# 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.

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how do you define the distance between a point and a set? – Jebruho Dec 6 '12 at 1:39
@Jebruho: I added the definition of the distance. Thanks! – Jenn Dec 6 '12 at 1:47
@BrianM.Scott: I thought that as $x\to \infty$ the distance between $x$ and the set $S$ goes to $0$, is that correct!? – Jenn Dec 6 '12 at 1:53
Yes; I when I wrote that I was thinking that the point’s $y$-coordinate was $-\sqrt n$ instead of the reciprocal. – Brian M. Scott Dec 6 '12 at 1:58
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? – Todd Wilcox Dec 6 '12 at 2:36

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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How did you find the formula for $\operatorname{dist}(x,S)^{2}$? – Jenn Dec 6 '12 at 14:59
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 $x$. – Jan Keersmaekers Dec 6 '12 at 19:17

(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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Since the limit is $0.\infty$ that doesn't mean that the limit must be doesn't exist! – Jenn Dec 6 '12 at 12:46
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. – Jan Keersmaekers Dec 6 '12 at 13:22