Limits!

Lets call the limit as $\mathfrak F$ .

$\mathfrak F=\displaystyle\lim_{x\to 0}\dfrac{2-\sqrt[3]{8+x}}{2x\underbrace{\sqrt[3]{8+x}}_2}=\displaystyle\lim_{x\to 0}\dfrac{1-\sqrt[3]{1+\frac x8}}{2x}$

Using Binomial approximation $\color{#3D99F6}{\small{\sqrt[3]{1+\frac x8}=1+\frac x{24}\text{ as }x\to 0}}$ and thus $\mathfrak F=\dfrac{-1}{48}$ $\Large \therefore\sqrt{1+48}=\boxed{\color{#007fff}{7}}$

Take x common from denominator and then rationalise the remaining expression to get the answer as $-\frac{1}{48}$