Question Number 105246 by moonanddolldoll last updated on 27/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$
Answered by bemath last updated on 27/Jul/20
$${another}\:{method} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\left(\mathrm{cos}\:{x}−\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt{{e}}}\: \\ $$
Commented by moonanddolldoll last updated on 27/Jul/20
$${thanks}! \\ $$
Answered by bemath last updated on 27/Jul/20
$${L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}{x}\:}{\mathrm{2}{x}\:\mathrm{cos}\:{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${L}\:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt{{e}}}\:\bigstar \\ $$
Answered by abdomathmax last updated on 27/Jul/20
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{cosx}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{ln}\left(\mathrm{cosx}\right)\sim\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{cosx}\right)\:\sim−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}} \\ $$
Answered by Dwaipayan Shikari last updated on 27/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{−\mathrm{2}}{{x}^{\mathrm{2}} }.\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} ={e}^{\frac{−\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$$${Another}\:{way} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{cosx}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} ={y} \\ $$$$\frac{{log}\left(\mathrm{1}+{cosx}−\mathrm{1}\right)}{{cosx}−\mathrm{1}}.\left(−\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }\right)={logy} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }={logy} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{2}\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}{{x}^{\mathrm{2}} }={logy}\Rightarrow{y}={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$