$\underset{x\to 0}{\lim }\frac{{\int }_{-x}^x f\left(t\right)dt}{{\int }_0^{2x} f(t+4)dt}$ is equal to

$$\begin{gathered} \underset{x\to 0}{lim} \frac{{\int }_{-x}^x f\left(t\right)dt}{{\int }_0^{2x} f(t+4)dt} \\ \begin{array}{l}=\underset{x\to 0}{lim} \frac{f\left(x\right)\times 1-f\left(-x\right)\times -1}{f\left(2x+4\right)\times 2-0} \left(apply L-H rule\right)\\ =\underset{x\to 0}{lim} \frac{f\left(x\right)+f(-x)}{2f(2x+4)}=\frac{2f\left(0\right)}{2f\left(4\right)}=\frac{f\left(0\right)}{f\left(4\right)}\end{array} \end{gathered}$$