Question Number 86603 by Ar Brandon last updated on 29/Mar/20
$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{{e}^{{x}} }{\left(\mathrm{1}+{x}\right)^{{n}} } \\ $$
Commented by abdomathmax last updated on 29/Mar/20
$${let}\:{f}\left({x}\right)=\frac{{e}^{{x}} }{\left(\mathrm{1}+{x}\right)^{{n}} }\:\:{changement}\:\mathrm{1}+{x}\:={u}\:{give} \\ $$$${f}\left({x}\right)=\frac{{e}^{{u}−\mathrm{1}} }{{u}^{{n}} }\:\:\:{so}\:{if}\:{n}>\mathrm{0}\:{lim}_{{x}\rightarrow−\mathrm{1}} {f}\left({x}\right)={lim}_{{u}\rightarrow\mathrm{0}} \:\:\frac{{e}^{{u}−\mathrm{1}} }{{u}^{{n}} } \\ $$$$=\frac{{e}^{−\mathrm{1}} }{\mathrm{0}^{+} }\:=+\infty \\ $$$${if}\:{n}<\mathrm{0}\:\:\:{lim}_{{x}\rightarrow−\mathrm{1}} {f}\left({x}\right)={lim}_{{u}\rightarrow\mathrm{0}} \:\:{u}^{−{n}} \:{e}^{{u}−\mathrm{1}} =\mathrm{0} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
$${great}! \\ $$
Answered by Rio Michael last updated on 29/Mar/20
$$+\infty? \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
$${how},\:\:{please}? \\ $$
Answered by Rio Michael last updated on 29/Mar/20
$$\mathrm{my}\:\mathrm{try} \\ $$$$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{{e}^{{x}} }{\left(\mathrm{1}+{x}\right)^{{n}} }\:=\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{{e}^{{x}} }{{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} }\:=\:+\infty\:\mathrm{for}\:{n}\:>\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{{e}^{{x}} }{{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} }\:=\:\mathrm{0}\:\mathrm{for}\:{n}\:<\:\mathrm{0} \\ $$