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Given a sequence:
$$
begin{cases}
{x_n} = left{1, frac{1}{10}, frac{2}{10},cdots,frac{9}{10}, frac{1}{10^2}, frac{2}{10^2},cdots,frac{99}{10^2}, cdots, frac{1}{10^n}, frac{2}{10^n}cdots frac{10^n-1}{10^n}, cdotsright} \
n in Bbb N
end{cases}
$$
Find the subsequential limits of ${x_n}$. Find $limsup{x_n}$ and $liminf{x_n}$.

For the second part it seems obvious since for the $liminf$ just take a subsequence:
$$
x_{n_k} = frac{1}{10}, frac{1}{10^2}, cdots, frac{1}{10^n}, cdots \
lim_{n_k toinfty} x_{n_k} = lim_{ntoinfty}inf x_n = 0
$$

To find limit supremum one may observer that:
$$
lim_{n_p to infty} x_{n_p} = left{1, frac{9}{10}, frac{99}{100}, cdots, frac{10^n-1}{10^n},cdotsright} =lim_{ntoinfty}sup x_n = 1
$$

But for the first part it's not obvious at a first glance. I believe there is a way to rearrange that sequence to find the set of subsequential limits.

The limit is not affected by cutting off a finite number of the terms from the sequence, and thus the subsequences should not be affected as well. And hence there is infinitely many starting points for the subsequences.

Based on that it feels like that set of subsequential limits will form a set of rational numbers in the range of $[0, 1]$ but I'm not quite sure.

How do I find the set of subsequential limits for $x_n$?

For any $r in [0, 1)$,
let
$r_n = dfrac{lfloor 10^nr rfloor}{10^n}
$.

Then
$10^nr-1 le lfloor 10^nr rfloor
le 10^nr$
so that
$r-10^{-n} le r_n le r$
and $r_n$ is in your sequence.

Therefore $r$ is the limit
of this subsequence.

Also,
$dfrac{10^n-1}{10^n}
to 1$
so $1$ is also a limit.

Therefore $[0, 1]$ is the limit set.

Find the set of partial limits.

First of all: As English is not my native language, I might misunderstand this question.

If I understand the term "partial limit" correctly, it is $limlimits_{ktoinfty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.

If my understanding is correct, the set of partial limits is $[0,1]$:

Using the sub-sequence $1,frac{1}{10}dotsfrac{1}{10^n}dots$ you can show that 0 is a partial limit.

And using the sub-sequence $frac{9}{10},frac{99}{100}dotsfrac{10^n-1}{10^n}dots$ you can show that 1 is a partial limit.

You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:

Example: $frac{sqrt{2}}{2}=0.70710678dots$

Now take the following sub-sequence: $0.7,0.70,0.707,0.7071dots$

... which can be written as: $frac{7}{10},frac{70}{100},frac{707}{1000},frac{7071}{10000},dots$

To prove that this sub-sequence really has the limit $frac{sqrt{2}}{2}$ you have to show that for each value $epsilon>0$ some value $k_0$ exists so that $|frac{sqrt{2}}{2}-x_{n_k}|<epsilon$ for all $k>k_0$.

Choose any value $k_0$ with $frac{1}{10^{k_0}}<epsiloniff k_0>-log_{10}epsilon$.