Determine the convergence Let $ x_{n} = \sqrt{n + 1} - \sqrt{n} $ for each $ n \in \mathbb{N} $. Show that $ (\sqrt{n} x_{n})_{n \in \mathbb{N}} $ is a convergent sequence and find its limit.
I have a problem, because by the criterion of the reason I can not conclude anything since I got $ L = 1 $.
I do not know if it could be developed by comparison.
Let the sequence defined by 
$$\left\lbrace
\begin{array}{l}
x_{1}>1& \\\\ 
x_{n+1}=2-\dfrac{1}{x_{n}}&\forall n\in\mathbb{N}.
\end{array}\right.
$$
Show that $ (x_{n})_{n \in \mathbb{N}} $ is bounded and monotone. How can I conclude that the sequence is convergent and find its limit?
 A: First sequence: $\sqrt n(\sqrt{n+1}-\sqrt n)=\sqrt{n^2+n}-n\to 1/2$. You can prove this by multiplying by $1=\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}$.
After you have shown that the second sequence is monotone and bounded, you can imply that it is convergent. The limit $a$ must satisfy $a=2-\frac1a$, which implies $a=1$.
A: For the first sequence, after rationalising the numerator, we get
$$x_n = \dfrac{\sqrt{n}}{\sqrt{n + 1} + \sqrt{n}} = \dfrac{1}{\sqrt{1 + \frac{1}{n}} + 1}$$
Hence, the sequence is convergent and in fact, the limit is $\dfrac{1}{2}$.
For the second sequence, we have $0 < 1 < x_1$ so that $1 < x_2 < 2$. Now, we shall prove that this is bounded by induction. After a few tries, we can guess the inequalities
$$1 < x_n < \dfrac{n + 1}{n}$$
for $n \geq 2$. The base case is already proved. Now, using induction hypothesis, let this inequality by true for $k < n$. Then, in particular we will have
$$1 < x_{n - 1} < \dfrac{n}{n - 1}$$
Then we have the following,
\begin{align}
&\therefore 1 > \dfrac{1}{x_{n - 1}} > \dfrac{n - 1}{n} \\
&\therefore -1 < - \dfrac{1}{x_{n - 1}} < - \dfrac{n - 1}{n} \\
&\therefore 1 < 2 - \dfrac{1}{x_{n - 1}} = x_n < 2 - \dfrac{n - 1}{n} = \dfrac{n + 1}{n}
\end{align}
For the monotone part, I have not come up with a solution yet. But as @st.math says, if you prove that it is monotone, the convergence follows and we can find the limit from this method.
Edit:-
To prove that the sequence is monotone, consider $x_{n + 1} - x_n$. We have
$$x_{n + 1} - x_n = 2 - \left( \dfrac{1}{x_n} + x_n \right)$$
Now, let $f \left( x \right) = \dfrac{1}{x} + x$. Then, $f' \left( x \right) = 1 - \dfrac{1}{x^2}$. For $x > 1$, we have $f' \left( x \right) > 0$, $f' \left( 1 \right) = 0$ and $f \left( 1 \right) = 2$. Hence, $f \left( 1 \right)$ is a local minimum and for $x > 1$, $f \left( x \right) > f \left( 1 \right) = 2$. Since $x_n > 1$, we can apply this to $x_{n + 1} - x_{n}$ to get $x_{n + 1} - x_n < 0$ so that the sequence is decreasing.
Now, being a bounded monotone sequence, it is convergent and the procedure to find the limit is already known.
A: Everytime you encounter something like $$ \lim_{n \to \infty} \sqrt{n+1} - \sqrt{n} $$ 
one of the thing you can try to do is to multiply by the conjugate:
$$\lim_{n \to \infty} \sqrt{n+1} - \sqrt{n} \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} = 0  $$
However, in your case, you have instead 
$$ \lim_{n \to \infty} \sqrt{n} \bigg(\sqrt{n+1} - \sqrt{n} \bigg) = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{2}$$ 
