Solution: We recall that a monotonic bounded sequence is convergent. This problem is an application of the Monotone Convergence Theorem. The steps are provided in the problem statement.
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First of all note that $a_n \geq 0$ for all $n\in \mathbb{N} $. We prove by induction that \( a_{n+1} \geq a_n \) for all \( n \geq 1 \). For \( n = 1 \),
\[
a_2 = \sqrt{2 + a_1} = \sqrt{2 + 1} = \sqrt{3} \geq 1 = a_1.
\]
Assume \( a_{k+1} \geq a_k \) for some \( k \geq 1 \). Now, we show that \( a_{k+2} \geq a_{k+1} \). Note that,
\[
a_{k+2} = \sqrt{2 + a_{k+1}} \geq \sqrt{2 + a_k} = a_{k+1}.
\]
Thus, \( \{a_n\} \) is monotonically increasing.
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For boundedness, we will again use the method of induction to show that \( a_n \leq 2 \) for all \( n \geq 1 \). For \( n = 1 \), \( a_1 = 1 \leq 2 \). Assume \( a_k \leq 2 \) for some \( k \geq 1 \). Then
\[
a_{k+1} = \sqrt{2 + a_k} \leq \sqrt{2 + 2} = \sqrt{4} = 2.
\]
Thus, \( \{a_n\} \) is bounded above by $2$.
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Since \( \{a_n\} \) is monotonically increasing and bounded above, it converges to a limit \( l \). Since $a_n \to l$ we have $a_{n+1} \to l$. Therefore,
\begin{align*}
& \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{2 + a_n} \\
\implies & l = \sqrt{2 + l} \\
\implies & l^2 = 2 + l \\
\implies & l^2 - l - 2 = 0 \\
\implies & l = 2, -1.
\end{align*}
Since, all the terms of the sequence is nonnegative, the limit must be nonnegative. Hence, \( \lim_{n \to \infty} a_n = 2 \).