I'm trying to understand a solution to a limit problem that uses conjugates. The textbook showed this problem, $$\lim\limits_{h \to 0} \frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h}$$ To get the limit, which is $-\frac{3}{2\sqrt{2-3x}}$, I know that I just can't do direct substitution since the denominator $h$ would become zero, and the conjugation method can be used. But the textbook indicated this conjugate: $$\frac{-3h}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$ I understand that the sign should be changed from $\sqrt{2-3x-3h}-\sqrt{2-3x}$ to $\sqrt{2-3x-3h}+\sqrt{2-3x}$. And I understand that having $-3h$ in the numerator of the conjugate would cancel the $h$ in the denominator of the original expression.
What I don't understand is that why and how did the $-3h$ got there in the first place. All that I learned about using conjugates is that I should change the sign between terms in a binomial that have square roots and multiply it to both numerator and denominator of the original expression. But I don't know how did the textbook got that $-3h$ in the numerator and placed what I know as the conjugate $\sqrt{2-3x-3h}+\sqrt{2-3x}$ in the denominator.
The $-3h$ term comes from expanding out the difference of two squares in the numerator. As follows:
$$\lim\limits_{h \to 0} \frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h}$$ $$=\lim\limits_{h \to 0} \frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h} \frac{\sqrt{2-3x-3h}+\sqrt{2-3x}}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$ $$=\lim\limits_{h \to 0} \frac{(2-3x-3h)-(2-3x)}{h(\sqrt{2-3x-3h}+\sqrt{2-3x})}$$ $$=\lim\limits_{h \to 0} \frac{-3}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$ $$= \frac{-3}{2\sqrt{2-3x}}$$