Suppose you had the function $$ p(x) = \frac{\sin(x)}{x} $$ I know, from other material online, that this point is analytic at the point $x=0$. However, my understanding was that a point of a function is analytic if it's differentiable at that point. The derivative of this function is $$ p'(x) = \frac{x \; \cos(x) - \sin(x)}{x^{2}} $$ Which is clearly undefined at the point $x=0$.
So why have I seen people saying that $x=0$ is an analytic point for this function?
As given the function $p$ is clearly not defined at $x=0$; and a fortiori it has no derivative there. But it is easily verified that $$p(x)=\int_0^1\cos(tx)\>dt=:q(x)\qquad\forall\>x\ne0\ ,$$ whereby the function $q$ is defined on all of ${\mathbb R}$, and is easily seen to be $C^\infty$ on ${\mathbb R}$. It is then only natural to define $p(0):=q(0)=1$, which then makes $p$ defined and smooth on all of ${\mathbb R}$. This function is called ${\rm sinc}$, and plays an important rôle in signal processing.