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Question Number 82800 by jagoll last updated on 24/Feb/20
lim_(x→0) ((1−(√(1+x^2 )) cos 2x)/x^2 )
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 24/Feb/20
let f(x)=((1−(√(1+x^2 ))cos(2x))/x^3 ) we have cos(2x)∼1−2x^2 (1+u)^α ∼1+αu +((α(α−1))/2)u^2 (u∼o) ⇒ (√(1+x^2 )) ∼1+(1/2)x^2 +(((1/2)(−(1/2)))/2)x^4 =1+(x^2 /2)−(1/8)x^4 cosu =Σ_(n=0) ^∞ (((−1)^n u^(2n) )/((2n)!)) ⇒cosu ∼1−(u^2 /2) +(u^4 /(4!)) ⇒ cos(2x)∼1−2x^2 +(((2x)^4 )/(4!)) =1−2x^2 +((16x^4 )/(4.6)) =1−2x^2 +(2/3)x^4 ⇒ (√(1+x^2 ))cos(2x)∼(1+(x^2 /2)−(x^4 /8))(1−2x^2 +(2/3)x^4 ) =1−2x^2 +(2/3)x^4 +(x^2 /2)−x^4 +(1/3)x^6 −(x^4 /8) +(1/4)x^6 −(x^8 /(12)) =1+(−2+(1/2))x^2 +((2/3)−1−(1/8))x^4 +((1/3)+(1/4))x^6 −(x^8 /(12)) =1−(3/2)x^2 +(5/(24))x^4 +(7/(12))x^6 −(x^8 /(12)) ⇒ f(x)∼(((3/2)x^2 +(5/(24))x^4 +(7/(12))x^6 −(x^8 /(12)))/x^3 ) =(3/(2x))+(5/(24))x+(7/(12))x^3 −(1/(12))x^5 ⇒ lim_(x→0) f(x) =∞
$${let}\:{f}\left({x}\right)=\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{3}} }\:\:{we}\:{have}\:{cos}\left(\mathrm{2}{x}\right)\sim\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{u}\right)^{\alpha} \:\sim\mathrm{1}+\alpha{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}}{u}^{\mathrm{2}} \:\:\:\:\left({u}\sim{o}\right)\:\Rightarrow \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}{x}^{\mathrm{4}} \:=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{4}} \\ $$$${cosu}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\:\Rightarrow{cosu}\:\sim\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\:\Rightarrow \\ $$$${cos}\left(\mathrm{2}{x}\right)\sim\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{4}} }{\mathrm{4}!}\:=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+\frac{\mathrm{16}{x}^{\mathrm{4}} }{\mathrm{4}.\mathrm{6}}\:=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{4}} \:\Rightarrow \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{cos}\left(\mathrm{2}{x}\right)\sim\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{4}} \right) \\ $$$$=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{4}} +\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{6}} −\frac{{x}^{\mathrm{4}} }{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{6}} −\frac{{x}^{\mathrm{8}} }{\mathrm{12}} \\ $$$$=\mathrm{1}+\left(−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right){x}^{\mathrm{2}} \:+\left(\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right){x}^{\mathrm{4}} +\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right){x}^{\mathrm{6}} \:−\frac{{x}^{\mathrm{8}} }{\mathrm{12}} \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\frac{\mathrm{5}}{\mathrm{24}}{x}^{\mathrm{4}} \:+\frac{\mathrm{7}}{\mathrm{12}}{x}^{\mathrm{6}} −\frac{{x}^{\mathrm{8}} }{\mathrm{12}}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{24}}{x}^{\mathrm{4}} \:+\frac{\mathrm{7}}{\mathrm{12}}{x}^{\mathrm{6}} −\frac{{x}^{\mathrm{8}} }{\mathrm{12}}}{{x}^{\mathrm{3}} }\:=\frac{\mathrm{3}}{\mathrm{2}{x}}+\frac{\mathrm{5}}{\mathrm{24}}{x}+\frac{\mathrm{7}}{\mathrm{12}}{x}^{\mathrm{3}} \:−\frac{\mathrm{1}}{\mathrm{12}}{x}^{\mathrm{5}} \:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:{f}\left({x}\right)\:=\infty \\ $$
Commented by jagoll last updated on 24/Feb/20
