Q:

Question about behaviour of $f(x)=\sin(x)/x$ at $x=0$

It's possible to show that the limit of $\sin(h)/h$ as $h$ tends to $0$ is $1$, famously.
Now, someone said to me the other day, "That means that $\sin(0)/0$ is 1", and I strongly disagreed. I said to them that a limit $\lim_{x\rightarrow a}f(x)$ equals $f(a)$ only if $f(x)$ is continuous at $a$ and we hadn't established that yet.
However, to establish whether $\sin(x)/x$ is continuous at $x=0$, one needs consider a sequence $f(a_0), f(a_1), f(a_2),...$, where $a_i$ is a sequence that tends to $0$, and see whether the limit of the sequence is the same as $f(0)$ - but that's exactly our problem! We need to know the function is continuous at $x=0$ to evaluate it there using our limit, but to know whether it is continuous at $x=0$, we need to know its evaluation there!
I apologise if my reasoning is highly fallacious, as I am only an amateur enthusiast, but what, if any, is the way out of this circular reasoning? Is our function continuous at $x=0$?

A:

It only makes sense to talk about continuity for points $x$ in the domain of  the function, so asking if $\sin x/x$ is continuous at $0$ is not well-posed.  However, we can define a function
$$
f(x)=\begin{cases}
\sin x/x&x \neq  0\\
1&x=0
\end{cases}
$$
which is continuous everywhere.

A:

First of all, I presume there is a typo in line 3 which should read "$\lim_{x\rightarrow a}f(x)$ equal to $f(a)$", not "$a$." Quite simply, $f(0)$ is undefined, and so this is what we call a removable discontinuity. Indeed, $\lim_{x\rightarrow 0}(sinx)/x$ does exist and, by L'Hopitals Rule is equal to $\lim_{x\rightarrow 0}cosx=1$. However, as $\lim_{x\rightarrow 0}f(x)$ exists but has a value different from $f(0)$ (in this case undefined), we have a removable discontinuity.
If you want a very interesting problem: Let $f$ be an arbitrary function from the $R$ to $R$ (reals). Consider $D_f$, the set of points of points at which f is discontinuous. Show that $D_f$ is a F-sigma set - that is, $D_f$ can be written as the countable union of closed sets (hint: first consider monotone functions, then generalise from there).
Speak spoon g
NOTE: functional limits require $a$ to be a limit point of the domain (as 0 is in this case), however, v. important subtle difference is that continuity requires that $a$ actually be contained in the domain, and of course in this case, 0 is not as $sin0/0$ is undefined.

