Answer
$$\frac{3}{32}.$$
Work Step by Step
Since we have $$\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2 \sqrt[8]{1+x}}{x^{2}}=\frac{0}{0}.$$
We can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2 \sqrt[8]{1+x}}{x^{2}}=\lim _{x \rightarrow 0} \frac{(1/2\sqrt{4+x})-(1/ 4(1+x)^{7/8})}{2x}=\frac{0}{0}.$$ Again, we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow 0} \frac{(1/2\sqrt{4+x})-(1/ 4(1+x)^{7/8})}{2x}=\lim _{x \rightarrow 0} \frac{(-1/4(4+x)^{3/2})+(7/ 32(1+x)^{15/8})}{2}=\frac{-(1/32)+(7/32)}{2}=\frac{3}{32}.$$
