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Question Number 59187 by maxmathsup by imad last updated on 05/May/19
calculate lim_(x→0) ((ln(arctan(1+x))−ln((π/4)))/x^2 )
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)}{{x}^{\mathrm{2}} } \\ $$
Commented by kaivan.ahmadi last updated on 05/May/19
hop lim_(x→0) (1/(2xarctg(1+x)(1+(1+x)^2 )))=∞
$${hop} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}{xarctg}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \right)}=\infty \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
let use hospital theorem let f(x)=ln(arctan(1+x))−ln((π/4)) and g(x)=x^2 ⇒f^′ (x) =((1/(1+(1+x)^2 ))/(arctan(1+x))) =(1/((1+(1+x)^2 )arctan(1+x))) =(1/((x^2 +2x+1+1) arctan(1+x))) =(1/((x^2 +2x+2)arctan(1+x))) =(1/(u(x))) ⇒ f^((2)) (x) =−((u^′ (x))/((u(x))^2 )) but u^′ (x) =(2x+2)arctan(1+x) +((x^2 +2x+2)/(1+(1+x)^2 )) =(2x+2) arctan(1+x) ⇒f^((2)) (x) =−(((2x+2)arctan(1+x))/((x^2 +2x+2)^2 (arctan(1+x))^2 )) ⇒ lim_(x→0) f^((2)) (x) =((2.(π/4))/(4((π/4))^2 )) =(1/(2.(π/4))) =(2/π) also we have g^′ (x)=2x and g^((2)) (x)=2 ⇒ lim_(x→0) ((ln(arctan(1+x))−ln((π/4)))/x^2 ) =(1/π) .
$${let}\:{use}\:{hospital}\:{theorem}\:\:{let}\:{f}\left({x}\right)={ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)\:{and} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} \:\:\:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}{{arctan}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{\left(\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \right){arctan}\left(\mathrm{1}+{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}+\mathrm{1}\right)\:{arctan}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right){arctan}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{{u}\left({x}\right)}\:\Rightarrow \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\frac{{u}^{'} \left({x}\right)}{\left({u}\left({x}\right)\right)^{\mathrm{2}} }\:\:{but}\:\:{u}^{'} \left({x}\right)\:=\left(\mathrm{2}{x}+\mathrm{2}\right){arctan}\left(\mathrm{1}+{x}\right)\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$$$=\left(\mathrm{2}{x}+\mathrm{2}\right)\:{arctan}\left(\mathrm{1}+{x}\right)\:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\frac{\left(\mathrm{2}{x}+\mathrm{2}\right){arctan}\left(\mathrm{1}+{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} \left({arctan}\left(\mathrm{1}+{x}\right)\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\frac{\mathrm{2}.\frac{\pi}{\mathrm{4}}}{\mathrm{4}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}.\frac{\pi}{\mathrm{4}}}\:=\frac{\mathrm{2}}{\pi}\:\:\:{also}\:{we}\:{have}\:{g}^{'} \left({x}\right)=\mathrm{2}{x}\:{and}\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\pi}\:. \\ $$$$ \\ $$

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