[The Order Limit Theorem] Assume $\lim a_n = a$ and $\lim b_n = a$. Then,

If $a_n \geq 0$ for all $n \in \mathbf{N}$, then $a \geq 0$;
If $a_n \leq b_n$ for all $n \in \mathbf{N}$, then $a \leq b$;
If there exists $c \in \mathbf{R}$ for which $c \leq b_n$ for all $n \in \mathbf{N}$, then $c \leq b$. Similarly, if $a_n \leq c$ for all $n \in \mathbf{N}$, then $a \leq c$

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Proof (i)

Assume for the sake of contradiction that $a < 0$. Since $\lim a_n = a$, then for $\epsilon > 0$, there exist some $n \geq N$ such that $|a_n - a| < \epsilon$. Choose $\epsilon = |a|$. Then, we must have an $n \geq N$ such that

$$ \begin{align*} |a_n - a| < |a| \\ |a_n - a| < |a| \\ |a_n| - |a| < |a|. \\ |a_n| < 0. \\ \end{align*} $$

But this contradicts the assumption that $a_n \geq 0$ for all $n \in \mathbf{N}$. Therefore, we must have $a \geq 0$. $\blacksquare$

Proof (ii)

Write $a_n \leq b_n$ as $b_n - a_n \geq 0$. Since we have both $\lim a_n = a$ and $\lim b_n = a$, then by the algebraic limit theorem, This limit will converge to $b - a$. By part (i), since $b_n - a_n \geq 0$, then $b-a \geq 0$ and so $a \leq b$ as required. $\blacksquare$

Proof (iii)

TODO

References:

Understanding Analysis by Stephen Abbott