Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
2  
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

2 Answers 2

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.

share|improve this answer
    
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 $0.5$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.