvert x-a\rvert<\delta$ and $x\in D_f$, then $\bigl\lvert f(x)-l\bigr\rvert<\varepsilon$. As you see, $f$ doesn't need to be defined at $a$.

Answer 2

Yes, you can: you can see this from the formal definition given in José's excellent answer. Or more intuitively, consider this "definition":

We say that the limit $\lim_{x\to a}f(x)=L$ if, as $x$ gets arbitrarily close to but is not equal to $a$, $f(x)$ gets arbitrarily close to $L$.

See, it doesn't really matter that $f(a)$ is not defined in the case where the domain of $f$ is $(a,b)$. All that matters is that we can choose a sequence of $x$s that get arbitrarily close to $a$, even if it does not necessarily ever equal to $a$. Then, we can look at what the value of $f(x)$ tends to, independently of what $f(a)$ is (or even if it is undefined). On the other hand, we can't consider something like $\lim_{x\to(a-1)}f(x)$, because we can't get arbitrarily close to $(a-1)$ while staying in the domain $(a,b)$ of the function (we can at most get to $1$ unit away).

Once you start to learn analysis, you will learn the concept of limit points of a set $S$, which are intuitively the set of points which we can get arbitrarily close to without leaving $S$. In this case, we say that $a$ and $b$ are limit points of the set $(a,b)$. In some sense, a point $p$ is a limit point if for a function $f$ defined over $S$, it makes sense to speak of $\lim_{x\to p}f(x)$.

Answer 3

Consider $f(x)=\sin(1/(x-1))$. Then $\lim_{x\to1}f(x)$ does not exist.