erweitere mit der dritten bin. Formel.
$$\sqrt{n^2+2n-8}-\sqrt{n^2+4n-6}=\frac{(\sqrt{n^2+2n-8}-\sqrt{n^2+4n-6})(\sqrt{n^2+2n-8}+\sqrt{n^2+4n-6})}{(\sqrt{n^2+2n-8}+\sqrt{n^2+4n-6})}\\ =\frac{n^2+2n-8-(n^2+4n-6)}{(\sqrt{n^2+2n-8}+\sqrt{n^2+4n-6})}=\frac{-2n-2}{(\sqrt{n^2+2n-8}+\sqrt{n^2+4n-6})}\\ =\frac{-2-2/n}{(\sqrt{1+2/n-8/n^2}+\sqrt{1+4/n-6/n^2})}\\$$
Schicke nun n gegen unendlich
