I did search for whether this question was already answered but couldn't find any.
Does a function have to be "continuous" at a point to be "defined" at the point?
For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.
Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.
I ask because at http://en.wikipedia.org/wiki/Continuous_function#Examples there is the text:
the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$.
and the caption of the associated graph reads:
The function is not defined for $x = -2$.
Do I interpret this to mean that a function can not be defined at a point of discontinuity or merely that the function is intentionally not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?
Best Answer
The most common definitions of continuity agree on the fact that a function can be continuous only on points of its domain.
Asking whether $f(x)=1/x$ is continuous is like asking what's the preferred food of unicorns.
You're being misled by the phrase "point of discontinuity". Well, the truth is that a continuous function can many points of discontinuity. It's just an unfortunate terminology that I find being an endless source of misunderstandings. The terminology is due to an old fashioned way of thinking to continuity: it marks a “break” in the graph. However, the concept that a function is continuous if “it can be drawn without lifting the pencil” is a wrong way to think to continuity. The function $$ f(x)= \begin{cases} 0 & \text{if $x=0$,}\\ x\sin(1/x) & \text{if $x\ne0$} \end{cases} $$ is everywhere continuous, but nobody can really think to draw its graph without lifting the pencil. Can you?
The fact that $1/x$ (defined on the real line except $0$) has a point of discontinuity doesn't mean that the function is not continuous somewhere. Indeed it is continuous at each point of its domain.
Prompted by a comment, I'll add that a function can be defined at a point an not be continuous at it. The easiest example is the Dirichlet function $$ D(x)= \begin{cases} 0 & \text{if $x$ is irrational,}\\ 1 & \text{if $x$ is rational} \end{cases} $$ which is continuous nowhere.
So a function can certainly be noncontinuous (I purposely avoid discontinuous) at a point where it is defined.
Returning to the function $f(x)=1/x$, one can specify any subset of the real numbers as its domain, so long as it doesn't contain $0$. When no domain is explicitly specified, it's customary to use the largest subset of the reals where the expression makes sense, in this case it is $\mathbb{R}\setminus\{0\}$.
It's surely possible to define a function $g$ that extends $f$ in $0$; the function $g$ cannot, however, be continuous, because the limit of $g$ at $0$ can't be the value $g(0)$.