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Question Number 89540 by M±th+et£s last updated on 17/Apr/20

lim_(x→0) ((cos(x^2 )−1+(x^4 /2))/(x^2 (x−sin(x))^2 ))

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{cos}\left({x}^{\mathrm{2}} \right)−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} \left({x}−{sin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$

Commented by abdomathmax last updated on 17/Apr/20

let f(x)=((cos(x^2 )−1+(x^4 /2))/(x^2 (x−sinx)^2 ))  we have cos(u) =Σ_(n=0) ^∞  (((−1)^n  u^(2n) )/((2n)!))  =1−(u^2 /(2!)) +(u^4 /(4!)) +...⇒cos(x^2 )∼1−(x^4 /2) +(x^8 /(4!))   (x ∼0)  sinx ∼x−(x^3 /6) ⇒x−sinx ∼(x^3 /6) ⇒  f(x)∼((1−(x^4 /2) +(x^8 /(4!))−1+(x^4 /2))/(x^2 ×(x^6 /6^2 ))) =(6^2 /(4!)) ⇒  lim_(x→0)  f(x)=((36)/(4.3.2)) =(3/2)

$${let}\:{f}\left({x}\right)=\frac{{cos}\left({x}^{\mathrm{2}} \right)−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} \left({x}−{sinx}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:{cos}\left({u}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}!}\:+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\:+...\Rightarrow{cos}\left({x}^{\mathrm{2}} \right)\sim\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}\:\:\:\left({x}\:\sim\mathrm{0}\right) \\ $$$${sinx}\:\sim{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow{x}−{sinx}\:\sim\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} ×\frac{{x}^{\mathrm{6}} }{\mathrm{6}^{\mathrm{2}} }}\:=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{4}!}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)=\frac{\mathrm{36}}{\mathrm{4}.\mathrm{3}.\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 17/Apr/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

Commented by abdomathmax last updated on 18/Apr/20

you are welcome sir.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

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