Limit at infinity I really want help in this problem: given a sequence of pairs $(x,y)$ in the $xy$-plane
$$S=\left\{\left(n, \frac{-1}{\sqrt{n}}\right)\right\}_{n=1}^{\infty}\;,$$
 how to find 
$$\lim_{x\to \infty} \frac{1}{\sqrt{1+|x|}} \, \frac{1}{\big(\operatorname{dist}(x,S)\big)^{2}}$$
where ''$\operatorname{dist}(x,S)$'' means the distance between the point $x$ and the set $S$, defined by $\operatorname{dist}(x,S)=\inf\limits_{a_{n}\in S}\operatorname{dist} (x,a_{n})$. And as it is known, the distance between any two points $P=(x_{1},y_{1})$, and $Q=(x_{2},y_{2})$ is  $\operatorname{dist}(P,Q)=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$.
I know that the limit of the first term is zero and the limit of the second term is $\infty$, but this does't help!! Any idea? 
EDIT: it was just a typo, a square should be on the distance.
 A: Set $a_x=\frac{1}{\sqrt{1+x}}\frac{1}{(\mathrm{dist}(x,S))^2}.$
We have $\mathrm{dist}(x,S)^2=\min\{(x-\lfloor x \rfloor)^2+(\lfloor x \rfloor)^{-1},(x-\lceil x \rceil)^2+(\lceil x \rceil)^{-1}\}$. In particular, if $n \in \mathbb{N}$ we have $\mathrm{dist}(n,S)^2=\frac{1}{n}$. But then $a_n=\frac{n}{\sqrt{1+n}}$ and $a_n \to \infty$. On the other hand, if $x_n=n+\frac{1}{2}$, we find $(\mathrm{dist}(x_n,S))^2=\frac{9}{4}$ and hence $a_{x_n} = \frac{4}{9\sqrt{1+x_n}}$. But then $\lim_{n\to \infty} a_{x_n}=0$. So the limit  $\lim_{x\rightarrow\infty}a_x$ does not exist.
A: (Assuming $\operatorname{dist}(x, S)$ means $\operatorname{dist}((x, 0), S)$)
The limit does not exist because $$\lim_{x\to \infty} \frac{1}{\big(\operatorname{dist}(x,S)\big)^{2}}$$
does not exist. For high $x$ we can approximate $$\operatorname{dist}(x,S) = \operatorname{dist}(x,S')$$
with $S' = \lbrace (n, 0) \rbrace_{n=1}^\infty$, and it is
$$\operatorname{dist}(x,S') = |\operatorname{frac}(x+0.5) - 0.5|$$ 
This function oscillates between $0$ and $0.5$.
