What is the "limit point of a function?" I am asked to prove that

Show that a sequence $x: \mathbb{N} \to \mathbb{R}$ has a limit point iff there $\lim_{n\to\infty} x(n)$ exists as a limit point of a function from a subset of metric space $(\mathbb{N}*, d)$ to metric space $\mathbb{R}$.  Note $\mathbb{N}* = \mathbb{N} \cup \infty$ and $d(m, n) = \vert \frac{1}{m} - \frac{1}{n} \vert$.

I am confused about what this problem is saying.  How can a function have a limit point?  Aren't those just for sets and sequences?
Thank you!
 A: A function certainly can have a limit as the variable approaches a certain quantity. We have from very early on dealt with such things as $\displaystyle\lim_{x\to 2} \:\:x^3$ and, more interestingly, $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}.$
Showing, for example, that $\displaystyle\lim_{x\to 2} \:\:x^3=8$ involves, essentially, showing that as the distance of $x$ from $2$ approaches $0$, the distance of $x^3$ from $8$ approaches $0$. 
We also have a familiar definition of what it means for the sequence $x(1), x(2), x(3), \dots$ to have a limit $b$ as $n \to \infty$.  Let's try to recast this definition in terms of distance.  Does $\displaystyle\lim_{n\to \infty} x(n)=b$ mean that as the distance of $n$ from "$\infty$" approaches $0$, the distance of $x(n)$ from $b$ approaches $0$?  
That sounds kind of absurd!  As $n$ increases without bound, how can its "distance" from $\infty$ be said to approach $0$?  First of all, there is no such thing as $\infty$. Secondly, even if we "add" an abstract object $\infty$ to the set of integers, the distance of any $n$ from $\infty$ will always be infinite.
However, that's for the ordinary intuitive notion of distance.   Add an "ideal" object $\infty$ to $\mathbb{N}$, to make a new set $\mathbb{N}^\ast$. On this set, define a metric as in your post, except that we need to add that the distance from $\infty$ to any ordinary positive integer $n$ is $\frac{1}{n}$. You can verify that this indeed gives a metric on $\mathbb{N}^\ast$.  Note also informally that as $n$ gets large, its distance from $\infty$ under this metric approaches $0$.  
You are being asked to show that the sequence $x(1), x(2), \dots$ has limit $b$ in the ordinary sense iff the limit of the function $x$, as the distance of $n$ from $\infty$ approaches $0$, is $b$.  In the part after the iff "distance" is in the sense of the metric that has just been defined on $\mathbb{N}^\ast$.
Writing out the solution is essentially a careful translation job.
A: Yes Functions can have limit points in Metric Space. It is the same as functions having a limit in an Euclidean Space. Here it is just abstracted and is applied over a Metric Space.
I think this Link will answer your question. Limit of a Sequence
A: I believe the following picture might help to get the intuition. (At least for people familiar with the topological spaces.)

This picture shows the space $\mathbb N\cup\{\infty\}$ (the sets on the picture form the base of the topology).
Let us define the function $f: \mathbb N\cup\{\infty\} \to \mathbb R$ by $f(n)=x_n$ and $f(\infty)=x$. (The function from the above picture.)
The following conditions are equivalent:


*

*$\lim_{n\to\infty} x_n=x$;

*the function $f$ is continuous;

*$\lim_{n\to\infty} f(n)=x$ holds.
(The same is true, if $\mathbb R$ is replaced by any topological space.)
