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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.

  • 1
    how do you define the distance between a point and a set?2012-12-06
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    @Jebruho: I added the definition of the distance. Thanks!2012-12-06
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    @BrianM.Scott: I thought that as $x\to \infty$ the distance between $x$ and the set $S$ goes to $0$, is that correct!?2012-12-06
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    Yes; I when I wrote that I was thinking that the point’s $y$-coordinate was $-\sqrt n$ instead of the reciprocal.2012-12-06
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    What is "the point $x$"? Does it have the coordinates $(x,0)$ and therefore as $x\rightarrow\infty$ the point goes to $(\infty ,0)$? Or does $x=(x_1,x_2)$ and if so then what does $x\rightarrow\infty$ mean?2012-12-06
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    @ToddWilcox: The point $x$ is a point on the real axis, $(x,0)\in \mathbb{R}^{2}$.2012-12-06

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