alternate form $\sqrt{x} - \sqrt{y}$

Question

For a part of one of my prove for limits I have to show that $$\sqrt x -\sqrt c = \frac{x-c}{\sqrt x + \sqrt c}$$

I understand that $\sqrt x = \frac{x}{\sqrt x}$ but I can not figure out why $$ \frac{x}{\sqrt x} - \frac{c}{\sqrt c} = \frac{x-c}{\sqrt x + \sqrt c}$$

If someone could explain how I can conclude that these two are the same that would be appreciated.

Answer 1

Take the fraction $$ \frac{\sqrt x-\sqrt c}{1} $$ and expand it by a factor of $\sqrt x+\sqrt c$. This doesn't change the value of the fraction, but it does change what it looks like.

Answer 2

Because$$x-c=\sqrt x^2-\sqrt c^2=\left(\sqrt x+\sqrt c\right)\left(\sqrt x-\sqrt c\right).$$

Answer 3

$$\sqrt x -\sqrt c = (\sqrt x -\sqrt c)\cdot \frac{\sqrt{x} + \sqrt c}{\sqrt x +\sqrt c} = \frac{(\sqrt x -\sqrt c)\cdot (\sqrt{x} + \sqrt c )}{\sqrt x +\sqrt c}\\ = \frac{\sqrt x \cdot \sqrt x - \sqrt c\cdot \sqrt c }{\sqrt x +\sqrt c} =\frac{x-c}{\sqrt x + \sqrt c}$$