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.
A: To some extent, this someone was right (though I bet many people here will disagree).
Because writing $\dfrac{\sin 0}0$ can be taken informally for an indeterminate expression where the limit of $\dfrac{\sin x}x$ is implied (otherwise one would have written $\dfrac{\sin 0}{0^2}$ or $\dfrac{\log(\sin0+1)}0$ or...), and because the natural choice for the value of the function when it is indeterminate is the limit.
So even if the statement lacks rigor, it can be understood as "the value of $\dfrac{\sin(0)}0$ should be $1$", and that makes the function continuous.
By the way the "official" cardinal sine function is so defined.
A: Since $f$ is undefined at 0, it is not continuous at 0.
