Question

${lim}_{x\to 0}\frac{{\int }_{-x}^{x}f\left(t\right)dt}{{\int }_0^{2x}f(t+4)dt}=$_________, where $f$ is a continuous fuuction and $f\left(x\right)>0$ for all $x$ satisfying $f\left(0\right)=48f\left(4\right)$.

Answer 1

${lim}_{x\to 0}\frac{{\int }_{-x}^{x}f\left(t\right)dt}{{\int }_0^{2x}f(t+4)dt}={lim}_{x\to 0}\frac{f\left(x\right)+f(-x)}{2f(2x+4)}=\frac{1}{2}\frac{{lim}_{x\to 0}(f\left(x\right)+f(-x))}{{lim}_{x\to 0}f(2x+4)}=\frac{1}{2}\frac{2f\left(0\right)}{f\left(4\right)}=48$