Question Number 173401 by DAVONG last updated on 11/Jul/22
$$\mathrm{1}.\mathrm{show}\:\mathrm{that}\:\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{2022}} \mathrm{n}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{n}} }=\mathrm{0}\:? \\ $$
Commented by alephzero last updated on 11/Jul/22
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{2022}} {n}^{\mathrm{2}} }{\mathrm{3}^{{n}} }\:=\:\mathrm{2}^{\mathrm{2022}} \:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}^{{n}} }\:=\:\mathrm{2}^{\mathrm{2022}} ×\mathrm{0} \\ $$$$=\:\mathrm{0} \\ $$$$\bigstar\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{a}} }{\alpha^{{n}} }\:=\:\mathrm{0}\:\forall{a}\:\forall\alpha \\ $$