If $\underset{x\to 0}{\lim }\frac{e^{2x}-1}{5x}=l , \underset{x\to 1}{\lim }\frac{2}{x-1}\log \left(x\right)=m$ , then a cubic equation whose roots are 5l, m and 1 is

$$\begin{gathered} \mathrm{l}=\underset{\mathrm{x}\to 0}{\mathrm{lt}}\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{5\mathrm{x}} \\ = \underset{\mathrm{x}\to 1}{\mathrm{lt}}\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}.\frac{2}{5}=\frac{2}{5} \\ \mathrm{m}=\underset{\mathrm{x}\to 0}{\lim 2} \frac{ \log \mathrm{x}}{\mathrm{x}-1} \\ =\underset{\mathrm{x}\to 0}{\lim } 2 \frac{\log (1+\mathrm{x}-1)}{\mathrm{x}-1}=2 \\ \mathrm{roots} \mathrm{are} 2, 2, 1 \\ \mathrm{eqn}. \mathrm{is} {\mathrm{x}}^3-5{\mathrm{x}}^2+8\mathrm{x}-4=0 \end{gathered}$$