Conjugation method in solving limits
Solution 1:
The $-3h$ term comes from expanding out the difference of two squares in the numerator. As follows:
$$\lim\limits_{h \to 0} \frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h}$$ $$=\lim\limits_{h \to 0} \frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h} \frac{\sqrt{2-3x-3h}+\sqrt{2-3x}}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$ $$=\lim\limits_{h \to 0} \frac{(2-3x-3h)-(2-3x)}{h(\sqrt{2-3x-3h}+\sqrt{2-3x})}$$ $$=\lim\limits_{h \to 0} \frac{-3}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$ $$= \frac{-3}{2\sqrt{2-3x}}$$