Question Number 136701 by mathlove last updated on 25/Mar/21
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}^{{n}} −{n}!}{\mathrm{2}^{{n}} −{n}!}=? \\ $$
Commented by yutytfjh67ihd last updated on 25/Mar/21
Commented by mathlove last updated on 25/Mar/21
$${help}\:{me} \\ $$
Answered by mathmax by abdo last updated on 25/Mar/21
$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{n}!−\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!−\mathrm{2}^{\mathrm{n}} }\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{1}−\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}}{\mathrm{1}−\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}!}} \\ $$$$\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\Rightarrow\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\sim\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}}\:\:\Rightarrow \\ $$$$\mathrm{log}\left(\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\right)\sim\mathrm{nlog3}−\mathrm{nlogn}+\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2n}\pi\right) \\ $$$$=\mathrm{n}\left(\mathrm{log3}−\mathrm{logn}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2n}\pi\right)\right)\rightarrow−\infty\:\Rightarrow\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\rightarrow\mathrm{0} \\ $$$$\mathrm{also}\:\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}!}\rightarrow\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{n}!−\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!−\mathrm{2}^{\mathrm{n}} }=\mathrm{1} \\ $$