$I_{1}(n)=\int_{0}^{1} \left \lfloor{a^n x} \right \rfloor dx=\int_{0}^{\frac{1}{a^n}} \left \lfloor{a^n x} \right \rfloor dx +\int_{\frac{1}{a^n}}^{\frac{2}{a^n}} \left \lfloor{a^n x} \right \rfloor dx +\int_{\frac{2}{a^n}}^{\frac{3}{a^n}} \left \lfloor{a^n x} \right \rfloor dx +...+\int_{\frac{a^n-1}{a^n}}^{\frac{a^n}{a^n}} \left \lfloor{a^n x} \right \rfloor dx =\int_{0}^{\frac{1}{a^n}}0 dx+\int_{\frac{1}{a^n}}^{\frac{2}{a^n}}1 dx+...+\int_{\frac{a^n-1}{a^n}}^{\frac{a^n}{a^n}}(a^n-1)dx$ or $I_1(n)=\sum_{i=1}^{a^n-1}i(\frac{i+1}{a^n}-\frac{i}{a^n})=\sum_{i=1}^{a^n-1}\frac{i}{a^n}=\frac{\sum_{i=1}^{a^n-1}i}{a^n}=\frac{a^n-1}{2}$ similarly, $I_2(n)=\frac{b^n-1}{2}$ so, $l=\lim_{n\to +\infty} \frac{I_1(n)}{I_2(n)}=\lim_{n\to +\infty} \frac{a^n-1}{b^n-1}$ If $a<b,$ $l=0$ If $a=b,$ $l=1$ If $a>b,$ $l=+\infty$ as @above