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Question Number 105246 by moonanddolldoll last updated on 27/Jul/20

lim_(x→0) (cosx)^(1/x^2 )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$

Answered by bemath last updated on 27/Jul/20

another method lim_(x→0) (1+(cos x−1))^(1/x^2 ) = e^(lim_(x→0) ((cos x−1)/x^2 )) =e^(lim_(x→0) (((1−(x^2 /2))−1)/x^2 )) = e^(−(1/2)) = (1/(√e))

$${another}\:{method} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\left(\mathrm{cos}\:{x}−\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\sqrt{{e}}}\: \\ $$

Commented by moonanddolldoll last updated on 27/Jul/20

thanks!

$${thanks}! \\ $$

Answered by bemath last updated on 27/Jul/20

L = lim_(x→0) (cos x)^(1/x^2 ) ln L = lim_(x→0) ((ln (cos x))/x^2 ) ln L = lim_(x→0) ((−sinx )/(2x cos x)) = −(1/2) L = e^(−(1/2)) = (1/(√e)) ★

$${L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}{x}\:}{\mathrm{2}{x}\:\mathrm{cos}\:{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${L}\:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\sqrt{{e}}}\:\bigstar \\ $$

Answered by abdomathmax last updated on 27/Jul/20

let f(x) =(cosx)^(1/x^2 ) ⇒f(x) =e^((1/x^2 )ln(cosx)) we have cosx∼1−(x^2 /2) ⇒ln(cosx)∼ln(1−(x^2 /2))∼−(x^2 /2) ⇒(1/x^2 )ln(cosx) ∼−(1/2) ⇒lim_(x→0) f(x) =e^(−(1/2)) =(1/(√e))

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{cosx}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{ln}\left(\mathrm{cosx}\right)\sim\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{cosx}\right)\:\sim−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\sqrt{\mathrm{e}}} \\ $$

Answered by Dwaipayan Shikari last updated on 27/Jul/20

lim_(x→0) (1−(x^2 /2))^(1/x^2 ) =lim_(x→0) (1−(x^2 /2))^(((−2)/x^2 ).(−(1/2))) =e^((−1)/2) =(1/(√e)) Another way lim_(x→0) (1+cosx−1)^(1/x^2 ) =y ((log(1+cosx−1))/(cosx−1)).(−((1−cosx)/x^2 ))=logy lim_(x→0) −((2sin^2 (x/2))/x^2 )=logy lim_(x→0) −((2(x^2 /4))/x^2 )=logy⇒y=e^(−(1/2)) =(1/(√e))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{−\mathrm{2}}{{x}^{\mathrm{2}} }.\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} ={e}^{\frac{−\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\sqrt{{e}}} \\ $$$${Another}\:{way} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{cosx}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} ={y} \\ $$$$\frac{{log}\left(\mathrm{1}+{cosx}−\mathrm{1}\right)}{{cosx}−\mathrm{1}}.\left(−\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }\right)={logy} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }={logy} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{2}\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}{{x}^{\mathrm{2}} }={logy}\Rightarrow{y}={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\sqrt{{e}}} \\ $$

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