Let, a sequence $a_n$, where Minimum value of $a_n=b$ and Maximum value of $a_n=c$.
If $a_{k+1}<a_k$, Then, it has a limit $L_1=b$
If $a_{k+1}>a_k$, Then, it has a limit $L_2=c$
It is quite obvious to me that the sequence is monotonically decreasing, and if the possible lower value of the sequence is $b$, then it will converge to $b$, equally true for $L_2=c$. However, I cannot find a way to prove it. Any help is highly solicited.
Consider the monotonically increasing case.
$a_{n+1}>a_n$
Exactly what do you mean the max is $c$? There could be many values greater than any $a_n$. For example Let $x_0=1$ and $x_{n+1}=\frac{x_n+2/x_n}{2}$. This is monotonically increasing and less than $\sqrt{2}$.
Since $x_n$ is always less than $\sqrt{2}$, it's less than $2,3, \pi, 100,$ etc.
$x_n$ will exceed any value less than $\sqrt{2}$, but never achieve it. So does it have a maximum?
As mentioned above, it has many upper bounds since it is always less than $2,3,...$
We can say $\sqrt{2}$ is a Least Upper Bound. Every $x_n$ is less than (or equal to) $\sqrt{2}$ and $\sqrt{2}$ is smaller than any other number having this property.
We can prove a monotonically increasing sequence has its l.u.b. as its limit.
Let $\epsilon>0$ be a real number and let l.u.b $=L$.
Suppose $x_n<L-\epsilon<L$ for all $n$. Then $L-\epsilon$ is a smaller upper bound than $L$. This contradicts the assumption that L is the least upper bound.
Since this works for any $\epsilon$, we can find an $N$ so $x_N$ is as close to $L$ as we want, i.e. $\forall \epsilon>0 \exists N $ so that $|x_N-L|< \epsilon$. That's formally the definition of a limit.