To find the limit we need to rationalize the function, we achieve this by multiplying by a one written in an appropiate way:
[tex]\begin{gathered} \lim_{h\to0}\frac{2-\sqrt{4+h}}{h}=\lim_{h\to0}\frac{2-\sqrt{4+h}}{h}\cdot\frac{2+\sqrt{4+h}}{2+\sqrt{4+h}} \\ =\lim_{h\to0}\frac{(4-(\sqrt{4+h})^2)}{h(2+\sqrt{4+h})} \\ =\lim_{h\to0}\frac{4-(4+h)}{h(2+\sqrt{4+h})} \\ =\lim_{h\to0}\frac{4-4-h}{h(2+\sqrt{4+h})} \\ =\lim_{h\to0}\frac{-h}{h(2+\sqrt{4+h})} \\ =\lim_{h\to0}\frac{-1}{2+\sqrt{4+h}} \\ =-\frac{1}{2+\sqrt{4+0}} \\ =-\frac{1}{2+\sqrt{4}} \\ =-\frac{1}{2+2} \\ =-\frac{1}{4} \end{gathered}[/tex]
Therefore:
[tex]\lim_{h\to0}\frac{2-\sqrt{4+h}}{h}=-\frac{1}{4}[/tex]
