How to find the limit of the sequence $x_n = 0.1\underbrace{00 \ldots0}_{n}1$? Given a sequence, $$(x_n) = (0.101,0.1001,0.10001,\ldots)$$How do I proceed with finding the limit of this sequence? Note that I'm asking for a "method" to solve/find the limit.

The best I can come up with is write $x_n = 0.1\underbrace{00 \ldots0}_{n}1$ and "guess" that the limit should be $0.1$ and prove that
$$\left|0.1\underbrace{00 \ldots0}_{n}1 - 0.1\right|= 10^{-(n+1)} < \epsilon, \quad \forall \epsilon > 0$$
when $n≥m=\lfloor \log_{10}{\epsilon}+1 \rfloor+1$. But that being said, I want to know how to correctly and/or mathematically find the limit without just "guessing".
And how do I find out such limits in general, for instance when the sequence is something like $(0.101001, 0.10010001,0.1000100001, \ldots)$ where my $x_n = 0.1\underbrace{00 \ldots0}_{n}1\underbrace{00 \ldots 0}_{n+1}1$

In general, this is not an easy task to do. There are a few tricks we can use though, especially if we have a closed form representation for the elements.
For example, for the first example, we can write $x_n = 0.1 + 10^{-(n+2)}$. Then:
\begin{align*}
\lim_{n \to \infty} x_n &= \lim_{n\to \infty} (0.1 + 10^{-(n+2)}) \\
&= \lim_{n \to \infty} (0.1) + \lim_{n \to \infty} (10^{-(n+2)}) \\
&= \lim_{n \to \infty} (0.1) + 10^{-2} \lim_{n \to \infty} \left(\left(\frac{1}{10}\right)^n\right)
\end{align*} We know the limit of a constant sequence is just the constant, and the limit of $a^{n}$ as $n \to \infty$ where $|a| < 1$ is $0$, so we can simplify to just $0.1 + 0 = 0.1$.
For the second example you gave, we again look for a nice representation of $x_n$. What I see here is $x_n = 0.1 + 10^{-(n+2)} + 10^{-(2n+4)}$. Again, both of the second terms vanish in the limit, so we are left with just the constant part.
In general, it will be helpful to know some rules about how limits work. For instance, you should try to find out (and prove!) when it is valid to write things like:

*

*$\lim(a_n+b_n) = \lim(a_n) + \lim(b_n)$

*$\lim(a_n b_n) = \lim(a_n)\lim(b_n)$

*$\lim(f(a_n)) = f(\lim(a_n))$
where $a_n, b_n$ are sequences and $f$ is a function.
