It's well known that a sequence of two dimensional Riemannian manifolds with uniform sectional curvature lower bound can Gromov-Hausdorff converge to a cone.
Let $y=|x|$, by rotating around the y-axis, we get a cone $X$. Now I want to construct functions $y=f_i(x)$ such that their rotations around the y-axis (denoted by $M_i$) are smooth Riemannian manifolds and Gromov-Hausdorff converge to the cone $X$. I also want to compute the Gaussian curvature of $M_i$.
I considered $f_i(x)=|x|^{a_i}$ such that $a_i>1$ decrease to 1, but $f_i''(0)=\infty$, so it's not smooth enough, and the curvature at $x=0$ can't be defined. I also considered mollification of $|x|$ by convolution of $\eta(x)=c\exp(\frac{1}{|x|^2-1})$, but we can't write down the explicit form of the integration.
Can any one give good approximations such that $f_i(x)$ can explicit be written down?
Thanks to Thomas Richard's answer, below we compute the Gaussian curvatures. Let $y=\sqrt{t^2+a^2}$, then the length of the curve $$ l(t)=\int_0^t \sqrt{1+[y'(t)]^2}=\int_0^t (2t^2+a^2)^{\frac12}(t^2+a^2)^{-\frac12}dt $$ The metric can be written as $dl^2+t^2 d\theta^2$, by the curvature formula of the warped products, and note that $t''(l)=-\frac{l''(t)}{(l'(t))^3}$, the Gaussian curvature $$ K=-\frac{t''(l)}{t}=\frac{l''(t)}{(l'(t))^3 t}=\frac{2a^2}{(2t^2+a^2)^2}. $$ At $t=0$, $K=\frac{2}{a^2}\to \infty$ as $a\to 0$.
