I was reading Thomas' calculus which said that $y=1/x$ is a continuous function because it was continuous at every point of its domain(it not being defined at $x=0$), but then goes on to show that $1/x^2$ is discontinuous saying it is an infinity discontinuity as $x$ approaches $0$. Why doesn't the same logic apply to $y=1/x$?
[Math] Why is $y=1/x$ a continuous function but not $y=(1/x)^2$
calculuscontinuity
Related Solutions
This is a real nightmare, for university teachers. Every student of mine comes to my class from high school and is sure that $x \mapsto 1/x$ is discontinuous at $0$. The reason why calculus textbooks are so ambiguous is that their authors do not like to leave something undiscussed. For some reason, the answer to any question should be "yes" or "no"; hence they tend to formally state that functions are discontinuous outside their domain of definition.
In my opinion, this is a very bad approach: it is a matter of fact that discontinuous should not be read as the negative of continuous. The domain of definition makes a difference, and the most useful idea is that of continuous extension.
Almost any mathematician would say that the tangent function is continuous inside its own domain of definition.
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 $g(0)$.
Best Answer
The notion of continuity at $x_0$ only applies to points of the domain.
By definition, a function is said continuous if it is continuous at each point of its domain.
The notion of discontinuity does not apply only to points of the domain but also to limit point of the domain that do not belong to the domain. So a continuous function can have discontinuities, but not in points of the domain.
As an example, the function $$ f(x)=\begin{cases} -1, & x<0 \\ +1, & x\geq0 \end{cases} $$ is not continuous because the limit in $0$ does not exist.
The function $$ f(x)=\begin{cases} -1, & x<0 \\ +1, & x>0 \end{cases} $$ is continuous, because it is continuous in each point of its domain (note that in this case $0$ does not belong to the domain). This function has however a discontinuity in $0$.