<s>[INST] How is $\lim_{x \to 0}\frac{\sqrt {2-x}-\sqrt {2}}{x} = \frac{-1}{2\sqrt{2}}$? How is
$$\displaystyle\lim_{x \to 0}\frac{\sqrt {2-x}-\sqrt {2}}{x} = \frac{-1}{2\sqrt{2}}$$ 
computable?
 [/INST] Multiplying$$\frac{\sqrt{2-x}-\sqrt 2}{x}$$
by$$\frac{\sqrt{2-x}+\sqrt 2}{\sqrt{2-x}+\sqrt 2}$$
gives us
$$\begin{align}\lim_{x\to 0}\frac{\sqrt{2-x}-\sqrt 2}{x}&=\lim_{x\to 0}\frac{(\sqrt{2-x}-\sqrt 2)(\sqrt{2-x}+\sqrt 2)}{x(\sqrt{2-x}+\sqrt 2)}\\&=\lim_{x\to 0}\frac{(2-x)-2}{x(\sqrt{2-x}+\sqrt 2)}\\&=\lim_{x\to 0}\frac{-1}{\sqrt{2-x}+\sqrt 2}\\&=\frac{-1}{\sqrt 2+\sqrt 2}\\&=-\frac{1}{2\sqrt 2}.\end{align}$$
</s>