Question Number 193117 by mustafazaheen last updated on 04/Jun/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$
Answered by Subhi last updated on 04/Jun/23
$${y}\:=\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left({cosx}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$${ln}\left({y}\right)\:=\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{ln}\left({cosx}\right)}{{x}} \\ $$$${apply}\:{L}\:{Hopital}'{s}\:{law} \\ $$$$=\:\frac{−{sin}\left({x}\right)}{{cos}\left({x}\right)}=−{tan}\left({x}\right)=−{tan}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${ln}\left({y}\right)=\mathrm{0} \\ $$$${y}\:=\:{e}^{\mathrm{0}} =\mathrm{1} \\ $$
Answered by horsebrand11 last updated on 04/Jun/23
$$\:\mathrm{L}=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right)} \\ $$$$\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)}\right)} =\mathrm{e}^{\mathrm{0}} =\mathrm{1}\: \\ $$
Answered by aba last updated on 04/Jun/23
$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{\frac{\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left(\mathrm{x}^{\mathrm{2}} \right)\right)}{\mathrm{x}}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{\frac{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{−\frac{\mathrm{x}}{\mathrm{2}}+{o}\left(\mathrm{x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{2}}+{o}\left(\mathrm{x}\right)\right)=\mathrm{1} \\ $$