Algebra with square roots and fractions.

Question

I have a problem that I simplify to equal $\frac{\sqrt{x + h - 9} -\sqrt{x - 9}}{h}$

I thought this would be as far as I could take it but the answer in the book shows $\frac{1}{\sqrt{x + h - 9} + \sqrt{x - 9}}$

Can someone please tell me how you get from $\frac{\sqrt{x + h - 9} -\sqrt{x - 9}}{h}$ to $\frac{1}{\sqrt{x + h - 9} + \sqrt{x - 9}}$ ?

To add more background. I am tasked with simplifying $\frac{F(x+h) - F(x)}{h}$ where F is a function defined as F(x) = $\sqrt{x - 9}$ so I get it to $\frac{\sqrt{x + h - 9} -\sqrt{x - 9}}{h}$ but the book gives $\frac{1}{\sqrt{x + h - 9} + \sqrt{x - 9}}$ But I don't know how to get the answer in the book.

Answer 1

$$\frac{\sqrt{x+h-9}-\sqrt{x-9}}{h}=\frac{\sqrt{x+h-9}-\sqrt{x-9}}{h}\cdot\frac{\sqrt{x+h-9}+\sqrt{x-9}}{\sqrt{x+h-9}+\sqrt{x-9}}= \\ =\frac{\sqrt{x+h-9}^2-\sqrt{x-9}^2}{h\cdot\sqrt{x+h-9}+\sqrt{x-9}} = \frac{(x+h-9)-(x-9)}{h\cdot\sqrt{x+h-9}+\sqrt{x-9}} = \\ = \frac{h}{h\cdot\sqrt{x+h-9}+\sqrt{x-9}} = \frac{1}{\sqrt{x+h-9}+\sqrt{x-9}} $$

Answer 2

it is $$\frac{\sqrt{x+h-9}-\sqrt{x-9}}{h}=\frac{x+h-9-x+9}{h(\sqrt{x+h-9}+\sqrt{x-9})}$$ so we get $$\lim_{h\to 0}\frac{1}{\sqrt{x+h-9}+\sqrt{x-9}}=\frac{1}{2\sqrt{x-9}}$$