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Question-72251

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Question Number 72251 by aliesam last updated on 26/Oct/19
Commented by mathmax by abdo last updated on 26/Oct/19
let f(x)=((ln(1+ax+bx^2 )−ln(1−ax+bx^2 ))/(1−cosx)) we have ln^′ (1+u)=(1/(1+u)) =1−u +o(u) (u→0) ⇒ ln(1+u)=u−(u^2 /2) +o(u^2 ) ⇒ln(1+ax+bx^2 )∼(ax+bx^2 )−(((ax+bx^2 )^2 )/2) =ax +bx^2 −(1/2)(a^2 x^2 +2abx^3 +b^2 x^4 ) also ln(1−ax +bx^2 )∼−ax+bx^2 −(1/2)(a^2 x^2 −2abx^3 +b^2 x^4 )(change x by−x) ⇒ln(1+ax+bx^2 )−ln(1−ax+bx^2 ) ∼ax +bx^2 −(1/2)a^2 x^2 −abx^3 −(b^2 /2)x^4 +ax−bx^2 +(1/2)a^2 x^2 −abx^3 +(1/2)b^2 x^4 =2ax−2abx^3 ⇒f(x)∼((2ax−2abx^3 )/(x^2 /2)) ( 1−cosx ∼(x^2 /2)) ⇒ f(x)∼ ((4ax−4abx^3 )/x^2 ) ⇒f(x)∼ ((4a)/x) −4abx ⇒lim_(x→0) f(x)=∞ another way ln(((1+ax+bx^2 )/(1−ax+bx^2 )))=ln(((1−ax+bx^2 +2ax)/(1−ax+bx^2 )))=ln(1+((2ax)/(1−ax+bx^2 ))) ∼((2ax)/(1−ax +bx^2 )) ⇒f(x)∼((2ax)/((1−ax+bx^2 )))×(1/(x^2 /2)) =((4ax)/(x^2 (1−ax+bx^2 ))) =((4a)/(x(1−ax+bx^2 ))) ⇒lim_(x→0) f(x)=∞
$${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)}{\mathrm{1}−{cosx}} \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}\right)\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow{ln}\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)\sim\left({ax}+{bx}^{\mathrm{2}} \right)−\frac{\left({ax}+{bx}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$={ax}\:+{bx}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{abx}^{\mathrm{3}} \:+{b}^{\mathrm{2}} {x}^{\mathrm{4}} \right)\:{also} \\ $$$${ln}\left(\mathrm{1}−{ax}\:+{bx}^{\mathrm{2}} \right)\sim−{ax}+{bx}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{abx}^{\mathrm{3}} +{b}^{\mathrm{2}} {x}^{\mathrm{4}} \right)\left({change}\:{x}\:{by}−{x}\right) \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right) \\ $$$$\sim{ax}\:+{bx}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} {x}^{\mathrm{2}} −{abx}^{\mathrm{3}} −\frac{{b}^{\mathrm{2}} }{\mathrm{2}}{x}^{\mathrm{4}} \:+{ax}−{bx}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} {x}^{\mathrm{2}} −{abx}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} {x}^{\mathrm{4}} \\ $$$$=\mathrm{2}{ax}−\mathrm{2}{abx}^{\mathrm{3}} \:\Rightarrow{f}\left({x}\right)\sim\frac{\mathrm{2}{ax}−\mathrm{2}{abx}^{\mathrm{3}} }{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\:\:\:\:\left(\:\mathrm{1}−{cosx}\:\sim\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\:\frac{\mathrm{4}{ax}−\mathrm{4}{abx}^{\mathrm{3}} }{{x}^{\mathrm{2}} }\:\Rightarrow{f}\left({x}\right)\sim\:\frac{\mathrm{4}{a}}{{x}}\:−\mathrm{4}{abx} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\infty \\ $$$${another}\:{way}\:\:{ln}\left(\frac{\mathrm{1}+{ax}+{bx}^{\mathrm{2}} }{\mathrm{1}−{ax}+{bx}^{\mathrm{2}} }\right)={ln}\left(\frac{\mathrm{1}−{ax}+{bx}^{\mathrm{2}} +\mathrm{2}{ax}}{\mathrm{1}−{ax}+{bx}^{\mathrm{2}} }\right)={ln}\left(\mathrm{1}+\frac{\mathrm{2}{ax}}{\mathrm{1}−{ax}+{bx}^{\mathrm{2}} }\right) \\ $$$$\sim\frac{\mathrm{2}{ax}}{\mathrm{1}−{ax}\:+{bx}^{\mathrm{2}} }\:\Rightarrow{f}\left({x}\right)\sim\frac{\mathrm{2}{ax}}{\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)}×\frac{\mathrm{1}}{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\:=\frac{\mathrm{4}{ax}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)}\:=\frac{\mathrm{4}{a}}{{x}\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\infty \\ $$

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