Recursive sequence, $x_{1} \geq 0, x_{n+1}=\sqrt{x_{n}+2}$

$$x_{n+1}=\sqrt{x_n+2}$$

$$x_{n+1}^2=x_n+2$$

$$x_{n+1}^2-4=x_n-2$$

$$(x_{n+1}-2)(x_{n+1}+2)=x_n-2$$

We have $x_{n+1}+2 \ge 2$ since $x_{n+1} \ge 0$.

$$x_{n+1}-2 = \frac{x_n-2}{x_{n+1}+2}$$

$$|x_{n+1}-2| \le \frac{|x_n-2|}{2}$$

Hence we have $$|x_{n+1}-2 | \le \frac{|x_1-2|}{2^n}$$

Hence as $n \to \infty$, we have $x_n \to 2$.

This answer is presenting a (quite) general method for tackling this kind of problem. For this particular sequence, the answer by the user Siong Thye Goh is much more elegant and elementary.

Preleminary inequality

Let $f(x) =\sqrt{x+2}-2$. We have $f'(x) = \frac 1 {2\sqrt {x+2}}$. By Mean Value Theorem :

$$|f(x)|=|f(x)-f(2)| \le \sup_{x \in \mathbb{R}^+} |f'(x)| |x-2|$$

$f'(x)$ is decreasing with $f'(0) = \frac 1 {2 \sqrt 2} < \frac 1 2$. Finally :

\begin{equation} f(x) < \frac {|x-2|} 2 \end{equation}

Proof of the statement 2 by induction

Let $x_1 \ge 0$.

We have : $$x_2=\sqrt{x_1+2}.$$

$$|x_2-2| = |\sqrt{x_1+2}-2|\overset{*}< \frac 1 2 |x_1-2|.$$

Where we have applied the preliminary inequality for step with a $*$.

Then, we suppose that, for some $n$ :

$$|x_{n}-2| < \frac 1 {2^{n-1}} |x_1-2|.$$

Thus : $$|x_{n+1}-2| = |\sqrt{x_n+2}-2|\overset{*}<\frac 1 2 |x_n-2|.$$

Where we have applied the preliminary inequality for step with a $*$.

By applying the inequality just above :

$$|x_{n+1}-2|<\frac 1 {2^{n}} |x_1-2|.$$

Since the inequality was true for $n=2$, by induction it is true for all $n$.

Conclusion

$\frac 1 {2^{n}} |x_1-2|$ is a decreasing sequence converging to $0$ as $|x_1-2|$ is just a constant and $\frac 1 {2^{n-1}}$ converges to $0$.

By squeeze theorem, we conclude : $|x_{n+1}-2| \to 0$ as $n \to \infty$.