Real Analysis proof, show there is a limit and find it. Assume$${ x }_{ 1 }=\sqrt { 3 } ,{ x }_{ 2 }=\sqrt { 3+\sqrt { 3 }  } ,..,{ x }_{ n }=\sqrt { 3+\sqrt { 3+..+\sqrt { 3 }  }  } $$this sequence has a limit and find the limit.
I know I need to show that is is bounded and then show $x_{ n }<{ x }_{ n+1 }$ before finding the limit but have no idea how to go about it. Please help!

You have
$$
x_{n+1} = \sqrt{3+x_n}
$$
so that
$$
x_{n+1}^2 = 3+x_n.
$$


*

*Assuming the sequence converges to a limit $\ell\in\mathbb{R}$, then $\ell$ must satisfy (by continuity)
$$
\ell^2 = 3+\ell
$$
hence the only possible values for $\ell$ are therefore $\frac{1\pm\sqrt{13}}{2}$. Since the limit of a positive sequence has to be non-negative, if the sequence has a limit then this limit is $\ell=\frac{1+\sqrt{13}}{2}$.

*Now, even if this does not seem to be requested by the question, let us also prove there is such a limit (that is, that the sequence converges).


*

*We can easily see that $x_n \geq x_{n-1} \geq \sqrt{3}$ for all $n\geq 2$. (e.g., by induction, from the recurrence relation: this is true for $x_2 \geq \sqrt{3} = x_1$, and then if $x_n\geq x_{n-1}$ then $3+x_n \geq 3+x_{n-1}$ by induction hypothesis)).

*We can show that it is bounded above by, say, $3$. Again, by induction: if $x_n \leq 3$, then $x_{n+1} = \sqrt{3+x_n} \leq \sqrt{3+3} < 3$.

*A bounded non-decreasing sequence converges.
