$x_{n+1}=\sqrt{a+x_{n}}, x_1=\sqrt{a}$ converges

Question

I have to show that the sequence $x_{n+1}=\sqrt{a+x_{n}}, x_1=\sqrt{a}$ has limit etc. This is part b) of a,b,c on the hw question, to show that this sequence is increasing. On part a) I proved $x_n<1+\sqrt{a}$ but it seems very useless. Is my way to show that $x_{n+1}>x_{n}$ Correct? Also $a>0$ $\text{By induction: } x_n<\frac{1+\sqrt{1+4a}}{2} \\ \text{Base case: }x_1=\sqrt{a}<\frac{1}{2}+\sqrt{a}=\frac{1+\sqrt{4a}}{2}<\frac{1+\sqrt{1+4a}}{2}\\ \text{Assume: }\forall n\in \mathbb{N}, x_n<\frac{1+\sqrt{4a}}{2} \\ 4a+4x_n<4a+4(\frac{1+\sqrt{1+4a}}{2}) \Rightarrow 4(a+x_n)<1+2\sqrt{1+4a}+1+4a \Rightarrow (a+x_n)<\frac{(1+\sqrt{1+4a})^2}{4} \Rightarrow \\x_{n+1}=\sqrt{a+x_n}<\frac{1+\sqrt{1+4a}}{2}\\ \text{Thus this inequality is proved. Now notice that:}\\ x_n-\frac{1+\sqrt{1+4a}}{2}<0 \text{ from the above inequality, and }\\ x_n-\frac{1-\sqrt{1+4a}}{2}>0 \text{ is obvious since the 1 inside the sqrt will kill the 1 outside for all a>0.} \\ \therefore 0>(x_n-\frac{1+\sqrt{1+4a}}{2})(x_n-\frac{1-\sqrt{1+4a}}{2})=x_{n}^{2}-x_n-a\Rightarrow x_n+a>x_n^2 \Rightarrow \sqrt{x_n+a}>x_n\Rightarrow x_{n+1}>x_n. \\ $