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.