$\lim\limits_{x \rightarrow 0}\frac{\int\limits_{-x}^x f(t) d t}{\int\limits_0^{2 x} f(t+4) d t}$ is equal to

$\frac{f(0)}{f(4)}$

$=\lim\limits_{x \rightarrow 0} \frac{\frac{d}{d x}(x) f(x)-\frac{d}{d x}(-x) f(-x)}{\frac{d}{d x}(2 x) f(2 x+4)-0}$

$=\lim\limits_{x \rightarrow 0} \frac{f(x)+f(-x)}{2 f(2 x+4)}=\frac{2 f(0)}{2 f(4)}=\frac{f(0)}{f(4)}$