Because both the numerator and denominator approach $0$ as $x$ approaches $0$, applying L'Hopital rule gives

$$\lim_{x\to 0}\frac{\int_0^x\sin(xt)^2dt}{x^5}=\lim_{x\to 0}\frac{\sin x^4}{5x^4}=\boxed{\frac{1}{5}}$$