Given $f\left(x\right)$ is continuous at $x=0$.

Therefore, $f\left(0\right) =\underset{x\to 0}{\lim } \frac{2-\sqrt{x+4}}{\sin 2x} ...\left(i\right)$

$f\left(x\right)$ forms a $\frac{0}{0}$ form at $x=0$.

Hence, using L'Hospital's Rule,

$\underset{x\to 0}{\lim } \frac{{\displaystyle \frac{-1}{2\sqrt{x+4}}}}{2\cos 2x}$

$\frac{{\displaystyle \frac{-1}{2\left(\sqrt{4}\right)}}}{2\left(1\right)}$ $=\frac{-1}{8}$

Using $\left(i\right)$, $f\left(0\right)=\frac{-1}{8}$