I've seen this definition of gaussian curvature on Wikipedia:
$$ K(p)=\lim\limits_{r \rightarrow 0^+} 12\frac{\pi r^2-A(r)}{\pi r^4} $$
where $A(r)$ is the geodesic disk of radius $r$
it is my understanding that on the unit sphere the geodesic disk is a spherical cap, meaning
$$ A(r)=2\pi(1-\cos(\frac{r}{\pi})) $$
but when i tried this case i got a curvature that diverges ($-\infty$) instead of $1$
NOTE:
as i was writing this question i realized that my problem was (probably): what i was thinking of as the radius of the disk is wrong, however if this is indeed the case i am not sure how to arrive at the right answer.
You should write $$A(r)=2 \pi\left(1-\cos \left(r\right)\right)$$ Remember that the geodesic distance between the poles is $\pi$ and the surface area of the sphere of radius one is $A(\pi)=4\pi$. \begin{align} K(p)&=\lim _{r \rightarrow 0^{+}} 12 \frac{\pi r^{2}-A(r)}{\pi r^{4}}\\ &=\lim _{r \rightarrow 0^{+}} 12 \frac{ r^{2}-2 \left(1-\cos \left(r\right)\right)}{r^{4}}\\ &=\lim _{r \rightarrow 0^{+}} 6 \frac{ r- \sin \left(r\right)}{r^{3}}\\ &=\lim _{r \rightarrow 0^{+}} 2 \frac{ 1- \cos \left(r\right)}{r^{2}}\\ &=\lim _{r \rightarrow 0^{+}} \frac{ \sin \left(r\right)}{r}=1 \end{align} The equality's are derived using L'Hôpital's rule.