Is the continuous extension of $\frac{\sin{x}}{x}$ differentiable at $0$?
Yes; actually, you can just divide through in the Maclaurin series of $\sin$ term by term to get $\frac{\sin(x)}{x}=\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n+1)!}$ for $x \neq 0$ (and the continuous extension agrees with this series everywhere). This reveals all the derivatives at zero, and in particular that the first derivative at zero is zero.