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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.

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    @ToddWilcox: The point $x$ is a point on the real axis, $(x,0)\in \mathbb{R}^{2}$.2012-12-06

2 Answers 2

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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.

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    Whenever $x \in \mathbb{R}$ the two points in $S$ that are closest are $x$ rounding down or up to the nearest integers, i.e. the floor or the ceiling of $x$.2012-12-06
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(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 $0.5$.

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    Simon S is saying that the limit isn't $0\cdot \infty$ but 'sometimes' $0\cdot \infty$ and sometimes $0$. Kind of like why $\lim_{x \to \infty} x\sin(x)$ doesn't exist.2012-12-06