Consider the set $M=\{ (1,x,x^2):x \in \mathbb{R}\}$.
We define $\text{cone}(M)$ as the conic hull of $M$, which admits conic combinations of its points, i.e. $\sum_i \lambda_i(1,x_i,x_i^2)\in M$ where $\lambda_i>=0\ \ \forall\ i$.
We can apparently see that $(0,0,1)\in\text{cl(cone(}M))$, the closure of the conic hull of $M$. The proof is simple:
$$ \begin{align} (0,0,1)&= \lim_{x\rightarrow \infty}\Big(\dfrac{1}{x^2},\dfrac{1}{x},1\Big)\\ &=\lim_{x\rightarrow \infty}\dfrac{1}{x^2}(1,x,x^2) \end{align}. $$
But I don't understand it. Can someone explain why the limit of $\dfrac{1}{x^2}\times(\text{an element of M})$ gives an element in the closure of $\text{cone}(M)$?
Let x $\in$ $\mathbb{R}$ and $\lambda$ = $\frac{1}{x^2}$. Then p = $\lambda(1,x,x^2) = (\frac{1}{x^2},\frac{1}{x},1)$. Note that p $\in$ cone(M) $\forall$ x $\in \mathbb{R}$. Now take the limit as x goes to 0. This is a sequence of point in cone(M); thus the limit point is in cl(cone(M)).