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Question Number 118427 by bramlexs22 last updated on 17/Oct/20

(1) lim_(x→0) ((cos x))^(1/(x^2 )) (2) lim_(x→1) ((1+cos πx)/(x^2 −2x+1))

$$\:\:\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{{x}^{\mathrm{2}} \:}]{\mathrm{cos}\:{x}}\: \\ $$$$\:\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cos}\:\pi{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}} \\ $$

Answered by Dwaipayan Shikari last updated on 17/Oct/20

lim_(x→0) (cosx)^(1/x^2 ) =lim_(x→0) (1−(x^2 /2))^(((−2)/x^2 ).(−(1/2))) =e^(−(1/2)) =(1/( (√e)))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({cosx}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{−\mathrm{2}}{{x}^{\mathrm{2}} }.\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} ={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

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

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

Answered by Dwaipayan Shikari last updated on 17/Oct/20

lim_(x→1) ((2cos^2 ((πx)/2))/((x−1)^2 ))=lim_(x→1) ((2sin^2 ((π/2)−(π/2)x))/((x−1)^2 )) =lim_(x→1) 2.((((π^2 /4))(1−x)^2 )/((x−1)^2 ))=(π^2 /2)

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\pi{x}}{\mathrm{2}}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{x}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}2}.\frac{\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by TANMAY PANACEA last updated on 17/Oct/20

2)lim_(x→1) ((2cos^2 ((πx)/2))/((x−1)^2 )) cos(((πx)/2))=sin((π/2)−((πx)/2))=sin(π/2)(1−x) lim_(x→1) ((2sin^2 (π/2)(1−x))/((1−x)^2 ×(π^2 /4)))×(π^2 /4) θ=(π/2)(1−x)→as x→1 θ→0 lim_(θ→0) ((2×sin^2 θ)/θ^2 )×(π^2 /4)=(π^2 /2)

$$\left.\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\pi{x}}{\mathrm{2}}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)={sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi{x}}{\mathrm{2}}\right)={sin}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} ×\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}×\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\theta=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{x}\right)\rightarrow{as}\:{x}\rightarrow\mathrm{1}\:\:\:\theta\rightarrow\mathrm{0} \\ $$$$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}×{sin}^{\mathrm{2}} \theta}{\theta^{\mathrm{2}} }×\frac{\pi^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by benjo_mathlover last updated on 17/Oct/20

(1) lim_(x→0) ((cos x))^(1/x^(2 ) ) = e^(lim_(x→0) (((cos x−1)/x^2 ))) =e^(lim_(x→0) (1−(1/2)x^2 −1).(1/x^2 )) = e^(−(1/2)) = (1/( (√e)))

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

Answered by benjo_mathlover last updated on 17/Oct/20

(2) set x=s+1 ∧ s→0 lim_(s→0) ((1+cos (πs+s))/s^2 ) = lim_(s→0) ((1−cos (πs))/s^2 ) lim_(s→0) ((1−(1−((π^2 s^2 )/2)))/s^2 ) = (π^2 /2)

$$\left(\mathrm{2}\right)\:{set}\:{x}={s}+\mathrm{1}\:\wedge\:{s}\rightarrow\mathrm{0} \\ $$$$\:\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cos}\:\left(\pi{s}+{s}\right)}{{s}^{\mathrm{2}} }\:=\:\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\pi{s}\right)}{{s}^{\mathrm{2}} } \\ $$$$\:\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\pi^{\mathrm{2}} {s}^{\mathrm{2}} }{\mathrm{2}}\right)}{{s}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by Bird last updated on 17/Oct/20

let f(x)=(cosx)^(1/x^2 ) =e^((1/x^2 )ln(cosx)) we jsve cosx∼1−(x^2 /2) ⇒ ln(1−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)))

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

Answered by mathmax by abdo last updated on 17/Oct/20

2) let f(x)=((1+cos(πx))/(x^2 −2x+1)) ⇒f(x)=((1+cos(πx))/((x−1)^2 )) we do the chang. x−1=t ⇒f(x)=f(t+1) =((1+cos(π(t+1)))/t^2 ) =((1−cos(πt))/t^2 ) /(x→1 ⇒t→0) so 1−cos(πt) ∼((π^2 t^2 )/2) ⇒ ((1−cos(πt))/t^2 )∼(π^2 /2) ⇒lim_(t→0) f(t+1)=(π^2 /2) =lim_(x→1) f(x)

$$\left.\mathrm{2}\right)\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}+\mathrm{cos}\left(\pi\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}+\mathrm{cos}\left(\pi\mathrm{x}\right)}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{chang}. \\ $$$$\mathrm{x}−\mathrm{1}=\mathrm{t}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)\:=\frac{\mathrm{1}+\mathrm{cos}\left(\pi\left(\mathrm{t}+\mathrm{1}\right)\right)}{\mathrm{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\left(\pi\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} }\:/\left(\mathrm{x}\rightarrow\mathrm{1}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\right)\:\mathrm{so}\:\mathrm{1}−\mathrm{cos}\left(\pi\mathrm{t}\right)\:\sim\frac{\pi^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\left(\pi\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} }\sim\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right) \\ $$

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