When $\beta>0$, the metric $$g=e^{-\frac{1}{|z|^\beta}}|dz|^2$$ has a conical singularity of angle $(\beta+2)(\beta+1)\pi$ at $z=0$ and non positive Gaussian curvature. Using polar coordinate $(r,\theta)\in (0,+\infty)\times \mathbb{S}^1$ we have $$g=e^{-\frac{1}{r^\beta\,}}\left((dr)^2+r^2(d\theta)^2\right).$$ Using the change of variable $$s=\int_0^r e^{-\frac{1}{2\,t^\beta}}dt=\int_0^r \frac{2t^{\beta+1}}{\beta+1} \frac{d}{dt}e^{-\frac{1}{2\,t^\beta}} dt=\frac{2r}{(\beta+2)(\beta+1)} e^{-\frac{1}{2\,r^\beta}}-\int_0^r \frac{2t^{\beta+2}}{(\beta+2)(\beta+1)}e^{-\frac{1}{2\,t^\beta}} dt$$ we get $$g=(ds)^2+J(s)^2(d\theta)^2$$ where $\lim_{s\to 0+}J(s)/s=\frac{(\beta+2)(\beta+1)}{2}$ hence the result.
The bad singularity at $z=0$ is an artefact of the coordinates that are not suitable to understand the geometry of $g$.
Sorry for my very stupid mistake ! I hope that this other computation is OK ! We have : $$s=\int_0^r e^{-\frac{1}{2\,t^\beta}}dt=\int_0^r \frac{2t^{\beta+1}}{\beta+1} \frac{d}{dt}e^{-\frac{1}{2\,t^\beta}} dt=\frac{2r^{\beta+1}}{\beta+1} e^{-\frac{1}{2\,r^\beta}}-\int_0^r 2t^{\beta}e^{-\frac{1}{2\,t^\beta}} dt.$$ Hence $$s=\frac{2r^{\beta+1}}{\beta+1} e^{-\frac{1}{2\,r^\beta}}\left(1+\mathcal{O}\left(r^\beta\right)\right).$$ Hence we have $$\log(s)=-\frac{1}{2r^{\beta}}+(\beta+1)\log(r)+\log\left(2/(\beta+1)\right)+ \mathcal{O}\left(r^\beta\right),$$ And $$J(s)=re^{-\frac{1}{2\,r^\beta}}=\frac{\beta+1}{2} s r^\beta\left(1+\mathcal{O}\left(r^\beta\right)\right)$$
Hence $$J(s)\simeq_{s\to 0+}\frac{\beta+1}{2} s \frac{1}{2\log(1/s)}.$$