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Question Number 59186 by maxmathsup by imad last updated on 05/May/19
calculate lim_(x→0) ((arctan{ln(1+x)})/x^2 )
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{arctan}\left\{{ln}\left(\mathrm{1}+{x}\right)\right\}}{{x}^{\mathrm{2}} } \\ $$
Commented by kaivan.ahmadi last updated on 05/May/19
hop lim_(x→0) (1/(2x(1+(ln(1+x))^2 )(1+x)))=∞
$${hop} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{1}+\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)}=\infty \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
let use hospital theorem let u(x)=arctan(ln(1+x)) and v(x)=x^2 we have u^′ (x) =((1/(1+x))/(1+ln^2 (1+x))) =(1/((1+x)(1+ln^2 (1+x)))) =(1/(f(x))) ⇒ u^((2)) (x) =−((f^′ (x))/(f^2 (x))) we have f(x)=(x+1)(1+ln^2 (1+x)) ⇒ f^′ (x) =1+ln^2 (1+x) +(x+1)2ln(1+x)(1/(1+x)) =1+ln^2 (1+x)+2ln(1+x) ⇒ u^((2)) (x) =−((1+ln^2 (1+x)+2ln(1+x))/((x+1)^2 {1+ln^2 (1+x)}^2 )) ⇒lim_(x→0) u^((2)) (x)=−1 also we have v^′ (x)=2x and v^((2)) (x)=2 ⇒lim_(x→0) v^((2)) (x) =2 ⇒lim_(x→0) ((arctan(ln(1+x)))/x^2 ) =−(1/2) .
$${let}\:{use}\:{hospital}\:{theorem}\:\:{let}\:{u}\left({x}\right)={arctan}\left({ln}\left(\mathrm{1}+{x}\right)\right)\:{and}\:{v}\left({x}\right)={x}^{\mathrm{2}} \\ $$$${we}\:{have}\:{u}^{'} \left({x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}}{\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right)}\:=\frac{\mathrm{1}}{{f}\left({x}\right)}\:\Rightarrow \\ $$$${u}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\frac{{f}^{'} \left({x}\right)}{{f}^{\mathrm{2}} \left({x}\right)}\:\:\:{we}\:{have}\:{f}\left({x}\right)=\left({x}+\mathrm{1}\right)\left(\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\:+\left({x}+\mathrm{1}\right)\mathrm{2}{ln}\left(\mathrm{1}+{x}\right)\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+\mathrm{2}{ln}\left(\mathrm{1}+{x}\right)\:\Rightarrow \\ $$$${u}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\frac{\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+\mathrm{2}{ln}\left(\mathrm{1}+{x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left\{\mathrm{1}+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right\}^{\mathrm{2}} }\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:{u}^{\left(\mathrm{2}\right)} \left({x}\right)=−\mathrm{1}\:\:{also}\:{we}\:{have} \\ $$$${v}^{'} \left({x}\right)=\mathrm{2}{x}\:{and}\:{v}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {v}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{arctan}\left({ln}\left(\mathrm{1}+{x}\right)\right)}{{x}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

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