sorry my question is typo. i will correct
$${sorry}\:{my}\:{question}\:{is}\:{typo}.\: \\ $$$${i}\:{will}\:{correct} \\ $$
Answered by TANMAY PANACEA last updated on 24/Feb/20
lim_(x→0) ((1−cos^2 2x(1+x^2 ))/x^3 )×(1/(1+(√(1+x^2 )) ×cos2x)) lim_(x→0) ((sin^2 2x−x^2 cos^2 2x)/x^3 )×(1/(1+1)) (1/2)lim_(x→0) (((sin2x+xcos2x))/1)×((sin2x−xcos2x)/x^3 ) =(1/2)×lim_(x→0) (((2x−((8x^3 )/(3!))+((32x^5 )/(5!))..)−x(1−((4x^2 )/(2!))+((16x^4 )/(4!))..))/x^3 ) in numinator x can not be eliminated
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}^{\mathrm{2}} \mathrm{2}{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{3}} }×\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:×{cos}\mathrm{2}{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}^{\mathrm{2}} \mathrm{2}{x}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \mathrm{2}{x}}{{x}^{\mathrm{3}} }×\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({sin}\mathrm{2}{x}+{xcos}\mathrm{2}{x}\right)}{\mathrm{1}}×\frac{{sin}\mathrm{2}{x}−{xcos}\mathrm{2}{x}}{{x}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}{x}−\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{5}!}..\right)−{x}\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{16}{x}^{\mathrm{4}} }{\mathrm{4}!}..\right)}{{x}^{\mathrm{3}} } \\ $$$${in}\:{numinator}\:{x}\:{can}\:{not}\:{be}\:{eliminated} \\ $$$$ \\ $$
Commented by jagoll last updated on 24/Feb/20
sir the originall question lim_(x→0) ((1−(√(1+x^2 ))cos 2x)/x^2 )
$${sir}\:{the}\:{originall}\:{question}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} } \\ $$
Commented by TANMAY PANACEA last updated on 24/Feb/20
ok..
$${ok}.. \\ $$
Answered by john santu last updated on 24/Feb/20
lim_(x→0) ((1−(1−((4x^2 )/(2!))+((16x^4 )/(4!))−o((2x)^4 ))(1+(1/2)x^2 +o(x^2 )))/x^2 ) lim_(x→0) ((1−(1+(1/2)x^2 +o(x^2 )−2x^2 −x^4 +...))/x^2 ) lim_(x→0) ((((3/2)x^2 −o(x^2 )+x^4 +...))/x^2 )= (3/2)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{16}{x}^{\mathrm{4}} }{\mathrm{4}!}−{o}\left(\left(\mathrm{2}{x}\right)^{\mathrm{4}} \right)\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)−\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{4}} +…\right)}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −{o}\left({x}^{\mathrm{2}} \right)+{x}^{\mathrm{4}} +…\right)}{{x}^{\mathrm{2}} }=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by john santu last updated on 24/Feb/20
Commented by jagoll last updated on 24/Feb/20
yess..thank you mister
$${yess}..{thank}\:{you}\:{mister} \\ $$
Answered by Henri Boucatchou last updated on 24/Feb/20
((1−(√(1+x^2 ))cos2x)/x^2 )=((1−(√(1+x^2 ))+2(√(1+x^2 ))sin^2 x)/x^2 ) =((−x^2 )/(x^2 (1+(√(1+x^2 )))))+2(√(1+x^2 )) (((sinx)/x))^2 =−(1/(1+(√(1+x^2 ))))+2(√(1+x^2 ))(((sinx)/x))^2 →−(1/2)+2=1,5 (x→0)
$$\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{cos}\mathrm{2}{x}}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\frac{−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\left(\frac{{sinx}}{{x}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\frac{{sinx}}{{x}}\right)^{\mathrm{2}} \:\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}=\mathrm{1},\mathrm{5}\:\:\left(\boldsymbol{{x}}\rightarrow\mathrm{0}\right) \\ $$

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