Question Number 41855 by 123456780 last updated on 14/Aug/18
$$\underset{{x}\rightarrow+\infty\:\:} {\mathrm{lim}}\frac{\mathrm{n}!}{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{n}!\right)} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limit} \\ $$
Commented by prof Abdo imad last updated on 14/Aug/18
$${let}\:{put}\:{p}={n}!\:\Rightarrow\:\:\frac{{n}!}{{ln}\left(\mathrm{1}+{n}!\right)}\:=\frac{{p}}{{ln}\left(\mathrm{1}+{p}\right)} \\ $$$$=\frac{{p}}{{lnp}\:+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)}\:\:\sim\:\frac{{p}}{{lnp}\:+\frac{\mathrm{1}}{{p}}}\:\sim\:\frac{{p}}{{ln}\left({p}\right)}\:\left({p}\rightarrow+\infty\right) \\ $$$${but}\:{lim}_{{p}\rightarrow+\infty} \frac{{p}}{{ln}\left({p}\right)}\:=+\infty\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\frac{{n}!}{{ln}\left(\mathrm{1}+{n}!\right)}\:=+\infty \\ $$$$ \\ $$