Question Number 165698 by azizmathhacker last updated on 06/Feb/22
Answered by alephzero last updated on 06/Feb/22
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{e}^{\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}}} \:=\:{e}^{\mathrm{ln}\:\mathrm{1}} \:=\:\mathrm{1} \\ $$
Commented by chrisbridge last updated on 06/Feb/22
$${lim}\:{e}^{\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}}\:=\:\:{lim}\:\frac{{x}+\mathrm{1}}{{x}}\:=\:\frac{{x}}{{x}}\:+\:\frac{\mathrm{1}}{{x}}} \\ $$$${x}\:\rightarrow\infty \\ $$$$\:\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1}\:+\frac{\mathrm{1}}{\infty}\:=\:\mathrm{1}\:+\:\mathrm{0}\:=\:\mathrm{1} \\ $$$$ \\ $$
Commented by alephzero last updated on 07/Feb/22
$$\mathrm{Oh},\:\mathrm{sorry}.\:\mathrm{I}\:\mathrm{forgot}\:\mathrm{about}\:\mathrm{that} \\ $$
Commented by TheSupreme last updated on 07/Feb/22
$${e}^{{ln}\left({f}\left({x}\right)\right)} ={f}\left({x}\right) \\ $$$${lim}\:{e}^{{ln}\left({f}\left({x}\right)\right)} ={lim}\:{f}\left({x}\right) \\ $$$$ \\ $$