Find $\limsup_n x_n$ and $\liminf_n x_n$ of $x_n=\sqrt{n+1}-\sqrt{n}$

Question

By trying I have found, that the sequence is falling and I'm fairly certain that it's going to $0$ for large n. So we know that $\liminf=\limsup=\lim$. But how do I show that the sequence converges to $0$? Also is my argumentation correct so far?

Answer 1

By multiplying by the conjugate we have

$$x_n=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$ can you find the limit?

Answer 2

We do this by multiplying $x_n$ by its conjugate, expressing the numerator as a difference of squares, and then taking the limit of the result:

$$\begin{align}\lim_{n \to \infty} x_n &= \lim_{n\to\infty} \frac{\sqrt{n+1}-\sqrt{n}}{1}\cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\ \\ & = \lim_{n\to \infty} \frac{n+1 - n}{\sqrt{n+1}+\sqrt{n}}\\ \\ &=\lim_{n \to \infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} \\ \\ & = 0\end{align}$$

The rest follows as you argue. Since the limit of $x_n$ is $0<\infty$, as $n\to \infty$, the $$\liminf x_n =\limsup x_n =\lim x_n, \;\;n \to \infty$$