Question Number 80670 by jagoll last updated on 05/Feb/20
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{{e}^{\mathrm{sin}\:{x}} −\mathrm{1}}{{x}−\pi}=? \\ $$
Commented by jagoll last updated on 05/Feb/20
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}.{e}^{\mathrm{sin}\:{x}} }{\mathrm{1}}=\:−\mathrm{1} \\ $$
Commented by abdomathmax last updated on 05/Feb/20
$${let}\:{f}\left({x}\right)=\frac{{e}^{{sinx}} −\mathrm{1}}{{x}−\pi}\:\:{changement}\:{x}−\pi={t}\:{give} \\ $$$${f}\left({x}\right)={g}\left({t}\right)=\frac{{e}^{{sin}\left(\pi+{t}\right)} −\mathrm{1}}{{t}}\:=\frac{{e}^{−{sint}} −\mathrm{1}}{{t}} \\ $$$$\sim\:\frac{{e}^{−{t}} −\mathrm{1}}{{t}}\:\sim\frac{\mathrm{1}−{t}−\mathrm{1}}{{t}}=−\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow\pi} \:\:{f}\left({x}\right)=−\mathrm{1} \\ $$
Answered by $@ty@m123 last updated on 05/Feb/20
$${Let}\:{x}−\pi={t} \\ $$$${x}=\pi+{t} \\ $$$${As}\:{x}\rightarrow\pi,\:{t}=\mathrm{0} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{\mathrm{sin}\:\left(\pi+{t}\right)} −\mathrm{1}}{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{−\mathrm{sin}\:{t}} −\mathrm{1}}{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}−\mathrm{sin}\:{t}+\frac{\mathrm{sin}\:^{\mathrm{2}} {t}}{\mathrm{2}!}+…..−\mathrm{1}}{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{−\mathrm{sin}{t}}{{t}} \\ $$$$=−\mathrm{1} \\ $$