${{{f}'}_{+}}(0)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{4-\left| x \right|}-2}{x}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{4-x}-2}{x}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{4-x-4}{x(\sqrt{4-x}+2)}=-\frac{1}{4}$  ${{{f}'}_{-}}(0)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{4-\left| x \right|}-2}{x}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{4+x}-2}{x}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{4+x-4}{x(\sqrt{4+x}+2)}=\frac{1}{4}$ $\Rightarrow {{{f}'}_{-}}(0)-{{{f}'}_{+}}(0)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$