Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$ I know that 
$$\lim_{n\to  \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$ 
But how can I prove this?  
I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.  
Edit:
Thank Y'all for the nice and fast answers!

Note that
$$( \sqrt{n^2+2}-\sqrt{n^2+1})(\sqrt{n^2+2}+\sqrt{n^2+1})=1$$
Thus 
$$\sqrt{n^2+2}+\sqrt{n^2+1} \to \infty \implies \sqrt{n^2+2}-\sqrt{n^2+1}\to 0$$
