Question Number 121726 by oustmuchiya@gmail.com last updated on 11/Nov/20
Answered by TANMAY PANACEA last updated on 11/Nov/20
$$\mathrm{6}{n}^{\mathrm{3}} >\mathrm{4}{n}^{\mathrm{3}} \:\:{also}\:\mathrm{2}^{{n}} >\mathrm{2} \\ $$$${so}\:\:\mathrm{6}{n}^{\mathrm{3}} +\mathrm{2}^{{n}} >\mathrm{4}{n}^{\mathrm{3}} +\mathrm{2} \\ $$$${so}\:\mathrm{li}\underset{{n}\rightarrow\infty} {\mathrm{m}}\:\frac{\mathrm{6}{n}^{\mathrm{3}} +\mathrm{2}^{{n}} }{\mathrm{4}{n}^{\mathrm{3}} +\mathrm{2}}=\infty \\ $$