Monotone Convergence Theorem - Bartle Elements of Real Analysis Book I am not sure how the author went from this $x - l \leq x_n \leq x + l$ to this $x - l \leq sup\{x_n\} \leq x + l$, I boulded this part on the theorem for clarity.
Let $X = (x_n)$ be a sequence of real numbers which is monotone increasing in the sense that $ x_1 \leq x_2 ...\leq x_n$
Then the sequence $X$ converges if and only if it is bounded, in which case $\lim(x_n) = \sup \{x_n\}$.
Proof. If  $x=\lim(x_n)$ and $l > 0$, then there exists a natural number $K(l)$ such that if $K(l)\leq$ n then
$\textbf{x - l $\leq$ x_n $\leq$ x + l}$
Since x is monotone, this relation yields
$x - l \leq\sup\{x_n\} \leq x + l$
....

 A: Let $(x_{n})_{n=0}^{\infty}$ be an increasing sequence of real numbers.
Let us also suppose that it is bounded.
Since the set $\{x_{n}\in\mathbb{R}\mid n\in\mathbb{N}\}$ is not empty and bounded, it admits a supremum, which we denote by $s$.
This means that for every $\varepsilon > 0$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ s.t.
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow s - \varepsilon < x_{n_{\varepsilon}} \leq x_{n} \leq s < s + \varepsilon \Rightarrow |x_{n} - s| < \varepsilon
\end{align*}
whence we conclude that $x_{n}\to s$.
Similar reasoning applies if the sequence $(x_{n})_{n=0}^{\infty}$ is decreasing and bounded below.
In such case, the sequence $(x_{n})_{n=0}^{\infty}$ converges to $\inf\{x_{n}\in\mathbb{R}\mid n\in\mathbb{N}\}$.
On the other hand, if the sequence converges to some $L\in\mathbb{R}$, then it is bounded.
Indeed, according to the definition of convergence, let $\varepsilon = 1$.
Then there corresponds $n_{1}\in\mathbb{N}$ s.t.
\begin{align*}
n\geq n_{1} \Rightarrow |x_{n} - L| < 1 \Rightarrow |x_{n}| < |L| + 1
\end{align*}
Then it suffices to take $M = \max\{|x_{0}|,|x_{1}|,\ldots,|x_{n_{1}}|,|L| + 1\}$, and we are done.
Hopefully this helps!
A: Let $U=\sup\{x_n\}$. Since $(x_n)$ is monotone increasing, then for $l>0$, there exists an $N\geq 1$, such that $x_N > U - l$. If this were not the case, then it would imply that $x_N\leq U-l < U$ for all $N\geq 1$, which contradicts that $U$ is the least upper bound on $(x_n)$. Thus, we have that
$$
n\geq N \implies U - l < X_N \leq x_n < U + l \implies \lvert x_n - U \rvert \leq l.
$$
So $x_n\to U = x$ as $n\to \infty$.
