Solutions

Let $f(x)=\sqrt{x}$ and consider the interval $[64,66]$. On this interval $f$ is continuous and differentiable so we can apply the mean value theorem which tells us that

$$\frac{\sqrt{66}-\sqrt{64}}{66-64}=\frac{\sqrt{66}-8}{2}=\frac{1}{2}x^{-1/2}$$

for some $x\in(64,66)$. But this means that

$$\frac{1}{9}=\frac{1}{2\sqrt{81}}<\frac{1}{2\sqrt{66}}<\frac{\sqrt{66}-8}{2}=\frac{1}{2\sqrt{x}}<\frac{1}{2\sqrt{64}}=\frac{1}{2\cdot 8}$$

and hence

$$\frac{1}{9}<\sqrt{66}-8<\frac{1}{8}.$$