Notion of cusp in algebraic geometry This relates to problem 3.14(b) of Hartshorne, wherein I am required to show that the projection of the twisted cubic curve (viewed as a subset of $\mathbb P^3$) from the point $(0,0,1,0)$ onto the hyperplane $\mathbb P^2$ defined by the equation $z=0$ is a cuspidal cubic curve. I have been able to show that the said projection is the zero set of the homogenous polynomial $f(x,y,z):= y^3-x^2z \in k[x,y,z]$ so it is a cubic curve. From some searching, I learnt that a cuspidal curve is one where all singularities are cusps and by using the characterization of singular points in projective curves defined by a single equation as points where all partial derivatives vanish, I have been able to show that the only singularity of $f$ occurs at the point $(0, 0, 1) \in \mathbb P^2$. Now I am not sure how to show that this point is a cusp, because I require a rigorous definition of "cusp" in this context which is easy to verify.
Most references I came across give a definition to the effect that a "cusp" is a point where two "branches" have a common semi-tangent, and the notion of "branch" is either handwaved or explained only for Euclidean spaces. The only rigorous definition I have found after searching of a cusp $P$ on a variety $Y$ involves completions of $\mathcal O_{P, Y}$. Now I am not familiar with the idea of completion in this context and would really appreciate a definition which is rigorous and can be used in the aforementioned context.
Edit: Here (https://arxiv.org/pdf/1511.02691.pdf) is a reference, where "cusp" is defined on pg. 6, but it uses the notion of a "branch", which has not been defined. I would really like to know a simple rigorous definition of that terminology.
 A: The plane curve $f(x,y)=x^2(x+1)-y^2=0\subset \Bbb{C}^2$ doesn't have a cusp at $(0,0)$ because we have  a parametrization $(x,x\sqrt{1+x}),(x,-x\sqrt{1+x})$ of two distinct "analytic curves" around $(0,0)$.
In contrary $x^2-y^3=0$ has a cusp because $(y\sqrt{y},y)$ is not an analytic curve around $(0,0)$.
Over $\Bbb{C}$ we have the ring of all functions that are analytic in some neighborhood of $x_0=0$, but this doesn't generalize well to arbitrary fields.
We may try to replace it by the ring of formal power series in $x$, which is $k[[x]]=\varprojlim k[x]/(x^n)$,
or in $y$, which is $k[[y]]=\varprojlim k[y]/(y^n)$, but
it appears that the most insightful ring is $\varprojlim k[x,y]/(f(x,y))/(x,y)^n=k[[x,y]]/(f(x,y))$.
If $char(k)\ne 2$ then $\sqrt{1+x}=\sum_{k\ge 0} {1/2\choose k}x^k$ is in this ring and $(y-x\sqrt{1+x}),(y+x\sqrt{1+x})$ are two distinct prime ideals. $k[[x,y]]/(f(x,y))$ is not an integral domain as $(y-x\sqrt{1+x})(y+x\sqrt{1+x})=0$. In contrary to $k[[x,y]]/(x^2-y^3)$ which is an integral domain.
For an arbitrary variety this generalizes to $\varprojlim O_P / \mathfrak{p}^n$ where $O_P$ is the ring of rational functions regular at $P$ (or the localization thereof for a non-closed point) and $\mathfrak{p}$ is the prime ideal of those vanishing at $P$ (its unique maximal ideal).
For a curve, when a parametrization $(x,h(x))$ exists with $h $ a formal power series then in fact $h$ is algebraic over $k[x]$ so that it can be seen as a rational function on a ramified covering of our curve.
