I was given the following exercise as homework: find the limit of $b_{n+1} = \sqrt{2 + b_n}$, $b_1 = \sqrt{2}$, with a hint that $b_n < 2 \forall n \in \mathbb{N}$. I have proven that $b_n$ is strictly monotonically increasing and strictly bounded above by $2$ by using induction over $\mathbb{N}$ (which means that it has a limit, by the monotone convergence principle). However, now I need to prove that
$$2 = \text{sup}\{b_n : b \in \mathbb{N}\}$$
In order to prove that the limit is indeed $2$. By definition, I would only have to prove that $\forall \epsilon > 0 \exists n \in \mathbb{N} : 2 - \epsilon < b_n$, which is intuitively true, but I have been so far unable to prove it.
Any hints will be appreciated.
You have proved that the limit exists, so let's set $$b:=\lim_{n\to\infty}b_n.$$ Given the recurrence relation $b_{n+1}^2=2+b_n$ we can take limits on both sides to find $$\lim_{n\to\infty}b_{n+1}^2=\lim_{n\to\infty}(2+b_n)=2+\lim_{n\to\infty}b_n=2+b.$$ Now on the left-hand side we have $$\lim_{n\to\infty}b_{n+1}^2=\left(\lim_{n\to\infty}b_{n+1}\right)^2=b^2.$$ This gives us the relation $b^2=2+b$. Solving for $b$ shows that either $b=-1$ or $b=2$. As you've shown that the sequence is increasing with $b_1=\sqrt{2}$, we conclude that $$\lim_{n\to\infty}b_n=2.$$